[Transcribers note: Many of the puzzles in this book assume afamiliarity with the currency of Great Britain in the early 1900s. Asthis is likely not common knowledge for those outside Britain (andpossibly many within, ) I am including a chart of relative values. 

The most common units used were:

 the Penny, abbreviated: d. (from the Roman penny, denarius) the Shilling, abbreviated: s. The Pound, abbreviated: £

There was 12 Pennies to a Shilling and 20 Shillings to a Pound, so therewas 240 Pennies in a Pound. 

To further complicate things, there were many coins which were variousfractional values of Pennies, Shillings or Pounds. 

 Farthing ¼d. 

 Half-penny ½d. 

 Penny 1d. 

 Three-penny 3d. 

 Sixpence (or tanner) 6d. 

 Shilling (or bob) 1s. 

 Florin or two shilling piece 2s. 

 Half-crown (or half-dollar) 2s. 6d. 

 Double-florin 4s. 

 Crown (or dollar) 5s. 

 Half-Sovereign 10s. 

 Sovereign (or Pound) £1 or 20s. 

This is by no means a comprehensive list, but it should be adequate tosolve the puzzles in this book. 

Exponents are represented in this text by ^, e. G. '3 squared' is 3^2. 

Numbers with fractional components (other than ¼, ½ and ¾) have a +symbol separating the whole number component from the fraction. It makesthe fraction look odd, but yeilds correct solutions no matter how it isinterpreted. E. G. , 4 and eleven twenty-thirds is 4+11/23, not 411/23 or4-11/23. 

]

AMUSEMENTS IN MATHEMATICS

 by

HENRY ERNEST DUDENEY

 In Mathematicks he was greater Than Tycho Brahe or Erra Pater: For he, by geometrick scale, Could take the size of pots of ale; Resolve, by sines and tangents, straight, If bread or butter wanted weight; And wisely tell what hour o' th' day The clock does strike by algebra. 

 BUTLER'S _Hudibras_. 

1917

PREFACE

In issuing this volume of my Mathematical Puzzles, of which some haveappeared in periodicals and others are given here for the first time, Imust acknowledge the encouragement that I have received from manyunknown correspondents, at home and abroad, who have expressed a desireto have the problems in a collected form, with some of the solutionsgiven at greater length than is possible in magazines and newspapers. Though I have included a few old puzzles that have interested the worldfor generations, where I felt that there was something new to be saidabout them, the problems are in the main original. It is true that someof these have become widely known through the press, and it is possiblethat the reader may be glad to know their source. 

On the question of Mathematical Puzzles in general there is, perhaps, little more to be said than I have written elsewhere. The history of thesubject entails nothing short of the actual story of the beginnings anddevelopment of exact thinking in man. The historian must start from thetime when man first succeeded in counting his ten fingers and individing an apple into two approximately equal parts. Every puzzle thatis worthy of consideration can be referred to mathematics and logic. Every man, woman, and child who tries to "reason out" the answer to thesimplest puzzle is working, though not of necessity consciously, onmathematical lines. Even those puzzles that we have no way of attackingexcept by haphazard attempts can be brought under a method of what hasbeen called "glorified trial"--a system of shortening our labours byavoiding or eliminating what our reason tells us is useless. It is, infact, not easy to say sometimes where the "empirical" begins and whereit ends. 

When a man says, "I have never solved a puzzle in my life, " it isdifficult to know exactly what he means, for every intelligentindividual is doing it every day. The unfortunate inmates of our lunaticasylums are sent there expressly because they cannot solvepuzzles--because they have lost their powers of reason. If there were nopuzzles to solve, there would be no questions to ask; and if there wereno questions to be asked, what a world it would be! We should all beequally omniscient, and conversation would be useless and idle. 

It is possible that some few exceedingly sober-minded mathematicians, who are impatient of any terminology in their favourite science but theacademic, and who object to the elusive x and y appearing under anyother names, will have wished that various problems had been presentedin a less popular dress and introduced with a less flippant phraseology. I can only refer them to the first word of my title and remind them thatwe are primarily out to be amused--not, it is true, without some hope ofpicking up morsels of knowledge by the way. If the manner is light, Ican only say, in the words of Touchstone, that it is "an ill-favouredthing, sir, but my own; a poor humour of mine, sir. "

As for the question of difficulty, some of the puzzles, especially inthe Arithmetical and Algebraical category, are quite easy. Yet some ofthose examples that look the simplest should not be passed over withouta little consideration, for now and again it will be found that there issome more or less subtle pitfall or trap into which the reader may beapt to fall. It is good exercise to cultivate the habit of being verywary over the exact wording of a puzzle. It teaches exactitude andcaution. But some of the problems are very hard nuts indeed, and notunworthy of the attention of the advanced mathematician. Readers willdoubtless select according to their individual tastes. 

In many cases only the mere answers are given. This leaves the beginnersomething to do on his own behalf in working out the method of solution, and saves space that would be wasted from the point of view of theadvanced student. On the other hand, in particular cases where it seemedlikely to interest, I have given rather extensive solutions and treatedproblems in a general manner. It will often be found that the notes onone problem will serve to elucidate a good many others in the book; sothat the reader's difficulties will sometimes be found cleared up as headvances. Where it is possible to say a thing in a manner that may be"understanded of the people" generally, I prefer to use this simplephraseology, and so engage the attention and interest of a largerpublic. The mathematician will in such cases have no difficulty inexpressing the matter under consideration in terms of his familiarsymbols. 

I have taken the greatest care in reading the proofs, and trust that anyerrors that may have crept in are very few. If any such should occur, Ican only plead, in the words of Horace, that "good Homer sometimesnods, " or, as the bishop put it, "Not even the youngest curate in mydiocese is infallible. "

I have to express my thanks in particular to the proprietors of _TheStrand Magazine_, _Cassell's Magazine_, _The Queen_, _Tit-Bits_, and_The Weekly Dispatch_ for their courtesy in allowing me to reprint someof the puzzles that have appeared in their pages. 

THE AUTHORS' CLUB _March_ 25, 1917

CONTENTS

 PREFACE v ARITHMETICAL AND ALGEBRAICAL PROBLEMS 1 Money Puzzles 1 Age and Kinship Puzzles 6 Clock Puzzles 9 Locomotion and Speed Puzzles 11 Digital Puzzles 13 Various Arithmetical and Algebraical Problems 17 GEOMETRICAL PROBLEMS 27 Dissection Puzzles 27 Greek Cross Puzzles 28 Various Dissection Puzzles 35 Patchwork Puzzles 46 Various Geometrical Puzzles 49 POINTS AND LINES PROBLEMS 56 MOVING COUNTER PROBLEMS 58 UNICURSAL AND ROUTE PROBLEMS 68 COMBINATION AND GROUP PROBLEMS 76 CHESSBOARD PROBLEMS 85 The Chessboard 85 Statical Chess Puzzles 88 The Guarded Chessboard 95 Dynamical Chess Puzzles 96 Various Chess Puzzles 105 MEASURING, WEIGHING, AND PACKING PUZZLES 109 CROSSING RIVER PROBLEMS 112 PROBLEMS CONCERNING GAMES 114 PUZZLE GAMES 117 MAGIC SQUARE PROBLEMS 119 Subtracting, Multiplying, and Dividing Magics 124 Magic Squares of Primes 125 MAZES AND HOW TO THREAD THEM 127 THE PARADOX PARTY 137 UNCLASSIFIED PROBLEMS 142 SOLUTIONS 148 INDEX 253

AMUSEMENTS IN MATHEMATICS. 

ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 

 "And what was he? Forsooth, a great arithmetician. " _Othello_, I. I. 

The puzzles in this department are roughly thrown together in classesfor the convenience of the reader. Some are very easy, others quitedifficult. But they are not arranged in any order of difficulty--andthis is intentional, for it is well that the solver should not be warnedthat a puzzle is just what it seems to be. It may, therefore, prove tobe quite as simple as it looks, or it may contain some pitfall intowhich, through want of care or over-confidence, we may stumble. 

Also, the arithmetical and algebraical puzzles are not separated in themanner adopted by some authors, who arbitrarily require certain problemsto be solved by one method or the other. The reader is left to make hisown choice and determine which puzzles are capable of being solved byhim on purely arithmetical lines. 

MONEY PUZZLES. 

 "Put not your trust in money, but put your money in trust. "

 OLIVER WENDELL HOLMES. 

1. --A POST-OFFICE PERPLEXITY. 

In every business of life we are occasionally perplexed by some chancequestion that for the moment staggers us. I quite pitied a young lady ina branch post-office when a gentleman entered and deposited a crown onthe counter with this request: "Please give me some twopenny stamps, sixtimes as many penny stamps, and make up the rest of the money intwopence-halfpenny stamps. " For a moment she seemed bewildered, then herbrain cleared, and with a smile she handed over stamps in exactfulfilment of the order. How long would it have taken you to think itout?

2. --YOUTHFUL PRECOCITY. 

The precocity of some youths is surprising. One is disposed to say onoccasion, "That boy of yours is a genius, and he is certain to do greatthings when he grows up;" but past experience has taught us that heinvariably becomes quite an ordinary citizen. It is so often the case, on the contrary, that the dull boy becomes a great man. You never cantell. Nature loves to present to us these queer paradoxes. It is wellknown that those wonderful "lightning calculators, " who now and againsurprise the world by their feats, lose all their mysterious powersdirectly they are taught the elementary rules of arithmetic. 

A boy who was demolishing a choice banana was approached by a youngfriend, who, regarding him with envious eyes, asked, "How much did youpay for that banana, Fred?" The prompt answer was quite remarkable inits way: "The man what I bought it of receives just half as manysixpences for sixteen dozen dozen bananas as he gives bananas for afiver. "

Now, how long will it take the reader to say correctly just how muchFred paid for his rare and refreshing fruit?

3. --AT A CATTLE MARKET. 

Three countrymen met at a cattle market. "Look here, " said Hodge toJakes, "I'll give you six of my pigs for one of your horses, and thenyou'll have twice as many animals here as I've got. " "If that's yourway of doing business, " said Durrant to Hodge, "I'll give you fourteenof my sheep for a horse, and then you'll have three times as manyanimals as I. " "Well, I'll go better than that, " said Jakes to Durrant;"I'll give you four cows for a horse, and then you'll have six times asmany animals as I've got here. "

No doubt this was a very primitive way of bartering animals, but it isan interesting little puzzle to discover just how many animals Jakes, Hodge, and Durrant must have taken to the cattle market. 

4. --THE BEANFEAST PUZZLE. 

A number of men went out together on a bean-feast. There were fourparties invited--namely, 25 cobblers, 20 tailors, 18 hatters, and 12glovers. They spent altogether £6, 13s. It was found that five cobblersspent as much as four tailors; that twelve tailors spent as much as ninehatters; and that six hatters spent as much as eight glovers. The puzzleis to find out how much each of the four parties spent. 

5. --A QUEER COINCIDENCE. 

Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards, Francis, and Gudgeon, were recently engaged in play. The name of theparticular game is of no consequence. They had agreed that whenever aplayer won a game he should double the money of each of the otherplayers--that is, he was to give the players just as much money as theyhad already in their pockets. They played seven games, and, strange tosay, each won a game in turn, in the order in which their names aregiven. But a more curious coincidence is this--that when they hadfinished play each of the seven men had exactly the same amount--twoshillings and eightpence--in his pocket. The puzzle is to find out howmuch money each man had with him before he sat down to play. 

6. --A CHARITABLE BEQUEST. 

A man left instructions to his executors to distribute once a yearexactly fifty-five shillings among the poor of his parish; but they wereonly to continue the gift so long as they could make it in differentways, always giving eighteenpence each to a number of women and half acrown each to men. During how many years could the charity beadministered? Of course, by "different ways" is meant a different numberof men and women every time. 

7. --THE WIDOW'S LEGACY. 

A gentleman who recently died left the sum of £8, 000 to be divided amonghis widow, five sons, and four daughters. He directed that every sonshould receive three times as much as a daughter, and that everydaughter should have twice as much as their mother. What was the widow'sshare?

8. --INDISCRIMINATE CHARITY. 

A charitable gentleman, on his way home one night, was appealed to bythree needy persons in succession for assistance. To the first person hegave one penny more than half the money he had in his pocket; to thesecond person he gave twopence more than half the money he then had inhis pocket; and to the third person he handed over threepence more thanhalf of what he had left. On entering his house he had only one penny inhis pocket. Now, can you say exactly how much money that gentleman hadon him when he started for home?

9. --THE TWO AEROPLANES. 

A man recently bought two aeroplanes, but afterwards found that theywould not answer the purpose for which he wanted them. So he sold themfor £600 each, making a loss of 20 per cent. On one machine and a profitof 20 per cent. On the other. Did he make a profit on the wholetransaction, or a loss? And how much?

10. --BUYING PRESENTS. 

"Whom do you think I met in town last week, Brother William?" said UncleBenjamin. "That old skinflint Jorkins. His family had been taking himaround buying Christmas presents. He said to me, 'Why cannot thegovernment abolish Christmas, and make the giving of presents punishableby law? I came out this morning with a certain amount of money in mypocket, and I find I have spent just half of it. In fact, if you willbelieve me, I take home just as many shillings as I had pounds, and halfas many pounds as I had shillings. It is monstrous!'" Can you sayexactly how much money Jorkins had spent on those presents?

11. --THE CYCLISTS' FEAST. 

 'Twas last Bank Holiday, so I've been told, Some cyclists rode abroad in glorious weather. Resting at noon within a tavern old, They all agreed to have a feast together. "Put it all in one bill, mine host, " they said, "For every man an equal share will pay. " The bill was promptly on the table laid, And four pounds was the reckoning that day. But, sad to state, when they prepared to square, 'Twas found that two had sneaked outside and fled. So, for two shillings more than his due share Each honest man who had remained was bled. They settled later with those rogues, no doubt. How many were they when they first set out?

12. --A QUEER THING IN MONEY. 

It will be found that £66, 6s. 6d. Equals 15, 918 pence. Now, the four6's added together make 24, and the figures in 15, 918 also add to 24. Itis a curious fact that there is only one other sum of money, in pounds, shillings, and pence (all similarly repetitions of one figure), of whichthe digits shall add up the same as the digits of the amount in pence. What is the other sum of money?

13. --A NEW MONEY PUZZLE. 

The largest sum of money that can be written in pounds, shillings, pence, and farthings, using each of the nine digits once and only once, is £98, 765, 4s. 3½d. Now, try to discover the smallest sum of moneythat can be written down under precisely the same conditions. There mustbe some value given for each denomination--pounds, shillings, pence, and farthings--and the nought may not be used. It requires just a littlejudgment and thought. 

14. --SQUARE MONEY. 

"This is queer, " said McCrank to his friend. "Twopence added to twopenceis fourpence, and twopence multiplied by twopence is also fourpence. " Ofcourse, he was wrong in thinking you can multiply money by money. Themultiplier must be regarded as an abstract number. It is true that twofeet multiplied by two feet will make four square feet. Similarly, twopence multiplied by two pence will produce four square pence! And itwill perplex the reader to say what a "square penny" is. But we willassume for the purposes of our puzzle that twopence multiplied bytwopence is fourpence. Now, what two amounts of money will produce thenext smallest possible result, the same in both cases, when added ormultiplied in this manner? The two amounts need not be alike, but theymust be those that can be paid in current coins of the realm. 

15. --POCKET MONEY. 

What is the largest sum of money--all in current silver coins and nofour-shilling piece--that I could have in my pocket without being ableto give change for a half-sovereign?

16. --THE MILLIONAIRE'S PERPLEXITY. 

Mr. Morgan G. Bloomgarten, the millionaire, known in the States as theClam King, had, for his sins, more money than he knew what to do with. It bored him. So he determined to persecute some of his poor but happyfriends with it. They had never done him any harm, but he resolved toinoculate them with the "source of all evil. " He therefore proposed todistribute a million dollars among them and watch them go rapidly to thebad. But he was a man of strange fancies and superstitions, and it wasan inviolable rule with him never to make a gift that was not either onedollar or some power of seven--such as 7, 49, 343, 2, 401, which numbersof dollars are produced by simply multiplying sevens together. Anotherrule of his was that he would never give more than six persons exactlythe same sum. Now, how was he to distribute the 1, 000, 000 dollars? Youmay distribute the money among as many people as you like, under theconditions given. 

17. --THE PUZZLING MONEY-BOXES. 

Four brothers--named John, William, Charles, and Thomas--had each amoney-box. The boxes were all given to them on the same day, and they atonce put what money they had into them; only, as the boxes were not verylarge, they first changed the money into as few coins as possible. Afterthey had done this, they told one another how much money they had saved, and it was found that if John had had 2s. More in his box than atpresent, if William had had 2s. Less, if Charles had had twice as much, and if Thomas had had half as much, they would all have had exactly thesame amount. 

Now, when I add that all four boxes together contained 45s. , and thatthere were only six coins in all in them, it becomes an entertainingpuzzle to discover just what coins were in each box. 

18. --THE MARKET WOMEN. 

A number of market women sold their various products at a certain priceper pound (different in every case), and each received the sameamount--2s. 2½d. What is the greatest number of women there couldhave been? The price per pound in every case must be such as could bepaid in current money. 

19. --THE NEW YEAR'S EVE SUPPERS. 

The proprietor of a small London café has given me some interestingfigures. He says that the ladies who come alone to his place forrefreshment spend each on an average eighteenpence, that theunaccompanied men spend half a crown each, and that when a gentlemanbrings in a lady he spends half a guinea. On New Year's Eve he suppliedsuppers to twenty-five persons, and took five pounds in all. Now, assuming his averages to have held good in every case, how was hiscompany made up on that occasion? Of course, only single gentlemen, single ladies, and pairs (a lady and gentleman) can be supposed to havebeen present, as we are not considering larger parties. 

20. --BEEF AND SAUSAGES. 

"A neighbour of mine, " said Aunt Jane, "bought a certain quantity ofbeef at two shillings a pound, and the same quantity of sausages ateighteenpence a pound. I pointed out to her that if she had divided thesame money equally between beef and sausages she would have gained twopounds in the total weight. Can you tell me exactly how much she spent?"

"Of course, it is no business of mine, " said Mrs. Sunniborne; "but alady who could pay such prices must be somewhat inexperienced indomestic economy. "

"I quite agree, my dear, " Aunt Jane replied, "but you see that is notthe precise point under discussion, any more than the name and morals ofthe tradesman. "

21. --A DEAL IN APPLES. 

I paid a man a shilling for some apples, but they were so small that Imade him throw in two extra apples. I find that made them cost just apenny a dozen less than the first price he asked. How many apples did Iget for my shilling?

22. --A DEAL IN EGGS. 

A man went recently into a dairyman's shop to buy eggs. He wanted themof various qualities. The salesman had new-laid eggs at the high priceof fivepence each, fresh eggs at one penny each, eggs at a halfpennyeach, and eggs for electioneering purposes at a greatly reduced figure, but as there was no election on at the time the buyer had no use for thelast. However, he bought some of each of the three other kinds andobtained exactly one hundred eggs for eight and fourpence. Now, as hebrought away exactly the same number of eggs of two of the threequalities, it is an interesting puzzle to determine just how many hebought at each price. 

23. --THE CHRISTMAS-BOXES. 

Some years ago a man told me he had spent one hundred English silvercoins in Christmas-boxes, giving every person the same amount, and itcost him exactly £1, 10s. 1d. Can you tell just how many personsreceived the present, and how he could have managed the distribution?That odd penny looks queer, but it is all right. 

24. --A SHOPPING PERPLEXITY. 

Two ladies went into a shop where, through some curious eccentricity, nochange was given, and made purchases amounting together to less thanfive shillings. "Do you know, " said one lady, "I find I shall require nofewer than six current coins of the realm to pay for what I havebought. " The other lady considered a moment, and then exclaimed: "By apeculiar coincidence, I am exactly in the same dilemma. " "Then we willpay the two bills together. " But, to their astonishment, they stillrequired six coins. What is the smallest possible amount of theirpurchases--both different?

25. --CHINESE MONEY. 

The Chinese are a curious people, and have strange inverted ways ofdoing things. It is said that they use a saw with an upward pressureinstead of a downward one, that they plane a deal board by pulling thetool toward them instead of pushing it, and that in building a housethey first construct the roof and, having raised that into position, proceed to work downwards. In money the currency of the country consistsof taels of fluctuating value. The tael became thinner and thinner until2, 000 of them piled together made less than three inches in height. Thecommon cash consists of brass coins of varying thicknesses, with around, square, or triangular hole in the centre, as in our illustration. 

[Illustration]

These are strung on wires like buttons. Supposing that eleven coins withround holes are worth fifteen ching-changs, that eleven with squareholes are worth sixteen ching-changs, and that eleven with triangularholes are worth seventeen ching-changs, how can a Chinaman give mechange for half a crown, using no coins other than the three mentioned?A ching-chang is worth exactly twopence and four-fifteenths of aching-chang. 

26. --THE JUNIOR CLERK'S PUZZLE. 

Two youths, bearing the pleasant names of Moggs and Snoggs, wereemployed as junior clerks by a merchant in Mincing Lane. They were bothengaged at the same salary--that is, commencing at the rate of £50 ayear, payable half-yearly. Moggs had a yearly rise of £10, and Snoggswas offered the same, only he asked, for reasons that do not concern ourpuzzle, that he might take his rise at £2, 10s. Half-yearly, to whichhis employer (not, perhaps, unnaturally!) had no objection. 

Now we come to the real point of the puzzle. Moggs put regularly intothe Post Office Savings Bank a certain proportion of his salary, whileSnoggs saved twice as great a proportion of his, and at the end of fiveyears they had together saved £268, 15s. How much had each saved? Thequestion of interest can be ignored. 

27. --GIVING CHANGE. 

Every one is familiar with the difficulties that frequently arise overthe giving of change, and how the assistance of a third person with afew coins in his pocket will sometimes help us to set the matter right. Here is an example. An Englishman went into a shop in New York andbought goods at a cost of thirty-four cents. The only money he had was adollar, a three-cent piece, and a two-cent piece. The tradesman had onlya half-dollar and a quarter-dollar. But another customer happened to bepresent, and when asked to help produced two dimes, a five-cent piece, atwo-cent piece, and a one-cent piece. How did the tradesman manage togive change? For the benefit of those readers who are not familiar withthe American coinage, it is only necessary to say that a dollar is ahundred cents and a dime ten cents. A puzzle of this kind should rarelycause any difficulty if attacked in a proper manner. 

28. --DEFECTIVE OBSERVATION. 

Our observation of little things is frequently defective, and ourmemories very liable to lapse. A certain judge recently remarked in acase that he had no recollection whatever of putting the wedding-ring onhis wife's finger. Can you correctly answer these questions withouthaving the coins in sight? On which side of a penny is the date given?Some people are so unobservant that, although they are handling the coinnearly every day of their lives, they are at a loss to answer thissimple question. If I lay a penny flat on the table, how many otherpennies can I place around it, every one also lying flat on the table, so that they all touch the first one? The geometrician will, of course, give the answer at once, and not need to make any experiment. He willalso know that, since all circles are similar, the same answer willnecessarily apply to any coin. The next question is a most interestingone to ask a company, each person writing down his answer on a slip ofpaper, so that no one shall be helped by the answers of others. What isthe greatest number of three-penny-pieces that may be laid flat on thesurface of a half-crown, so that no piece lies on another or overlapsthe surface of the half-crown? It is amazing what a variety of differentanswers one gets to this question. Very few people will be found to givethe correct number. Of course the answer must be given without lookingat the coins. 

29. --THE BROKEN COINS. 

A man had three coins--a sovereign, a shilling, and a penny--and hefound that exactly the same fraction of each coin had been broken away. Now, assuming that the original intrinsic value of these coins was thesame as their nominal value--that is, that the sovereign was worth apound, the shilling worth a shilling, and the penny worth a penny--whatproportion of each coin has been lost if the value of the threeremaining fragments is exactly one pound?

30. --TWO QUESTIONS IN PROBABILITIES. 

There is perhaps no class of puzzle over which people so frequentlyblunder as that which involves what is called the theory ofprobabilities. I will give two simple examples of the sort of puzzle Imean. They are really quite easy, and yet many persons are tripped up bythem. A friend recently produced five pennies and said to me: "Inthrowing these five pennies at the same time, what are the chances thatat least four of the coins will turn up either all heads or all tails?"His own solution was quite wrong, but the correct answer ought not to behard to discover. Another person got a wrong answer to the followinglittle puzzle which I heard him propound: "A man placed three sovereignsand one shilling in a bag. How much should be paid for permission todraw one coin from it?" It is, of course, understood that you are aslikely to draw any one of the four coins as another. 

31. --DOMESTIC ECONOMY. 

Young Mrs. Perkins, of Putney, writes to me as follows: "I should bevery glad if you could give me the answer to a little sum that has beenworrying me a good deal lately. Here it is: We have only been married ashort time, and now, at the end of two years from the time when we setup housekeeping, my husband tells me that he finds we have spent a thirdof his yearly income in rent, rates, and taxes, one-half in domesticexpenses, and one-ninth in other ways. He has a balance of £190remaining in the bank. I know this last, because he accidentally leftout his pass-book the other day, and I peeped into it. Don't you thinkthat a husband ought to give his wife his entire confidence in his moneymatters? Well, I do; and--will you believe it?--he has never told mewhat his income really is, and I want, very naturally, to find out. Canyou tell me what it is from the figures I have given you?"

Yes; the answer can certainly be given from the figures contained inMrs. Perkins's letter. And my readers, if not warned, will bepractically unanimous in declaring the income to be--something absurdlyin excess of the correct answer!

32. --THE EXCURSION TICKET PUZZLE. 

When the big flaming placards were exhibited at the little provincialrailway station, announcing that the Great ---- Company would run cheapexcursion trains to London for the Christmas holidays, the inhabitantsof Mudley-cum-Turmits were in quite a flutter of excitement. Half anhour before the train came in the little booking office was crowded withcountry passengers, all bent on visiting their friends in the greatMetropolis. The booking clerk was unaccustomed to dealing with crowds ofsuch a dimension, and he told me afterwards, while wiping his manlybrow, that what caused him so much trouble was the fact that theserustics paid their fares in such a lot of small money. 

He said that he had enough farthings to supply a West End draper withchange for a week, and a sufficient number of threepenny pieces for thecongregations of three parish churches. "That excursion fare, " said he, "is nineteen shillings and ninepence, and I should like to know in justhow many different ways it is possible for such an amount to be paid inthe current coin of this realm. "

Here, then, is a puzzle: In how many different ways may nineteenshillings and ninepence be paid in our current coin? Remember that thefourpenny-piece is not now current. 

33. --PUZZLE IN REVERSALS. 

Most people know that if you take any sum of money in pounds, shillings, and pence, in which the number of pounds (less than £12) exceeds that ofthe pence, reverse it (calling the pounds pence and the pence pounds), find the difference, then reverse and add this difference, the result isalways £12, 18s. 11d. But if we omit the condition, "less than £12, " andallow nought to represent shillings or pence--(1) What is the lowestamount to which the rule will not apply? (2) What is the highest amountto which it will apply? Of course, when reversing such a sum as £14, 15s. 3d. It may be written £3, 16s. 2d. , which is the same as £3, 15s. 14d. 

34. --THE GROCER AND DRAPER. 

A country "grocer and draper" had two rival assistants, who pridedthemselves on their rapidity in serving customers. The young man on thegrocery side could weigh up two one-pound parcels of sugar per minute, while the drapery assistant could cut three one-yard lengths of cloth inthe same time. Their employer, one slack day, set them a race, givingthe grocer a barrel of sugar and telling him to weigh up forty-eightone-pound parcels of sugar While the draper divided a roll offorty-eight yards of cloth into yard pieces. The two men wereinterrupted together by customers for nine minutes, but the draper wasdisturbed seventeen times as long as the grocer. What was the result ofthe race?

35. --JUDKINS'S CATTLE. 

Hiram B. Judkins, a cattle-dealer of Texas, had five droves of animals, consisting of oxen, pigs, and sheep, with the same number of animals ineach drove. One morning he sold all that he had to eight dealers. Eachdealer bought the same number of animals, paying seventeen dollars foreach ox, four dollars for each pig, and two dollars for each sheep; andHiram received in all three hundred and one dollars. What is thegreatest number of animals he could have had? And how many would therebe of each kind?

36. --BUYING APPLES. 

As the purchase of apples in small quantities has always presentedconsiderable difficulties, I think it well to offer a few remarks onthis subject. We all know the story of the smart boy who, on being toldby the old woman that she was selling her apples at four for threepence, said: "Let me see! Four for threepence; that's three for twopence, twofor a penny, one for nothing--I'll take _one_!"

There are similar cases of perplexity. For example, a boy once picked upa penny apple from a stall, but when he learnt that the woman's pearswere the same price he exchanged it, and was about to walk off. "Stop!"said the woman. "You haven't paid me for the pear!" "No, " said the boy, "of course not. I gave you the apple for it. " "But you didn't pay forthe apple!" "Bless the woman! You don't expect me to pay for the appleand the pear too!" And before the poor creature could get out of thetangle the boy had disappeared. 

Then, again, we have the case of the man who gave a boy sixpence andpromised to repeat the gift as soon as the youngster had made it intoninepence. Five minutes later the boy returned. "I have made it intoninepence, " he said, at the same time handing his benefactor threepence. "How do you make that out?" he was asked. "I bought threepennyworth ofapples. " "But that does not make it into ninepence!" "I should ratherthink it did, " was the boy's reply. "The apple woman has threepence, hasn't she? Very well, I have threepennyworth of apples, and I have justgiven you the other threepence. What's that but ninepence?"

I cite these cases just to show that the small boy really stands in needof a little instruction in the art of buying apples. So I will give asimple poser dealing with this branch of commerce. 

An old woman had apples of three sizes for sale--one a penny, two apenny, and three a penny. Of course two of the second size and three ofthe third size were respectively equal to one apple of the largest size. Now, a gentleman who had an equal number of boys and girls gave hischildren sevenpence to be spent amongst them all on these apples. Thepuzzle is to give each child an equal distribution of apples. How wasthe sevenpence spent, and how many children were there?

37. --BUYING CHESTNUTS. 

Though the following little puzzle deals with the purchase of chestnuts, it is not itself of the "chestnut" type. It is quite new. At first sightit has certainly the appearance of being of the "nonsense puzzle"character, but it is all right when properly considered. 

A man went to a shop to buy chestnuts. He said he wanted a pennyworth, and was given five chestnuts. "It is not enough; I ought to have asixth, " he remarked! "But if I give you one chestnut more. " the shopmanreplied, "you will have five too many. " Now, strange to say, they wereboth right. How many chestnuts should the buyer receive for half acrown?

38. --THE BICYCLE THIEF. 

Here is a little tangle that is perpetually cropping up in variousguises. A cyclist bought a bicycle for £15 and gave in payment a chequefor £25. The seller went to a neighbouring shopkeeper and got him tochange the cheque for him, and the cyclist, having received his £10change, mounted the machine and disappeared. The cheque proved to bevalueless, and the salesman was requested by his neighbour to refund theamount he had received. To do this, he was compelled to borrow the £25from a friend, as the cyclist forgot to leave his address, and could notbe found. Now, as the bicycle cost the salesman £11, how much money didhe lose altogether?

39. --THE COSTERMONGER'S PUZZLE. 

"How much did yer pay for them oranges, Bill?"

"I ain't a-goin' to tell yer, Jim. But I beat the old cove downfourpence a hundred. "

"What good did that do yer?"

"Well, it meant five more oranges on every ten shillin's-worth. "

Now, what price did Bill actually pay for the oranges? There is only onerate that will fit in with his statements. 

AGE AND KINSHIP PUZZLES. 

 "The days of our years are threescore years and ten. "

 --_Psalm_ xc. 10. 

For centuries it has been a favourite method of propounding arithmeticalpuzzles to pose them in the form of questions as to the age of anindividual. They generally lend themselves to very easy solution by theuse of algebra, though often the difficulty lies in stating themcorrectly. They may be made very complex and may demand considerableingenuity, but no general laws can well be laid down for their solution. The solver must use his own sagacity. As for puzzles in relationship orkinship, it is quite curious how bewildering many people find thesethings. Even in ordinary conversation, some statement as torelationship, which is quite clear in the mind of the speaker, willimmediately tie the brains of other people into knots. Such expressionsas "He is my uncle's son-in-law's sister" convey absolutely nothing tosome people without a detailed and laboured explanation. In such casesthe best course is to sketch a brief genealogical table, when the eyecomes immediately to the assistance of the brain. In these days, when wehave a growing lack of respect for pedigrees, most people have got outof the habit of rapidly drawing such tables, which is to be regretted, as they would save a lot of time and brain racking on occasions. 

40. --MAMMA'S AGE. 

Tommy: "How old are you, mamma?"

Mamma: "Let me think, Tommy. Well, our three ages add up to exactlyseventy years. "

Tommy: "That's a lot, isn't it? And how old are you, papa?"

Papa: "Just six times as old as you, my son. "

Tommy: "Shall I ever be half as old as you, papa?"

Papa: "Yes, Tommy; and when that happens our three ages will add up toexactly twice as much as to-day. "

Tommy: "And supposing I was born before you, papa; and supposing mammahad forgot all about it, and hadn't been at home when I came; andsupposing--"

Mamma: "Supposing, Tommy, we talk about bed. Come along, darling. You'llhave a headache. "

Now, if Tommy had been some years older he might have calculated theexact ages of his parents from the information they had given him. Canyou find out the exact age of mamma?

41. --THEIR AGES. 

"My husband's age, " remarked a lady the other day, "is represented bythe figures of my own age reversed. He is my senior, and the differencebetween our ages is one-eleventh of their sum. "

42. --THE FAMILY AGES. 

When the Smileys recently received a visit from the favourite uncle, thefond parents had all the five children brought into his presence. Firstcame Billie and little Gertrude, and the uncle was informed that the boywas exactly twice as old as the girl. Then Henrietta arrived, and it waspointed out that the combined ages of herself and Gertrude equalledtwice the age of Billie. Then Charlie came running in, and somebodyremarked that now the combined ages of the two boys were exactly twicethe combined ages of the two girls. The uncle was expressing hisastonishment at these coincidences when Janet came in. "Ah! uncle, " sheexclaimed, "you have actually arrived on my twenty-first birthday!" Tothis Mr. Smiley added the final staggerer: "Yes, and now the combinedages of the three girls are exactly equal to twice the combined ages ofthe two boys. " Can you give the age of each child?

43. --MRS. TIMPKINS'S AGE. 

Edwin: "Do you know, when the Timpkinses married eighteen years agoTimpkins was three times as old as his wife, and to-day he is just twiceas old as she?"

Angelina: "Then how old was Mrs. Timpkins on the wedding day?"

Can you answer Angelina's question?

44--A CENSUS PUZZLE. 

Mr. And Mrs. Jorkins have fifteen children, all born at intervals of oneyear and a half. Miss Ada Jorkins, the eldest, had an objection to stateher age to the census man, but she admitted that she was just seventimes older than little Johnnie, the youngest of all. What was Ada'sage? Do not too hastily assume that you have solved this little poser. You may find that you have made a bad blunder!

45. --MOTHER AND DAUGHTER. 

"Mother, I wish you would give me a bicycle, " said a girl of twelve theother day. 

"I do not think you are old enough yet, my dear, " was the reply. "When Iam only three times as old as you are you shall have one. "

Now, the mother's age is forty-five years. When may the young ladyexpect to receive her present?

46. --MARY AND MARMADUKE. 

Marmaduke: "Do you know, dear, that in seven years' time our combinedages will be sixty-three years?"

Mary: "Is that really so? And yet it is a fact that when you were mypresent age you were twice as old as I was then. I worked it out lastnight. "

Now, what are the ages of Mary and Marmaduke?

47--ROVER'S AGE. 

"Now, then, Tommy, how old is Rover?" Mildred's young man asked herbrother. 

"Well, five years ago, " was the youngster's reply, "sister was fourtimes older than the dog, but now she is only three times as old. "

Can you tell Rover's age?

48. --CONCERNING TOMMY'S AGE. 

Tommy Smart was recently sent to a new school. On the first day of hisarrival the teacher asked him his age, and this was his curious reply:"Well, you see, it is like this. At the time I was born--I forget theyear--my only sister, Ann, happened to be just one-quarter the age ofmother, and she is now one-third the age of father. " "That's all verywell, " said the teacher, "but what I want is not the age of your sisterAnn, but your own age. " "I was just coming to that, " Tommy answered; "Iam just a quarter of mother's present age, and in four years' time Ishall be a quarter the age of father. Isn't that funny?"

This was all the information that the teacher could get out of TommySmart. Could you have told, from these facts, what was his precise age?It is certainly a little puzzling. 

49. --NEXT-DOOR NEIGHBOURS. 

There were two families living next door to one another at TootingBec--the Jupps and the Simkins. The united ages of the four Juppsamounted to one hundred years, and the united ages of the Simkins alsoamounted to the same. It was found in the case of each family that thesum obtained by adding the squares of each of the children's ages to thesquare of the mother's age equalled the square of the father's age. Inthe case of the Jupps, however, Julia was one year older than herbrother Joe, whereas Sophy Simkin was two years older than her brotherSammy. What was the age of each of the eight individuals?

50. --THE BAG OF NUTS. 

Three boys were given a bag of nuts as a Christmas present, and it wasagreed that they should be divided in proportion to their ages, whichtogether amounted to 17½ years. Now the bag contained 770 nuts, andas often as Herbert took four Robert took three, and as often as Herberttook six Christopher took seven. The puzzle is to find out how many nutseach had, and what were the boys' respective ages. 

51. --HOW OLD WAS MARY?

Here is a funny little age problem, by the late Sam Loyd, which has beenvery popular in the United States. Can you unravel the mystery?

The combined ages of Mary and Ann are forty-four years, and Mary istwice as old as Ann was when Mary was half as old as Ann will be whenAnn is three times as old as Mary was when Mary was three times as oldas Ann. How old is Mary? That is all, but can you work it out? If not, ask your friends to help you, and watch the shadow of bewilderment creepover their faces as they attempt to grip the intricacies of thequestion. 

52. --QUEER RELATIONSHIPS. 

"Speaking of relationships, " said the Parson at a certain dinner-party, "our legislators are getting the marriage law into a frightful tangle, Here, for example, is a puzzling case that has come under my notice. Twobrothers married two sisters. One man died and the other man's wife alsodied. Then the survivors married. "

"The man married his deceased wife's sister under the recent Act?" putin the Lawyer. 

"Exactly. And therefore, under the civil law, he is legally married andhis child is legitimate. But, you see, the man is the woman's deceasedhusband's brother, and therefore, also under the civil law, she is notmarried to him and her child is illegitimate. "

"He is married to her and she is not married to him!" said the Doctor. 

"Quite so. And the child is the legitimate son of his father, but theillegitimate son of his mother. "

"Undoubtedly 'the law is a hass, '" the Artist exclaimed, "if I may bepermitted to say so, " he added, with a bow to the Lawyer. 

"Certainly, " was the reply. "We lawyers try our best to break in thebeast to the service of man. Our legislators are responsible for thebreed. "

"And this reminds me, " went on the Parson, "of a man in my parish whomarried the sister of his widow. This man--"

"Stop a moment, sir, " said the Professor. "Married the sister of hiswidow? Do you marry dead men in your parish?"

"No; but I will explain that later. Well, this man has a sister of hisown. Their names are Stephen Brown and Jane Brown. Last week a youngfellow turned up whom Stephen introduced to me as his nephew. Naturally, I spoke of Jane as his aunt, but, to my astonishment, the youthcorrected me, assuring me that, though he was the nephew of Stephen, hewas not the nephew of Jane, the sister of Stephen. This perplexed me agood deal, but it is quite correct. "

The Lawyer was the first to get at the heart of the mystery. What washis solution?

53. --HEARD ON THE TUBE RAILWAY. 

First Lady: "And was he related to you, dear?"

Second Lady: "Oh, yes. You see, that gentleman's mother was my mother'smother-in-law, but he is not on speaking terms with my papa. "

First Lady: "Oh, indeed!" (But you could see that she was not muchwiser. )

How was the gentleman related to the Second Lady?

54. --A FAMILY PARTY. 

A certain family party consisted of 1 grandfather, 1 grandmother, 2fathers, 2 mothers, 4 children, 3 grandchildren, 1 brother, 2 sisters, 2sons, 2 daughters, 1 father-in-law, 1 mother-in-law, and 1daughter-in-law. Twenty-three people, you will say. No; there were onlyseven persons present. Can you show how this might be?

55. --A MIXED PEDIGREE. 

Joseph Bloggs: "I can't follow it, my dear boy. It makes me dizzy!"

John Snoggs: "It's very simple. Listen again! You happen to be myfather's brother-in-law, my brother's father-in-law, and also myfather-in-law's brother. You see, my father was--"

But Mr. Bloggs refused to hear any more. Can the reader show how thisextraordinary triple relationship might have come about?

56. --WILSON'S POSER. 

"Speaking of perplexities--" said Mr. Wilson, throwing down a magazineon the table in the commercial room of the Railway Hotel. 

"Who was speaking of perplexities?" inquired Mr. Stubbs. 

"Well, then, reading about them, if you want to be exact--it justoccurred to me that perhaps you three men may be interested in a littlematter connected with myself. "

It was Christmas Eve, and the four commercial travellers were spendingthe holiday at Grassminster. Probably each suspected that the others hadno homes, and perhaps each was conscious of the fact that he was in thatpredicament himself. In any case they seemed to be perfectlycomfortable, and as they drew round the cheerful fire the conversationbecame general. 

"What is the difficulty?" asked Mr. Packhurst. 

"There's no difficulty in the matter, when you rightly understand it. Itis like this. A man named Parker had a flying-machine that would carrytwo. He was a venturesome sort of chap--reckless, I should call him--andhe had some bother in finding a man willing to risk his life in makingan ascent with him. However, an uncle of mine thought he would chanceit, and one fine morning he took his seat in the machine and she startedoff well. When they were up about a thousand feet, my nephewsuddenly--"

"Here, stop, Wilson! What was your nephew doing there? You said youruncle, " interrupted Mr. Stubbs. 

"Did I? Well, it does not matter. My nephew suddenly turned to Parkerand said that the engine wasn't running well, so Parker called out to myuncle--"

"Look here, " broke in Mr. Waterson, "we are getting mixed. Was it youruncle or your nephew? Let's have it one way or the other. "

"What I said is quite right. Parker called out to my uncle to dosomething or other, when my nephew--"

"There you are again, Wilson, " cried Mr. Stubbs; "once for all, are weto understand that both your uncle and your nephew were on the machine?"

"Certainly. I thought I made that clear. Where was I? Well, my nephewshouted back to Parker--"

"Phew! I'm sorry to interrupt you again, Wilson, but we can't get onlike this. Is it true that the machine would only carry two?"

"Of course. I said at the start that it only carried two. "

"Then what in the name of aerostation do you mean by saying that therewere three persons on board?" shouted Mr. Stubbs. 

"Who said there were three?"

"You have told us that Parker, your uncle, and your nephew went up onthis blessed flying-machine. "

"That's right. "

"And the thing would only carry two!"

"Right again. "

"Wilson, I have known you for some time as a truthful man and atemperate man, " said Mr. Stubbs, solemnly. "But I am afraid since youtook up that new line of goods you have overworked yourself. "

"Half a minute, Stubbs, " interposed Mr. Waterson. "I see clearly wherewe all slipped a cog. Of course, Wilson, you meant us to understand thatParker is either your uncle or your nephew. Now we shall be all right ifyou will just tell us whether Parker is your uncle or nephew. "

"He is no relation to me whatever. "

The three men sighed and looked anxiously at one another. Mr. Stubbs gotup from his chair to reach the matches, Mr. Packhurst proceeded to windup his watch, and Mr. Waterson took up the poker to attend to the fire. It was an awkward moment, for at the season of goodwill nobody wished totell Mr. Wilson exactly what was in his mind. 

"It's curious, " said Mr. Wilson, very deliberately, "and it's rathersad, how thick-headed some people are. You don't seem to grip the facts. It never seems to have occurred to either of you that my uncle and mynephew are one and the same man. "

"What!" exclaimed all three together. 

"Yes; David George Linklater is my uncle, and he is also my nephew. Consequently, I am both his uncle and nephew. Queer, isn't it? I'llexplain how it comes about. "

Mr. Wilson put the case so very simply that the three men saw how itmight happen without any marriage within the prohibited degrees. Perhapsthe reader can work it out for himself. 

CLOCK PUZZLES. 

 "Look at the clock!"

 _Ingoldsby Legends_. 

In considering a few puzzles concerning clocks and watches, and thetimes recorded by their hands under given conditions, it is well that aparticular convention should always be kept in mind. It is frequentlythe case that a solution requires the assumption that the hands canactually record a time involving a minute fraction of a second. Such atime, of course, cannot be really indicated. Is the puzzle, therefore, impossible of solution? The conclusion deduced from a logical syllogismdepends for its truth on the two premises assumed, and it is the same inmathematics. Certain things are antecedently assumed, and the answerdepends entirely on the truth of those assumptions. 

"If two horses, " says Lagrange, "can pull a load of a certain weight, itis natural to suppose that four horses could pull a load of double thatweight, six horses a load of three times that weight. Yet, strictlyspeaking, such is not the case. For the inference is based on theassumption that the four horses pull alike in amount and direction, which in practice can scarcely ever be the case. It so happens that weare frequently led in our reckonings to results which diverge widelyfrom reality. But the fault is not the fault of mathematics; formathematics always gives back to us exactly what we have put into it. The ratio was constant according to that supposition. The result isfounded upon that supposition. If the supposition is false the result isnecessarily false. "

If one man can reap a field in six days, we say two men will reap it inthree days, and three men will do the work in two days. We here assume, as in the case of Lagrange's horses, that all the men are exactlyequally capable of work. But we assume even more than this. For whenthree men get together they may waste time in gossip or play; or, on theother hand, a spirit of rivalry may spur them on to greater diligence. We may assume any conditions we like in a problem, provided they beclearly expressed and understood, and the answer will be in accordancewith those conditions. 

57. --WHAT WAS THE TIME?

"I say, Rackbrane, what is the time?" an acquaintance asked our friendthe professor the other day. The answer was certainly curious. 

"If you add one quarter of the time from noon till now to half the timefrom now till noon to-morrow, you will get the time exactly. "

What was the time of day when the professor spoke?

58. --A TIME PUZZLE. 

How many minutes is it until six o'clock if fifty minutes ago it wasfour times as many minutes past three o'clock?

59. --A PUZZLING WATCH. 

A friend pulled out his watch and said, "This watch of mine does notkeep perfect time; I must have it seen to. I have noticed that theminute hand and the hour hand are exactly together every sixty-fiveminutes. " Does that watch gain or lose, and how much per hour?

60. --THE WAPSHAW'S WHARF MYSTERY. 

There was a great commotion in Lower Thames Street on the morning ofJanuary 12, 1887. When the early members of the staff arrived atWapshaw's Wharf they found that the safe had been broken open, aconsiderable sum of money removed, and the offices left in greatdisorder. The night watchman was nowhere to be found, but nobody who hadbeen acquainted with him for one moment suspected him to be guilty ofthe robbery. In this belief the proprietors were confirmed when, laterin the day, they were informed that the poor fellow's body had beenpicked up by the River Police. Certain marks of violence pointed to thefact that he had been brutally attacked and thrown into the river. Awatch found in his pocket had stopped, as is invariably the case in suchcircumstances, and this was a valuable clue to the time of the outrage. But a very stupid officer (and we invariably find one or two stupidindividuals in the most intelligent bodies of men) had actually amusedhimself by turning the hands round and round, trying to set the watchgoing again. After he had been severely reprimanded for this seriousindiscretion, he was asked whether he could remember the time that wasindicated by the watch when found. He replied that he could not, but herecollected that the hour hand and minute hand were exactly together, one above the other, and the second hand had just passed the forty-ninthsecond. More than this he could not remember. 

What was the exact time at which the watchman's watch stopped? The watchis, of course, assumed to have been an accurate one. 

61. --CHANGING PLACES. 

[Illustration]

The above clock face indicates a little before 42 minutes past 4. Thehands will again point at exactly the same spots a little after 23minutes past 8. In fact, the hands will have changed places. How manytimes do the hands of a clock change places between three o'clock p. M. And midnight? And out of all the pairs of times indicated by thesechanges, what is the exact time when the minute hand will be nearest tothe point IX?

62. --THE CLUB CLOCK. 

One of the big clocks in the Cogitators' Club was found the other nightto have stopped just when, as will be seen in the illustration, thesecond hand was exactly midway between the other two hands. One of themembers proposed to some of his friends that they should tell him theexact time when (if the clock had not stopped) the second hand wouldnext again have been midway between the minute hand and the hour hand. Can you find the correct time that it would happen?

[Illustration]

63. --THE STOP-WATCH. 

[Illustration]

We have here a stop-watch with three hands. The second hand, whichtravels once round the face in a minute, is the one with the little ringat its end near the centre. Our dial indicates the exact time when itsowner stopped the watch. You will notice that the three hands are nearlyequidistant. The hour and minute hands point to spots that are exactly athird of the circumference apart, but the second hand is a little tooadvanced. An exact equidistance for the three hands is not possible. Now, we want to know what the time will be when the three hands are nextat exactly the same distances as shown from one another. Can you statethe time?

64. --THE THREE CLOCKS. 

On Friday, April 1, 1898, three new clocks were all set going preciselyat the same time--twelve noon. At noon on the following day it was foundthat clock A had kept perfect time, that clock B had gained exactly oneminute, and that clock C had lost exactly one minute. Now, supposingthat the clocks B and C had not been regulated, but all three allowed togo on as they had begun, and that they maintained the same rates ofprogress without stopping, on what date and at what time of day wouldall three pairs of hands again point at the same moment at twelveo'clock?

65. --THE RAILWAY STATION CLOCK. 

A clock hangs on the wall of a railway station, 71 ft. 9 in. Long and 10ft. 4 in. High. Those are the dimensions of the wall, not of the clock!While waiting for a train we noticed that the hands of the clock werepointing in opposite directions, and were parallel to one of thediagonals of the wall. What was the exact time?

66. --THE VILLAGE SIMPLETON. 

A facetious individual who was taking a long walk in the country cameupon a yokel sitting on a stile. As the gentleman was not quite sure ofhis road, he thought he would make inquiries of the local inhabitant;but at the first glance he jumped too hastily to the conclusion that hehad dropped on the village idiot. He therefore decided to test thefellow's intelligence by first putting to him the simplest question hecould think of, which was, "What day of the week is this, my good man?"The following is the smart answer that he received:--

"When the day after to-morrow is yesterday, to-day will be as far fromSunday as to-day was from Sunday when the day before yesterday wasto-morrow. "

Can the reader say what day of the week it was? It is pretty evidentthat the countryman was not such a fool as he looked. The gentleman wenton his road a puzzled but a wiser man. 

LOCOMOTION AND SPEED PUZZLES. 

"The race is not to the swift. "--_Ecclesiastes_ ix. II. 

67. --AVERAGE SPEED. 

In a recent motor ride it was found that we had gone at the rate of tenmiles an hour, but we did the return journey over the same route, owingto the roads being more clear of traffic, at fifteen miles an hour. Whatwas our average speed? Do not be too hasty in your answer to this simplelittle question, or it is pretty certain that you will be wrong. 

68. --THE TWO TRAINS. 

I put this little question to a stationmaster, and his correct answerwas so prompt that I am convinced there is no necessity to seek talentedrailway officials in America or elsewhere. 

Two trains start at the same time, one from London to Liverpool, theother from Liverpool to London. If they arrive at their destinations onehour and four hours respectively after passing one another, how muchfaster is one train running than the other?

69. --THE THREE VILLAGES. 

I set out the other day to ride in a motor-car from Acrefield toButterford, but by mistake I took the road going _via_ Cheesebury, whichis nearer Acrefield than Butterford, and is twelve miles to the left ofthe direct road I should have travelled. After arriving at Butterford Ifound that I had gone thirty-five miles. What are the three distancesbetween these villages, each being a whole number of miles? I maymention that the three roads are quite straight. 

70. --DRAWING HER PENSION. 

"Speaking of odd figures, " said a gentleman who occupies some post in aGovernment office, "one of the queerest characters I know is an old lamewidow who climbs up a hill every week to draw her pension at the villagepost office. She crawls up at the rate of a mile and a half an hour andcomes down at the rate of four and a half miles an hour, so that ittakes her just six hours to make the double journey. Can any of you tellme how far it is from the bottom of the hill to the top?"

[Illustration]

71. --SIR EDWYN DE TUDOR. 

In the illustration we have a sketch of Sir Edwyn de Tudor going torescue his lady-love, the fair Isabella, who was held a captive by aneighbouring wicked baron. Sir Edwyn calculated that if he rode fifteenmiles an hour he would arrive at the castle an hour too soon, while ifhe rode ten miles an hour he would get there just an hour too late. Now, it was of the first importance that he should arrive at the exact timeappointed, in order that the rescue that he had planned should be asuccess, and the time of the tryst was five o'clock, when the captivelady would be taking her afternoon tea. The puzzle is to discoverexactly how far Sir Edwyn de Tudor had to ride. 

72. --THE HYDROPLANE QUESTION. 

The inhabitants of Slocomb-on-Sea were greatly excited over the visit ofa certain flying man. All the town turned out to see the flight of thewonderful hydroplane, and, of course, Dobson and his family were there. Master Tommy was in good form, and informed his father that Englishmenmade better airmen than Scotsmen and Irishmen because they are not soheavy. "How do you make that out?" asked Mr. Dobson. "Well, you see, "Tommy replied, "it is true that in Ireland there are men of Cork and inScotland men of Ayr, which is better still, but in England there arelightermen. " Unfortunately it had to be explained to Mrs. Dobson, andthis took the edge off the thing. The hydroplane flight was from Slocombto the neighbouring watering-place Poodleville--five miles distant. Butthere was a strong wind, which so helped the airman that he made theoutward journey in the short time of ten minutes, though it took him anhour to get back to the starting point at Slocomb, with the wind deadagainst him. Now, how long would the ten miles have taken him if therehad been a perfect calm? Of course, the hydroplane's engine workeduniformly throughout. 

73. --DONKEY RIDING. 

During a visit to the seaside Tommy and Evangeline insisted on having adonkey race over the mile course on the sands. Mr. Dobson and some ofhis friends whom he had met on the beach acted as judges, but, as thedonkeys were familiar acquaintances and declined to part company thewhole way, a dead heat was unavoidable. However, the judges, beingstationed at different points on the course, which was marked off inquarter-miles, noted the following results:--The first three-quarterswere run in six and three-quarter minutes, the first half-mile took thesame time as the second half, and the third quarter was run in exactlythe same time as the last quarter. From these results Mr. Dobson amusedhimself in discovering just how long it took those two donkeys to runthe whole mile. Can you give the answer?

74. --THE BASKET OF POTATOES. 

A man had a basket containing fifty potatoes. He proposed to his son, asa little recreation, that he should place these potatoes on the groundin a straight line. The distance between the first and second potatoeswas to be one yard, between the second and third three yards, betweenthe third and fourth five yards, between the fourth and fifth sevenyards, and so on--an increase of two yards for every successive potatolaid down. Then the boy was to pick them up and put them in the basketone at a time, the basket being placed beside the first potato. How farwould the boy have to travel to accomplish the feat of picking them allup? We will not consider the journey involved in placing the potatoes, so that he starts from the basket with them all laid out. 

75. --THE PASSENGER'S FARE. 

At first sight you would hardly think there was matter for dispute inthe question involved in the following little incident, yet it took thetwo persons concerned some little time to come to an agreement. Mr. Smithers hired a motor-car to take him from Addleford to Clinkervilleand back again for £3. At Bakenham, just midway, he picked up anacquaintance, Mr. Tompkins, and agreed to take him on to Clinkervilleand bring him back to Bakenham on the return journey. How much should hehave charged the passenger? That is the question. What was a reasonablefare for Mr. Tompkins?

DIGITAL PUZZLES. 

 "Nine worthies were they called. " DRYDEN: _The Flower and the Leaf. _

I give these puzzles, dealing with the nine digits, a class tothemselves, because I have always thought that they deserve moreconsideration than they usually receive. Beyond the mere trick of"casting out nines, " very little seems to be generally known of the lawsinvolved in these problems, and yet an acquaintance with the propertiesof the digits often supplies, among other uses, a certain number ofarithmetical checks that are of real value in the saving of labour. Letme give just one example--the first that occurs to me. 

If the reader were required to determine whether or not15, 763, 530, 163, 289 is a square number, how would he proceed? If thenumber had ended with a 2, 3, 7, or 8 in the digits place, of course hewould know that it could not be a square, but there is nothing in itsapparent form to prevent its being one. I suspect that in such a case hewould set to work, with a sigh or a groan, at the laborious task ofextracting the square root. Yet if he had given a little attention tothe study of the digital properties of numbers, he would settle thequestion in this simple way. The sum of the digits is 59, the sum ofwhich is 14, the sum of which is 5 (which I call the "digital root"), and therefore I know that the number cannot be a square, and for thisreason. The digital root of successive square numbers from 1 upwards isalways 1, 4, 7, or 9, and can never be anything else. In fact, theseries, 1, 4, 9, 7, 7, 9, 4, 1, 9, is repeated into infinity. Theanalogous series for triangular numbers is 1, 3, 6, 1, 6, 3, 1, 9, 9. Sohere we have a similar negative check, for a number cannot be triangular(that is, (n² + n)/2) if its digital root be 2, 4, 5, 7, or 8. 

76. --THE BARREL OF BEER. 

A man bought an odd lot of wine in barrels and one barrel containingbeer. These are shown in the illustration, marked with the number ofgallons that each barrel contained. He sold a quantity of the wine toone man and twice the quantity to another, but kept the beer to himself. The puzzle is to point out which barrel contains beer. Can you say whichone it is? Of course, the man sold the barrels just as he bought them, without manipulating in any way the contents. 

[Illustration:

 ( 15 Gals )

 (31 Gals) (19 Gals)

 (20 Gals) (16 Gals) (18 Gals)

]

77. --DIGITS AND SQUARES. 

[Illustration:

 +---+---+---+ | 1 | 9 | 2 | +---+---+---+ | 3 | 8 | 4 | +---+---+---+ | 5 | 7 | 6 | +---+---+---+

]

It will be seen in the diagram that we have so arranged the nine digitsin a square that the number in the second row is twice that in the firstrow, and the number in the bottom row three times that in the top row. There are three other ways of arranging the digits so as to produce thesame result. Can you find them?

78. --ODD AND EVEN DIGITS. 

The odd digits, 1, 3, 5, 7, and 9, add up 25, while the even figures, 2, 4, 6, and 8, only add up 20. Arrange these figures so that the odd onesand the even ones add up alike. Complex and improper fractions andrecurring decimals are not allowed. 

79. --THE LOCKERS PUZZLE. 

[Illustration:

 A B C ================== ================== ================== | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | ================== ================== ================== | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | ------------------ ------------------ ------------------

]

A man had in his office three cupboards, each containing nine lockers, as shown in the diagram. He told his clerk to place a differentone-figure number on each locker of cupboard A, and to do the same inthe case of B, and of C. As we are here allowed to call nought a digit, and he was not prohibited from using nought as a number, he clearly hadthe option of omitting any one of ten digits from each cupboard. 

Now, the employer did not say the lockers were to be numbered in anynumerical order, and he was surprised to find, when the work was done, that the figures had apparently been mixed up indiscriminately. Callingupon his clerk for an explanation, the eccentric lad stated that thenotion had occurred to him so to arrange the figures that in each casethey formed a simple addition sum, the two upper rows of figuresproducing the sum in the lowest row. But the most surprising point wasthis: that he had so arranged them that the addition in A gave thesmallest possible sum, that the addition in C gave the largest possiblesum, and that all the nine digits in the three totals were different. The puzzle is to show how this could be done. No decimals are allowedand the nought may not appear in the hundreds place. 

80. --THE THREE GROUPS. 

There appeared in "Nouvelles Annales de Mathématiques" the followingpuzzle as a modification of one of my "Canterbury Puzzles. " Arrange thenine digits in three groups of two, three, and four digits, so that thefirst two numbers when multiplied together make the third. Thus, 12 ×483 = 5, 796. I now also propose to include the cases where there areone, four, and four digits, such as 4 × 1, 738 = 6, 952. Can you find allthe possible solutions in both cases?

81. --THE NINE COUNTERS. 

[Illustration:

 (1)(5)(8) (7)(9) (2)(3) (4)(6)

]

I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8 and 9. I arranged them on the table in two groups, as shownin the illustration, so as to form two multiplication sums, and foundthat both sums gave the same product. You will find that 158 multipliedby 23 is 3, 634, and that 79 multiplied by 46 is also 3, 634. Now, thepuzzle I propose is to rearrange the counters so as to get as large aproduct as possible. What is the best way of placing them? Remember bothgroups must multiply to the same amount, and there must be threecounters multiplied by two in one case, and two multiplied by twocounters in the other, just as at present. 

82. --THE TEN COUNTERS. 

In this case we use the nought in addition to the 1, 2, 3, 4, 5, 6, 7, 8, 9. The puzzle is, as in the last case, so to arrange the ten countersthat the products of the two multiplications shall be the same, and youmay here have one or more figures in the multiplier, as you choose. Theabove is a very easy feat; but it is also required to find the twoarrangements giving pairs of the highest and lowest products possible. Of course every counter must be used, and the cipher may not be placedto the left of a row of figures where it would have no effect. Vulgarfractions or decimals are not allowed. 

83. --DIGITAL MULTIPLICATION. 

Here is another entertaining problem with the nine digits, the noughtbeing excluded. Using each figure once, and only once, we can form twomultiplication sums that have the same product, and this may be done inmany ways. For example, 7 × 658 and 14 × 329 contain all the digitsonce, and the product in each case is the same--4, 606. Now, it will beseen that the sum of the digits in the product is 16, which is neitherthe highest nor the lowest sum so obtainable. Can you find the solutionof the problem that gives the lowest possible sum of digits in thecommon product? Also that which gives the highest possible sum?

84. --THE PIERROT'S PUZZLE. 

[Illustration]

The Pierrot in the illustration is standing in a posture that representsthe sign of multiplication. He is indicating the peculiar fact that 15multiplied by 93 produces exactly the same figures (1, 395), differentlyarranged. The puzzle is to take any four digits you like (all different)and similarly arrange them so that the number formed on one side of thePierrot when multiplied by the number on the other side shall producethe same figures. There are very few ways of doing it, and I shall giveall the cases possible. Can you find them all? You are allowed to puttwo figures on each side of the Pierrot as in the example shown, or toplace a single figure on one side and three figures on the other. If weonly used three digits instead of four, the only possible ways arethese: 3 multiplied by 51 equals 153, and 6 multiplied by 21 equals 126. 

85. --THE CAB NUMBERS. 

A London policeman one night saw two cabs drive off in oppositedirections under suspicious circumstances. This officer was aparticularly careful and wide-awake man, and he took out his pocket-bookto make an entry of the numbers of the cabs, but discovered that he hadlost his pencil. Luckily, however, he found a small piece of chalk, withwhich he marked the two numbers on the gateway of a wharf close by. Whenhe returned to the same spot on his beat he stood and looked again atthe numbers, and noticed this peculiarity, that all the nine digits (nonought) were used and that no figure was repeated, but that if hemultiplied the two numbers together they again produced the nine digits, all once, and once only. When one of the clerks arrived at the wharf inthe early morning, he observed the chalk marks and carefully rubbed themout. As the policeman could not remember them, certain mathematicianswere then consulted as to whether there was any known method fordiscovering all the pairs of numbers that have the peculiarity that theofficer had noticed; but they knew of none. The investigation, however, was interesting, and the following question out of many was proposed:What two numbers, containing together all the nine digits, will, whenmultiplied together, produce another number (the _highest possible_)containing also all the nine digits? The nought is not allowed anywhere. 

86. --QUEER MULTIPLICATION. 

If I multiply 51, 249, 876 by 3 (thus using all the nine digits once, andonce only), I get 153, 749, 628 (which again contains all the nine digitsonce). Similarly, if I multiply 16, 583, 742 by 9 the result is149, 253, 678, where in each case all the nine digits are used. Now, take6 as your multiplier and try to arrange the remaining eight digits so asto produce by multiplication a number containing all nine once, and onceonly. You will find it far from easy, but it can be done. 

87. --THE NUMBER-CHECKS PUZZLE. 

[Illustration]

Where a large number of workmen are employed on a building it iscustomary to provide every man with a little disc bearing his number. These are hung on a board by the men as they arrive, and serve as acheck on punctuality. Now, I once noticed a foreman remove a number ofthese checks from his board and place them on a split-ring which hecarried in his pocket. This at once gave me the idea for a good puzzle. In fact, I will confide to my readers that this is just how ideas forpuzzles arise. You cannot really create an idea: it happens--and youhave to be on the alert to seize it when it does so happen. 

It will be seen from the illustration that there are ten of thesechecks on a ring, numbered 1 to 9 and 0. The puzzle is to divide theminto three groups without taking any off the ring, so that the firstgroup multiplied by the second makes the third group. For example, wecan divide them into the three groups, 2--8 9 0 7--1 5 4 6 3, bybringing the 6 and the 3 round to the 4, but unfortunately the firsttwo when multiplied together do not make the third. Can you separatethem correctly? Of course you may have as many of the checks as youlike in any group. The puzzle calls for some ingenuity, unless youhave the luck to hit on the answer by chance. 

88. --DIGITAL DIVISION. 

It is another good puzzle so to arrange the nine digits (the noughtexcluded) into two groups so that one group when divided by the otherproduces a given number without remainder. For example, 1 3 4 5 8divided by 6 7 2 9 gives 2. Can the reader find similar arrangementsproducing 3, 4, 5, 6, 7, 8, and 9 respectively? Also, can he find thepairs of smallest possible numbers in each case? Thus, 1 4 6 5 8 dividedby 7 3 2 9 is just as correct for 2 as the other example we have given, but the numbers are higher. 

89. --ADDING THE DIGITS. 

If I write the sum of money, £987, 5s. 4½d. , and add up the digits, they sum to 36. No digit has thus been used a second time in the amountor addition. This is the largest amount possible under the conditions. Now find the smallest possible amount, pounds, shillings, pence, andfarthings being all represented. You need not use more of the ninedigits than you choose, but no digit may be repeated throughout. Thenought is not allowed. 

90. --THE CENTURY PUZZLE. 

Can you write 100 in the form of a mixed number, using all the ninedigits once, and only once? The late distinguished French mathematician, Edouard Lucas, found seven different ways of doing it, and expressed hisdoubts as to there being any other ways. As a matter of fact there arejust eleven ways and no more. Here is one of them, 91+5742/638. Nine ofthe other ways have similarly two figures in the integral part of thenumber, but the eleventh expression has only one figure there. Can thereader find this last form?

91. --MORE MIXED FRACTIONS. 

When I first published my solution to the last puzzle, I was led toattempt the expression of all numbers in turn up to 100 by a mixedfraction containing all the nine digits. Here are twelve numbers for thereader to try his hand at: 13, 14, 15, 16, 18, 20, 27, 36, 40, 69, 72, 94. Use every one of the nine digits once, and only once, in every case. 

92. --DIGITAL SQUARE NUMBERS. 

Here are the nine digits so arranged that they form four square numbers:9, 81, 324, 576. Now, can you put them all together so as to form asingle square number--(I) the smallest possible, and (II) the largestpossible?

93. --THE MYSTIC ELEVEN. 

Can you find the largest possible number containing any nine of the tendigits (calling nought a digit) that can be divided by 11 without aremainder? Can you also find the smallest possible number produced inthe same way that is divisible by 11? Here is an example, where thedigit 5 has been omitted: 896743012. This number contains nine of thedigits and is divisible by 11, but it is neither the largest nor thesmallest number that will work. 

94. --THE DIGITAL CENTURY. 

1 2 3 4 5 6 7 8 9 = 100. 

It is required to place arithmetical signs between the nine figures sothat they shall equal 100. Of course, you must not alter the presentnumerical arrangement of the figures. Can you give a correct solutionthat employs (1) the fewest possible signs, and (2) the fewest possibleseparate strokes or dots of the pen? That is, it is necessary to use asfew signs as possible, and those signs should be of the simplest form. The signs of addition and multiplication (+ and ×) will thus count astwo strokes, the sign of subtraction (-) as one stroke, the sign ofdivision (÷) as three, and so on. 

95. --THE FOUR SEVENS. 

[Illustration]

In the illustration Professor Rackbrane is seen demonstrating one of thelittle posers with which he is accustomed to entertain his class. Hebelieves that by taking his pupils off the beaten tracks he is thebetter able to secure their attention, and to induce original andingenious methods of thought. He has, it will be seen, just shown howfour 5's may be written with simple arithmetical signs so as torepresent 100. Every juvenile reader will see at a glance that hisexample is quite correct. Now, what he wants you to do is this: Arrangefour 7's (neither more nor less) with arithmetical signs so that theyshall represent 100. If he had said we were to use four 9's we might atonce have written 99+9/9, but the four 7's call for rather moreingenuity. Can you discover the little trick?

96. --THE DICE NUMBERS. 

[Illustration]

I have a set of four dice, not marked with spots in the ordinary way, but with Arabic figures, as shown in the illustration. Each die, ofcourse, bears the numbers 1 to 6. When put together they will form agood many, different numbers. As represented they make the number 1246. Now, if I make all the different four-figure numbers that are possiblewith these dice (never putting the same figure more than once in anynumber), what will they all add up to? You are allowed to turn the 6upside down, so as to represent a 9. I do not ask, or expect, the readerto go to all the labour of writing out the full list of numbers and thenadding them up. Life is not long enough for such wasted energy. Can youget at the answer in any other way?

VARIOUS ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 

 "Variety's the very spice of life, That gives it all its flavour. "

 COWPER: _The Task. _

97. --THE SPOT ON THE TABLE. 

A boy, recently home from school, wished to give his father anexhibition of his precocity. He pushed a large circular table into thecorner of the room, as shown in the illustration, so that it touchedboth walls, and he then pointed to a spot of ink on the extreme edge. 

[Illustration]

"Here is a little puzzle for you, pater, " said the youth. "That spot isexactly eight inches from one wall and nine inches from the other. Canyou tell me the diameter of the table without measuring it?"

The boy was overheard to tell a friend, "It fairly beat the guv'nor;"but his father is known to have remarked to a City acquaintance that hesolved the thing in his head in a minute. I often wonder which spoke thetruth. 

98. --ACADEMIC COURTESIES. 

In a certain mixed school, where a special feature was made of theinculcation of good manners, they had a curious rule on assembling everymorning. There were twice as many girls as boys. Every girl made a bowto every other girl, to every boy, and to the teacher. Every boy made abow to every other boy, to every girl, and to the teacher. In all therewere nine hundred bows made in that model academy every morning. Now, can you say exactly how many boys there were in the school? If you arenot very careful, you are likely to get a good deal out in yourcalculation. 

99. --THE THIRTY-THREE PEARLS. 

[Illustration]

"A man I know, " said Teddy Nicholson at a certain family party, "possesses a string of thirty-three pearls. The middle pearl is thelargest and best of all, and the others are so selected and arrangedthat, starting from one end, each successive pearl is worth £100 morethan the preceding one, right up to the big pearl. From the other endthe pearls increase in value by £150 up to the large pearl. The wholestring is worth £65, 000. What is the value of that large pearl?"

"Pearls and other articles of clothing, " said Uncle Walter, when theprice of the precious gem had been discovered, "remind me of Adam andEve. Authorities, you may not know, differ as to the number of applesthat were eaten by Adam and Eve. It is the opinion of some that Eve 8(ate) and Adam 2 (too), a total of 10 only. But certain mathematicianshave figured it out differently, and hold that Eve 8 and Adam a total of16. Yet the most recent investigators think the above figures entirelywrong, for if Eve 8 and Adam 8 2, the total must be 90. "

"Well, " said Harry, "it seems to me that if there were giants in thosedays, probably Eve 8 1 and Adam 8 2, which would give a total of 163. "

"I am not at all satisfied, " said Maud. "It seems to me that if Eve 8 1and Adam 8 1 2, they together consumed 893. "

"I am sure you are all wrong, " insisted Mr. Wilson, "for I consider thatEve 8 1 4 Adam, and Adam 8 1 2 4 Eve, so we get a total of 8, 938. "

"But, look here, " broke in Herbert. "If Eve 8 1 4 Adam and Adam 8 1 2 42 oblige Eve, surely the total must have been 82, 056!"

At this point Uncle Walter suggested that they might let the matterrest. He declared it to be clearly what mathematicians call anindeterminate problem. 

100. --THE LABOURER'S PUZZLE. 

Professor Rackbrane, during one of his rambles, chanced to come upon aman digging a deep hole. 

"Good morning, " he said. "How deep is that hole?"

"Guess, " replied the labourer. "My height is exactly five feet teninches. "

"How much deeper are you going?" said the professor. 

"I am going twice as deep, " was the answer, "and then my head will betwice as far below ground as it is now above ground. "

Rackbrane now asks if you could tell how deep that hole would be whenfinished. 

101. --THE TRUSSES OF HAY. 

Farmer Tompkins had five trusses of hay, which he told his man Hodge toweigh before delivering them to a customer. The stupid fellow weighedthem two at a time in all possible ways, and informed his master thatthe weights in pounds were 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121. Now, how was Farmer Tompkins to find out from these figures howmuch every one of the five trusses weighed singly? The reader may atfirst think that he ought to be told "which pair is which pair, " orsomething of that sort, but it is quite unnecessary. Can you give thefive correct weights?

102. --MR. GUBBINS IN A FOG. 

Mr. Gubbins, a diligent man of business, was much inconvenienced by aLondon fog. The electric light happened to be out of order and he had tomanage as best he could with two candles. His clerk assured him thatthough both were of the same length one candle would burn for four hoursand the other for five hours. After he had been working some time he putthe candles out as the fog had lifted, and he then noticed that whatremained of one candle was exactly four times the length of what wasleft of the other. 

When he got home that night Mr. Gubbins, who liked a good puzzle, saidto himself, "Of course it is possible to work out just how long thosetwo candles were burning to-day. I'll have a shot at it. " But he soonfound himself in a worse fog than the atmospheric one. Could you haveassisted him in his dilemma? How long were the candles burning?

103. --PAINTING THE LAMP-POSTS. 

Tim Murphy and Pat Donovan were engaged by the local authorities topaint the lamp-posts in a certain street. Tim, who was an early riser, arrived first on the job, and had painted three on the south side whenPat turned up and pointed out that Tim's contract was for the northside. So Tim started afresh on the north side and Pat continued on thesouth. When Pat had finished his side he went across the street andpainted six posts for Tim, and then the job was finished. As there wasan equal number of lamp-posts on each side of the street, the simplequestion is: Which man painted the more lamp-posts, and just how manymore?

104. --CATCHING THE THIEF. 

"Now, constable, " said the defendant's counsel in cross-examination, "you say that the prisoner was exactly twenty-seven steps ahead of youwhen you started to run after him?"

"Yes, sir. "

"And you swear that he takes eight steps to your five?"

"That is so. "

"Then I ask you, constable, as an intelligent man, to explain how youever caught him, if that is the case?"

"Well, you see, I have got a longer stride. In fact, two of my steps areequal in length to five of the prisoner's. If you work it out, you willfind that the number of steps I required would bring me exactly to thespot where I captured him. "

Here the foreman of the jury asked for a few minutes to figure out thenumber of steps the constable must have taken. Can you also say how manysteps the officer needed to catch the thief?

105. --THE PARISH COUNCIL ELECTION. 

Here is an easy problem for the novice. At the last election of theparish council of Tittlebury-in-the-Marsh there were twenty-threecandidates for nine seats. Each voter was qualified to vote for nine ofthese candidates or for any less number. One of the electors wants toknow in just how many different ways it was possible for him to vote. 

106. --THE MUDDLETOWN ELECTION. 

At the last Parliamentary election at Muddletown 5, 473 votes werepolled. The Liberal was elected by a majority of 18 over theConservative, by 146 over the Independent, and by 575 over theSocialist. Can you give a simple rule for figuring out how many voteswere polled for each candidate?

107. --THE SUFFRAGISTS' MEETING. 

At a recent secret meeting of Suffragists a serious difference ofopinion arose. This led to a split, and a certain number left themeeting. "I had half a mind to go myself, " said the chair-woman, "and ifI had done so, two-thirds of us would have retired. " "True, " saidanother member; "but if I had persuaded my friends Mrs. Wild andChristine Armstrong to remain we should only have lost half our number. "Can you tell how many were present at the meeting at the start?

108. --THE LEAP-YEAR LADIES. 

Last leap-year ladies lost no time in exercising the privilege of makingproposals of marriage. If the figures that reached me from an occultsource are correct, the following represents the state of affairs inthis country. 

A number of women proposed once each, of whom one-eighth were widows. Inconsequence, a number of men were to be married of whom one-eleventhwere widowers. Of the proposals made to widowers, one-fifth weredeclined. All the widows were accepted. Thirty-five forty-fourths of thewidows married bachelors. One thousand two hundred and twenty-onespinsters were declined by bachelors. The number of spinsters acceptedby bachelors was seven times the number of widows accepted by bachelors. Those are all the particulars that I was able to obtain. Now, how manywomen proposed?

109. --THE GREAT SCRAMBLE. 

After dinner, the five boys of a household happened to find a parcel ofsugar-plums. It was quite unexpected loot, and an exciting scrambleensued, the full details of which I will recount with accuracy, as itforms an interesting puzzle. 

You see, Andrew managed to get possession of just two-thirds of theparcel of sugar-plums. Bob at once grabbed three-eighths of these, andCharlie managed to seize three-tenths also. Then young David dashed uponthe scene, and captured all that Andrew had left, except one-seventh, which Edgar artfully secured for himself by a cunning trick. Now the funbegan in real earnest, for Andrew and Charlie jointly set upon Bob, whostumbled against the fender and dropped half of all that he had, whichwere equally picked up by David and Edgar, who had crawled under a tableand were waiting. Next, Bob sprang on Charlie from a chair, and upsetall the latter's collection on to the floor. Of this prize Andrew gotjust a quarter, Bob gathered up one-third, David got two-sevenths, whileCharlie and Edgar divided equally what was left of that stock. 

[Illustration]

They were just thinking the fray was over when David suddenly struck outin two directions at once, upsetting three-quarters of what Bob andAndrew had last acquired. The two latter, with the greatest difficulty, recovered five-eighths of it in equal shares, but the three others eachcarried off one-fifth of the same. Every sugar-plum was now accountedfor, and they called a truce, and divided equally amongst them theremainder of the parcel. What is the smallest number of sugar-plumsthere could have been at the start, and what proportion did each boyobtain?

110. --THE ABBOT'S PUZZLE. 

The first English puzzlist whose name has come down to us was aYorkshireman--no other than Alcuin, Abbot of Canterbury (A. D. 735-804). Here is a little puzzle from his works, which is at least interesting onaccount of its antiquity. "If 100 bushels of corn were distributed among100 people in such a manner that each man received three bushels, eachwoman two, and each child half a bushel, how many men, women, andchildren were there?"

Now, there are six different correct answers, if we exclude a case wherethere would be no women. But let us say that there were just five timesas many women as men, then what is the correct solution?

111. --REAPING THE CORN. 

A farmer had a square cornfield. The corn was all ripe for reaping, and, as he was short of men, it was arranged that he and his son should sharethe work between them. The farmer first cut one rod wide all round thesquare, thus leaving a smaller square of standing corn in the middle ofthe field. "Now, " he said to his son, "I have cut my half of the field, and you can do your share. " The son was not quite satisfied as to theproposed division of labour, and as the village schoolmaster happened tobe passing, he appealed to that person to decide the matter. He foundthe farmer was quite correct, provided there was no dispute as to thesize of the field, and on this point they were agreed. Can you tell thearea of the field, as that ingenious schoolmaster succeeded in doing?

112. --A PUZZLING LEGACY. 

A man left a hundred acres of land to be divided among his threesons--Alfred, Benjamin, and Charles--in the proportion of one-third, one-fourth, and one-fifth respectively. But Charles died. How was theland to be divided fairly between Alfred and Benjamin?

113. --THE TORN NUMBER. 

[Illustration]

I had the other day in my possession a label bearing the number 3 0 2 5in large figures. This got accidentally torn in half, so that 3 0 was onone piece and 2 5 on the other, as shown on the illustration. On lookingat these pieces I began to make a calculation, scarcely conscious ofwhat I was doing, when I discovered this little peculiarity. If we addthe 3 0 and the 2 5 together and square the sum we get as the result thecomplete original number on the label! Thus, 30 added to 25 is 55, and55 multiplied by 55 is 3025. Curious, is it not? Now, the puzzle is tofind another number, composed of four figures, all different, which maybe divided in the middle and produce the same result. 

114. --CURIOUS NUMBERS. 

The number 48 has this peculiarity, that if you add 1 to it the resultis a square number (49, the square of 7), and if you add 1 to its half, you also get a square number (25, the square of 5). Now, there is nolimit to the numbers that have this peculiarity, and it is aninteresting puzzle to find three more of them--the smallest possiblenumbers. What are they?

115. --A PRINTER'S ERROR. 

In a certain article a printer had to set up the figures 5^4×2^3, which, of course, means that the fourth power of 5 (625) is to be multiplied bythe cube of 2 (8), the product of which is 5, 000. But he printed 5^4×2^3as 5 4 2 3, which is not correct. Can you place four digits in themanner shown, so that it will be equally correct if the printer sets itup aright or makes the same blunder?

116. --THE CONVERTED MISER. 

Mr. Jasper Bullyon was one of the very few misers who have ever beenconverted to a sense of their duty towards their less fortunatefellow-men. One eventful night he counted out his accumulated wealth, and resolved to distribute it amongst the deserving poor. 

He found that if he gave away the same number of pounds every day in theyear, he could exactly spread it over a twelvemonth without there beinganything left over; but if he rested on the Sundays, and only gave awaya fixed number of pounds every weekday, there would be one sovereignleft over on New Year's Eve. Now, putting it at the lowest possible, what was the exact number of pounds that he had to distribute?

Could any question be simpler? A sum of pounds divided by one number ofdays leaves no remainder, but divided by another number of days leaves asovereign over. That is all; and yet, when you come to tackle thislittle question, you will be surprised that it can become so puzzling. 

117. --A FENCE PROBLEM. 

[Illustration]

The practical usefulness of puzzles is a point that we are liable tooverlook. Yet, as a matter of fact, I have from time to time receivedquite a large number of letters from individuals who have found that themastering of some little principle upon which a puzzle was built hasproved of considerable value to them in a most unexpected way. Indeed, it may be accepted as a good maxim that a puzzle is of little real valueunless, as well as being amusing and perplexing, it conceals someinstructive and possibly useful feature. It is, however, very curioushow these little bits of acquired knowledge dovetail into theoccasional requirements of everyday life, and equally curious to whatstrange and mysterious uses some of our readers seem to apply them. What, for example, can be the object of Mr. Wm. Oxley, who writes to meall the way from Iowa, in wishing to ascertain the dimensions of a fieldthat he proposes to enclose, containing just as many acres as thereshall be rails in the fence?

The man wishes to fence in a perfectly square field which is to containjust as many acres as there are rails in the required fence. Eachhurdle, or portion of fence, is seven rails high, and two lengths wouldextend one pole (16½ ft. ): that is to say, there are fourteen railsto the pole, lineal measure. Now, what must be the size of the field?

118. --CIRCLING THE SQUARES. 

[Illustration]

The puzzle is to place a different number in each of the ten squares sothat the sum of the squares of any two adjacent numbers shall be equalto the sum of the squares of the two numbers diametrically opposite tothem. The four numbers placed, as examples, must stand as they are. Thesquare of 16 is 256, and the square of 2 is 4. Add these together, andthe result is 260. Also--the square of 14 is 196, and the square of 8 is64. These together also make 260. Now, in precisely the same way, B andC should be equal to G and H (the sum will not necessarily be 260), Aand K to F and E, H and I to C and D, and so on, with any two adjoiningsquares in the circle. 

All you have to do is to fill in the remaining six numbers. Fractionsare not allowed, and I shall show that no number need contain more thantwo figures. 

119. --RACKBRANE'S LITTLE LOSS. 

Professor Rackbrane was spending an evening with his old friends, Mr. And Mrs. Potts, and they engaged in some game (he does not say whatgame) of cards. The professor lost the first game, which resulted indoubling the money that both Mr. And Mrs. Potts had laid on the table. The second game was lost by Mrs. Potts, which doubled the money thenheld by her husband and the professor. Curiously enough, the third gamewas lost by Mr. Potts, and had the effect of doubling the money thenheld by his wife and the professor. It was then found that each personhad exactly the same money, but the professor had lost five shillings inthe course of play. Now, the professor asks, what was the sum of moneywith which he sat down at the table? Can you tell him?

120. --THE FARMER AND HIS SHEEP. 

[Illustration]

Farmer Longmore had a curious aptitude for arithmetic, and was known inhis district as the "mathematical farmer. " The new vicar was not awareof this fact when, meeting his worthy parishioner one day in the lane, he asked him in the course of a short conversation, "Now, how many sheephave you altogether?" He was therefore rather surprised at Longmore'sanswer, which was as follows: "You can divide my sheep into twodifferent parts, so that the difference between the two numbers is thesame as the difference between their squares. Maybe, Mr. Parson, youwill like to work out the little sum for yourself. "

Can the reader say just how many sheep the farmer had? Supposing he hadpossessed only twenty sheep, and he divided them into the two parts 12and 8. Now, the difference between their squares, 144 and 64, is 80. Sothat will not do, for 4 and 80 are certainly not the same. If you canfind numbers that work out correctly, you will know exactly how manysheep Farmer Longmore owned. 

121. --HEADS OR TAILS. 

Crooks, an inveterate gambler, at Goodwood recently said to a friend, "I'll bet you half the money in my pocket on the toss of a coin--heads Iwin, tails I lose. " The coin was tossed and the money handed over. Herepeated the offer again and again, each time betting half the moneythen in his possession. We are not told how long the game went on, orhow many times the coin was tossed, but this we know, that the number oftimes that Crooks lost was exactly equal to the number of times that hewon. Now, did he gain or lose by this little venture?

122. --THE SEE-SAW PUZZLE. 

Necessity is, indeed, the mother of invention. I was amused the otherday in watching a boy who wanted to play see-saw and, in his failure tofind another child to share the sport with him, had been driven backupon the ingenious resort of tying a number of bricks to one end of theplank to balance his weight at the other. 

As a matter of fact, he just balanced against sixteen bricks, when thesewere fixed to the short end of plank, but if he fixed them to the longend of plank he only needed eleven as balance. 

Now, what was that boy's weight, if a brick weighs equal to athree-quarter brick and three-quarters of a pound?

123. --A LEGAL DIFFICULTY. 

"A client of mine, " said a lawyer, "was on the point of death when hiswife was about to present him with a child. I drew up his will, in whichhe settled two-thirds of his estate upon his son (if it should happen tobe a boy) and one-third on the mother. But if the child should be agirl, then two-thirds of the estate should go to the mother andone-third to the daughter. As a matter of fact, after his death twinswere born--a boy and a girl. A very nice point then arose. How was theestate to be equitably divided among the three in the closest possibleaccordance with the spirit of the dead man's will?"

124. --A QUESTION OF DEFINITION. 

"My property is exactly a mile square, " said one landowner to another. 

"Curiously enough, mine is a square mile, " was the reply. 

"Then there is no difference?"

Is this last statement correct?

125. --THE MINERS' HOLIDAY. 

Seven coal-miners took a holiday at the seaside during a big strike. Sixof the party spent exactly half a sovereign each, but Bill Harris wasmore extravagant. Bill spent three shillings more than the average ofthe party. What was the actual amount of Bill's expenditure?

126. --SIMPLE MULTIPLICATION. 

If we number six cards 1, 2, 4, 5, 7, and 8, and arrange them on thetable in this order:--

 1 4 2 8 5 7

We can demonstrate that in order to multiply by 3 all that is necessaryis to remove the 1 to the other end of the row, and the thing is done. The answer is 428571. Can you find a number that, when multiplied by 3and divided by 2, the answer will be the same as if we removed the firstcard (which in this case is to be a 3) From the beginning of the row tothe end?

127. --SIMPLE DIVISION. 

Sometimes a very simple question in elementary arithmetic will cause agood deal of perplexity. For example, I want to divide the four numbers, 701, 1, 059, 1, 417, and 2, 312, by the largest number possible that willleave the same remainder in every case. How am I to set to work Ofcourse, by a laborious system of trial one can in time discover theanswer, but there is quite a simple method of doing it if you can onlyfind it. 

128. --A PROBLEM IN SQUARES. 

We possess three square boards. The surface of the first contains fivesquare feet more than the second, and the second contains five squarefeet more than the third. Can you give exact measurements for the sidesof the boards? If you can solve this little puzzle, then try to findthree squares in arithmetical progression, with a common difference of 7and also of 13. 

129. --THE BATTLE OF HASTINGS. 

All historians know that there is a great deal of mystery anduncertainty concerning the details of the ever-memorable battle on thatfatal day, October 14, 1066. My puzzle deals with a curious passage inan ancient monkish chronicle that may never receive the attention thatit deserves, and if I am unable to vouch for the authenticity of thedocument it will none the less serve to furnish us with a problem thatcan hardly fail to interest those of my readers who have arithmeticalpredilections. Here is the passage in question. 

"The men of Harold stood well together, as their wont was, and formedsixty and one squares, with a like number of men in every squarethereof, and woe to the hardy Norman who ventured to enter theirredoubts; for a single blow of a Saxon war-hatchet would break his lanceand cut through his coat of mail. . . . When Harold threw himself into thefray the Saxons were one mighty square of men, shouting thebattle-cries, 'Ut!' 'Olicrosse!' 'Godemitè!'"

Now, I find that all the contemporary authorities agree that the Saxonsdid actually fight in this solid order. For example, in the "Carmen deBello Hastingensi, " a poem attributed to Guy, Bishop of Amiens, livingat the time of the battle, we are told that "the Saxons stood fixed in adense mass, " and Henry of Huntingdon records that "they were like unto acastle, impenetrable to the Normans;" while Robert Wace, a centuryafter, tells us the same thing. So in this respect my newly-discoveredchronicle may not be greatly in error. But I have reason to believe thatthere is something wrong with the actual figures. Let the reader seewhat he can make of them. 

The number of men would be sixty-one times a square number; but whenHarold himself joined in the fray they were then able to form one largesquare. What is the smallest possible number of men there could havebeen?

In order to make clear to the reader the simplicity of the question, Iwill give the lowest solutions in the case of 60 and 62, the numbersimmediately preceding and following 61. They are 60 × 4² + 1 = 31², and 62 × 8² + 1 = 63². That is, 60 squares of 16 men each would be 960men, and when Harold joined them they would be 961 in number, and soform a square with 31 men on every side. Similarly in the case of thefigures I have given for 62. Now, find the lowest answer for 61. 

130. --THE SCULPTOR'S PROBLEM. 

An ancient sculptor was commissioned to supply two statues, each on acubical pedestal. It is with these pedestals that we are concerned. Theywere of unequal sizes, as will be seen in the illustration, and when thetime arrived for payment a dispute arose as to whether the agreement wasbased on lineal or cubical measurement. But as soon as they came tomeasure the two pedestals the matter was at once settled, because, curiously enough, the number of lineal feet was exactly the same as thenumber of cubical feet. The puzzle is to find the dimensions for twopedestals having this peculiarity, in the smallest possible figures. Yousee, if the two pedestals, for example, measure respectively 3 ft. And 1ft. On every side, then the lineal measurement would be 4 ft. And thecubical contents 28 ft. , which are not the same, so these measurementswill not do. 

[Illustration]

131. --THE SPANISH MISER. 

There once lived in a small town in New Castile a noted miser named DonManuel Rodriguez. His love of money was only equalled by a strongpassion for arithmetical problems. These puzzles usually dealt in someway or other with his accumulated treasure, and were propounded by himsolely in order that he might have the pleasure of solving them himself. Unfortunately very few of them have survived, and when travellingthrough Spain, collecting material for a proposed work on "The SpanishOnion as a Cause of National Decadence, " I only discovered a very few. One of these concerns the three boxes that appear in the accompanyingauthentic portrait. 

[Illustration]

Each box contained a different number of golden doubloons. Thedifference between the number of doubloons in the upper box and thenumber in the middle box was the same as the difference between thenumber in the middle box and the number in the bottom box. And if thecontents of any two of the boxes were united they would form a squarenumber. What is the smallest number of doubloons that there could havebeen in any one of the boxes?

132. --THE NINE TREASURE BOXES. 

The following puzzle will illustrate the importance on occasions ofbeing able to fix the minimum and maximum limits of a required number. This can very frequently be done. For example, it has not yet beenascertained in how many different ways the knight's tour can beperformed on the chess board; but we know that it is fewer than thenumber of combinations of 168 things taken 63 at a time and is greaterthan 31, 054, 144--for the latter is the number of routes of a particulartype. Or, to take a more familiar case, if you ask a man how many coinshe has in his pocket, he may tell you that he has not the slightestidea. But on further questioning you will get out of him some suchstatement as the following: "Yes, I am positive that I have more thanthree coins, and equally certain that there are not so many astwenty-five. " Now, the knowledge that a certain number lies between 2and 12 in my puzzle will enable the solver to find the exact answer;without that information there would be an infinite number of answers, from which it would be impossible to select the correct one. 

This is another puzzle received from my friend Don Manuel Rodriguez, thecranky miser of New Castile. On New Year's Eve in 1879 he showed me ninetreasure boxes, and after informing me that every box contained a squarenumber of golden doubloons, and that the difference between the contentsof A and B was the same as between B and C, D and E, E and F, G and H, or H and I, he requested me to tell him the number of coins in every oneof the boxes. At first I thought this was impossible, as there would bean infinite number of different answers, but on consideration I foundthat this was not the case. I discovered that while every box containedcoins, the contents of A, B, C increased in weight in alphabeticalorder; so did D, E, F; and so did G, H, I; but D or E need not beheavier than C, nor G or H heavier than F. It was also perfectly certainthat box A could not contain more than a dozen coins at the outside;there might not be half that number, but I was positive that there werenot more than twelve. With this knowledge I was able to arrive at thecorrect answer. 

In short, we have to discover nine square numbers such that A, B, C; andD, E, F; and G, H, I are three groups in arithmetical progression, thecommon difference being the same in each group, and A being less than12. How many doubloons were there in every one of the nine boxes?

133. --THE FIVE BRIGANDS. 

The five Spanish brigands, Alfonso, Benito, Carlos, Diego, and Esteban, were counting their spoils after a raid, when it was found that they hadcaptured altogether exactly 200 doubloons. One of the band pointed outthat if Alfonso had twelve times as much, Benito three times as much, Carlos the same amount, Diego half as much, and Esteban one-third asmuch, they would still have altogether just 200 doubloons. How manydoubloons had each?

There are a good many equally correct answers to this question. Here isone of them:

 A 6 × 12 = 72 B 12 × 3 = 36 C 17 × 1 = 17 D 120 × ½ = 60 E 45 × 1/3 = 15 ___ ___ 200 200

The puzzle is to discover exactly how many different answers there are, it being understood that every man had something and that there is to beno fractional money--only doubloons in every case. 

This problem, worded somewhat differently, was propounded by Tartaglia(died 1559), and he flattered himself that he had found one solution;but a French mathematician of note (M. A. Labosne), in a recent work, says that his readers will be astonished when he assures them that thereare 6, 639 different correct answers to the question. Is this so? Howmany answers are there?

134. --THE BANKER'S PUZZLE. 

A banker had a sporting customer who was always anxious to wager onanything. Hoping to cure him of his bad habit, he proposed as a wagerthat the customer would not be able to divide up the contents of a boxcontaining only sixpences into an exact number of equal piles ofsixpences. The banker was first to put in one or more sixpences (as manyas he liked); then the customer was to put in one or more (but in hiscase not more than a pound in value), neither knowing what the other putin. Lastly, the customer was to transfer from the banker's counter tothe box as many sixpences as the banker desired him to put in. Thepuzzle is to find how many sixpences the banker should first put in andhow many he should ask the customer to transfer, so that he may have thebest chance of winning. 

135. --THE STONEMASON'S PROBLEM. 

A stonemason once had a large number of cubic blocks of stone in hisyard, all of exactly the same size. He had some very fanciful littleways, and one of his queer notions was to keep these blocks piled incubical heaps, no two heaps containing the same number of blocks. He haddiscovered for himself (a fact that is well known to mathematicians)that if he took all the blocks contained in any number of heaps inregular order, beginning with the single cube, he could always arrangethose on the ground so as to form a perfect square. This will be clearto the reader, because one block is a square, 1 + 8 = 9 is a square, 1 +8 + 27 = 36 is a square, 1 + 8 + 27 + 64 = 100 is a square, and so on. In fact, the sum of any number of consecutive cubes, beginning alwayswith 1, is in every case a square number. 

One day a gentleman entered the mason's yard and offered him a certainprice if he would supply him with a consecutive number of these cubicalheaps which should contain altogether a number of blocks that could belaid out to form a square, but the buyer insisted on more than threeheaps and _declined to take the single block_ because it contained aflaw. What was the smallest possible number of blocks of stone that themason had to supply?

136. --THE SULTAN'S ARMY. 

A certain Sultan wished to send into battle an army that could be formedinto two perfect squares in twelve different ways. What is the smallestnumber of men of which that army could be composed? To make it clear tothe novice, I will explain that if there were 130 men, they could beformed into two squares in only two different ways--81 and 49, or 121and 9. Of course, all the men must be used on every occasion. 

137. --A STUDY IN THRIFT. 

Certain numbers are called triangular, because if they are taken torepresent counters or coins they may be laid out on the table so as toform triangles. The number 1 is always regarded as triangular, just as 1is a square and a cube number. Place one counter on the table--that is, the first triangular number. Now place two more counters beneath it, andyou have a triangle of three counters; therefore 3 is triangular. Nextplace a row of three more counters, and you have a triangle of sixcounters; therefore 6 is triangular. We see that every row of countersthat we add, containing just one more counter than the row above it, makes a larger triangle. 

Now, half the sum of any number and its square is always a triangularnumber. Thus half of 2 + 2² = 3; half of 3 + 3² = 6; half of 4 +4² = 10; half of 5 + 5²= 15; and so on. So if we want to form atriangle with 8 counters on each side we shall require half of 8 +8², or 36 counters. This is a pretty little property of numbers. Before going further, I will here say that if the reader refers to the"Stonemason's Problem" (No. 135) he will remember that the sum of anynumber of consecutive cubes beginning with 1 is always a square, andthese form the series 1², 3², 6², 10², etc. It will now be understoodwhen I say that one of the keys to the puzzle was the fact that theseare always the squares of triangular numbers--that is, the squares of 1, 3, 6, 10, 15, 21, 28, etc. , any of which numbers we have seen will forma triangle. 

Every whole number is either triangular, or the sum of two triangularnumbers or the sum of three triangular numbers. That is, if we take anynumber we choose we can always form one, two, or three triangles withthem. The number 1 will obviously, and uniquely, only form one triangle;some numbers will only form two triangles (as 2, 4, 11, etc. ); somenumbers will only form three triangles (as 5, 8, 14, etc. ). Then, again, some numbers will form both one and two triangles (as 6), others bothone and three triangles (as 3 and 10), others both two and threetriangles (as 7 and 9), while some numbers (like 21) will form one, two, or three triangles, as we desire. Now for a little puzzle in triangularnumbers. 

Sandy McAllister, of Aberdeen, practised strict domestic economy, andwas anxious to train his good wife in his own habits of thrift. He toldher last New Year's Eve that when she had saved so many sovereigns thatshe could lay them all out on the table so as to form a perfect square, or a perfect triangle, or two triangles, or three triangles, just as hemight choose to ask he would add five pounds to her treasure. Soon shewent to her husband with a little bag of £36 in sovereigns and claimedher reward. It will be found that the thirty-six coins will form asquare (with side 6), that they will form a single triangle (with side8), that they will form two triangles (with sides 5 and 6), and thatthey will form three triangles (with sides 3, 5, and 5). In each of thefour cases all the thirty-six coins are used, as required, and Sandytherefore made his wife the promised present like an honest man. 

The Scotsman then undertook to extend his promise for five more years, so that if next year the increased number of sovereigns that she hassaved can be laid out in the same four different ways she will receive asecond present; if she succeeds in the following year she will get athird present, and so on until she has earned six presents in all. Now, how many sovereigns must she put together before she can win the sixthpresent?

What you have to do is to find five numbers, the smallest possible, higher than 36, that can be displayed in the four ways--to form asquare, to form a triangle, to form two triangles, and to form threetriangles. The highest of your five numbers will be your answer. 

138. --THE ARTILLERYMEN'S DILEMMA. 

[Illustration: [Pyramid of cannon-balls]]

 MMMMMMMr MM MM: M 0 rWZX M : MWM aX, BM M 0M M aMMMM2MW 02 MMWMMr ZM. M@M 8MM 7 XM2 MS2 M. MMMWMMMM MM M MX iMM M7W 8 . M r W M@ Z;M M 0r ; M M M W 22 W M @ M M M. M2WMMMMZ ;MM@X:7MMMB; MMM ZM M:MM0;8:, MS Ma 8 MMMMMMMi rM 2MMMMMM MB M 7 XM, ,: BMM: r7S . , MM MM MB M i, M, 2 ; aMMMMMMMMM XM; MZM M . M 7 M . Z M M M8 M M S M . 0 M 8MM aMi: MMMM7M, 7 . IM X M @ aZ M M 8, @MMMMBMMMa SMW 7M, XZ@MM M 8M M . M MMMM@X MMr Ma MMMMMMMMM@ M . WM M @WM7WMM . WX MZS M 8M :MMMWMMMM 8X MMMBMMM7 7aM 2MM r, 8r ZM2 Mr2 aMM; Mai :MS :iM ZiM @MX M M . M Wr. MMMaBMMMB M M MZ. , M MMZ Mr M M B0 Z 2S iM S XM 7 WMM MM @. M M M W M. M M 0;M2M;MMMM: WW8aMM M S@ M M M : MaMMMMMM MM0W;MZM: M i M M MM MMMZMBZa0ar B20rMMM Si i BW MMM02 7MM0 2MMM MMMMMMMM . M SM@aiMM20BWM XM 0ZMMM:MMMMW; r. 0WMBM2XrB: . 

"All cannon-balls are to be piled in square pyramids, " was the orderissued to the regiment. This was done. Then came the further order, "Allpyramids are to contain a square number of balls. " Whereupon the troublearose. "It can't be done, " said the major. "Look at this pyramid, forexample; there are sixteen balls at the base, then nine, then four, thenone at the top, making thirty balls in all. But there must be six moreballs, or five fewer, to make a square number. " "It _must_ be done, "insisted the general. "All you have to do is to put the right number ofballs in your pyramids. " "I've got it!" said a lieutenant, themathematical genius of the regiment. "Lay the balls out singly. " "Bosh!"exclaimed the general. "You can't _pile_ one ball into a pyramid!" Is itreally possible to obey both orders?

139. --THE DUTCHMEN'S WIVES. 

I wonder how many of my readers are acquainted with the puzzle of the"Dutchmen's Wives"--in which you have to determine the names of threemen's wives, or, rather, which wife belongs to each husband. Some thirtyyears ago it was "going the rounds, " as something quite new, but Irecently discovered it in the _Ladies' Diary_ for 1739-40, so it wasclearly familiar to the fair sex over one hundred and seventy years ago. How many of our mothers, wives, sisters, daughters, and aunts couldsolve the puzzle to-day? A far greater proportion than then, let ushope. 

Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives, Gurtrün, Katrün, and Anna, purchase hogs. Each buys as many as he (orshe) gives shillings for one. Each husband pays altogether three guineasmore than his wife. Hendrick buys twenty-three more hogs than Katrün, and Elas eleven more than Gurtrün. Now, what was the name of each man'swife?

[Illustration]

140. --FIND ADA'S SURNAME. 

This puzzle closely resembles the last one, my remarks on the solutionof which the reader may like to apply in another case. It was recentlysubmitted to a Sydney evening newspaper that indulges in "intellectsharpeners, " but was rejected with the remark that it is childish andthat they only published problems capable of solution! Five ladies, accompanied by their daughters, bought cloth at the same shop. Each ofthe ten paid as many farthings per foot as she bought feet, and eachmother spent 8s. 5¼d. More than her daughter. Mrs. Robinson spent 6s. More than Mrs. Evans, who spent about a quarter as much as Mrs. Jones. Mrs. Smith spent most of all. Mrs. Brown bought 21 yards more thanBessie--one of the girls. Annie bought 16 yards more than Mary and spent£3, 0s. 8d. More than Emily. The Christian name of the other girl wasAda. Now, what was her surname?

141. --SATURDAY MARKETING. 

Here is an amusing little case of marketing which, although it dealswith a good many items of money, leads up to a question of a totallydifferent character. Four married couples went into their village on arecent Saturday night to do a little marketing. They had to be veryeconomical, for among them they only possessed forty shilling coins. Thefact is, Ann spent 1s. , Mary spent 2s. , Jane spent 3s. , and Kate spent4s. The men were rather more extravagant than their wives, for Ned Smithspent as much as his wife, Tom Brown twice as much as his wife, BillJones three times as much as his wife, and Jack Robinson four times asmuch as his wife. On the way home somebody suggested that they shoulddivide what coin they had left equally among them. This was done, andthe puzzling question is simply this: What was the surname of eachwoman? Can you pair off the four couples?

GEOMETRICAL PROBLEMS. 

 "God geometrizes continually. "

 PLATO. 

"There is no study, " said Augustus de Morgan, "which presents so simplea beginning as that of geometry; there is none in which difficultiesgrow more rapidly as we proceed. " This will be found when the readercomes to consider the following puzzles, though they are not arranged instrict order of difficulty. And the fact that they have interested andgiven pleasure to man for untold ages is no doubt due in some measure tothe appeal they make to the eye as well as to the brain. Sometimes analgebraical formula or theorem seems to give pleasure to themathematician's eye, but it is probably only an intellectual pleasure. But there can be no doubt that in the case of certain geometricalproblems, notably dissection or superposition puzzles, the æstheticfaculty in man contributes to the delight. For example, there areprobably few readers who will examine the various cuttings of the Greekcross in the following pages without being in some degree stirred by asense of beauty. Law and order in Nature are always pleasing tocontemplate, but when they come under the very eye they seem to make aspecially strong appeal. Even the person with no geometrical knowledgewhatever is induced after the inspection of such things to exclaim, "Howvery pretty!" In fact, I have known more than one person led on to astudy of geometry by the fascination of cutting-out puzzles. I have, therefore, thought it well to keep these dissection puzzles distinctfrom the geometrical problems on more general lines. 

DISSECTION PUZZLES. 

 "Take him and cut him out in little stars. "

 _Romeo and Juliet_, iii. 2. 

Puzzles have infinite variety, but perhaps there is no class moreancient than dissection, cutting-out, or superposition puzzles. Theywere certainly known to the Chinese several thousand years before theChristian era. And they are just as fascinating to-day as they can havebeen at any period of their history. It is supposed by those who haveinvestigated the matter that the ancient Chinese philosophers used thesepuzzles as a sort of kindergarten method of imparting the principles ofgeometry. Whether this was so or not, it is certain that all gooddissection puzzles (for the nursery type of jig-saw puzzle, which merelyconsists in cutting up a picture into pieces to be put together again, is not worthy of serious consideration) are really based on geometricallaws. This statement need not, however, frighten off the novice, for itmeans little more than this, that geometry will give us the "reasonwhy, " if we are interested in knowing it, though the solutions may oftenbe discovered by any intelligent person after the exercise of patience, ingenuity, and common sagacity. 

If we want to cut one plane figure into parts that by readjustment willform another figure, the first thing is to find a way of doing it atall, and then to discover how to do it in the fewest possible pieces. Often a dissection problem is quite easy apart from this limitation ofpieces. At the time of the publication in the _Weekly Dispatch_, in1902, of a method of cutting an equilateral triangle into four partsthat will form a square (see No. 26, "Canterbury Puzzles"), nogeometrician would have had any difficulty in doing what is required infive pieces: the whole point of the discovery lay in performing thelittle feat in four pieces only. 

Mere approximations in the case of these problems are valueless; thesolution must be geometrically exact, or it is not a solution at all. Fallacies are cropping up now and again, and I shall have occasion torefer to one or two of these. They are interesting merely as fallacies. But I want to say something on two little points that are always arisingin cutting-out puzzles--the questions of "hanging by a thread" and"turning over. " These points can best be illustrated by a puzzle that isfrequently to be found in the old books, but invariably with a falsesolution. The puzzle is to cut the figure shown in Fig. 1 into threepieces that will fit together and form a half-square triangle. Theanswer that is invariably given is that shown in Figs. 1 and 2. Now, itis claimed that the four pieces marked C are really only one piece, because they may be so cut that they are left "hanging together by amere thread. " But no serious puzzle lover will ever admit this. If thecut is made so as to leave the four pieces joined in one, then it cannotresult in a perfectly exact solution. If, on the other hand, thesolution is to be exact, then there will be four pieces--or six piecesin all. It is, therefore, not a solution in three pieces. 

[Illustration: Fig. 1]

[Illustration: Fig. 2]

If, however, the reader will look at the solution in Figs. 3 and 4, hewill see that no such fault can be found with it. There is no questionwhatever that there are three pieces, and the solution is in thisrespect quite satisfactory. But another question arises. It will befound on inspection that the piece marked F, in Fig. 3, is turned overin Fig. 4--that is to say, a different side has necessarily to bepresented. If the puzzle were merely to be cut out of cardboard or wood, there might be no objection to this reversal, but it is quite possiblethat the material would not admit of being reversed. There might be apattern, a polish, a difference of texture, that prevents it. But it isgenerally understood that in dissection puzzles you are allowed to turnpieces over unless it is distinctly stated that you may not do so. Andvery often a puzzle is greatly improved by the added condition, "nopiece may be turned over. " I have often made puzzles, too, in which thediagram has a small repeated pattern, and the pieces have then so to becut that not only is there no turning over, but the pattern has to bematched, which cannot be done if the pieces are turned round, even withthe proper side uppermost. 

[Illustration: Fig. 3]

[Illustration: Fig. 4]

Before presenting a varied series of cutting-out puzzles, some very easyand others difficult, I propose to consider one family alone--thoseproblems involving what is known as the Greek cross with the square. This will exhibit a great variety of curious transpositions, and, byhaving the solutions as we go along, the reader will be saved thetrouble of perpetually turning to another part of the book, and willhave everything under his eye. It is hoped that in this way the articlemay prove somewhat instructive to the novice and interesting to others. 

GREEK CROSS PUZZLES. 

 "To fret thy soul with crosses. "

 SPENSER. 

 "But, for my part, it was Greek to me. "

 _Julius Cæsar_, i. 2. 

Many people are accustomed to consider the cross as a wholly Christiansymbol. This is erroneous: it is of very great antiquity. The ancientEgyptians employed it as a sacred symbol, and on Greek sculptures wefind representations of a cake (the supposed real origin of our hotcross buns) bearing a cross. Two such cakes were discovered atHerculaneum. Cecrops offered to Jupiter Olympus a sacred cake or _boun_of this kind. The cross and ball, so frequently found on Egyptianfigures, is a circle and the _tau_ cross. The circle signified theeternal preserver of the world, and the T, named from the Greek letter_tau_, is the monogram of Thoth, the Egyptian Mercury, meaning wisdom. This _tau_ cross is also called by Christians the cross of St. Anthony, and is borne on a badge in the bishop's palace at Exeter. As for theGreek or mundane cross, the cross with four equal arms, we are told bycompetent antiquaries that it was regarded by ancient occultists forthousands of years as a sign of the dual forces of Nature--the male andfemale spirit of everything that was everlasting. 

[Illustration: Fig. 5. ]

The Greek cross, as shown in Fig. 5, is formed by the assemblingtogether of five equal squares. We will start with what is known as theHindu problem, supposed to be upwards of three thousand years old. Itappears in the seal of Harvard College, and is often given in old worksas symbolical of mathematical science and exactitude. Cut the cross intofive pieces to form a square. Figs. 6 and 7 show how this is done. Itwas not until the middle of the nineteenth century that we found thatthe cross might be transformed into a square in only four pieces. Figs. 8 and 9 will show how to do it, if we further require the four pieces tobe all of the same size and shape. This Fig. 9 is remarkable because, according to Dr. Le Plongeon and others, as expounded in a work byProfessor Wilson of the Smithsonian Institute, here we have the greatSwastika, or sign, of "good luck to you "--the most ancient symbol ofthe human race of which there is any record. Professor Wilson's workgives some four hundred illustrations of this curious sign as found inthe Aztec mounds of Mexico, the pyramids of Egypt, the ruins of Troy, and the ancient lore of India and China. One might almost say there is acurious affinity between the Greek cross and Swastika! If, however, werequire that the four pieces shall be produced by only two clips of thescissors (assuming the puzzle is in paper form), then we must cut as inFig. 10 to form Fig. 11, the first clip of the scissors being from ato b. Of course folding the paper, or holding the pieces togetherafter the first cut, would not in this case be allowed. But there is aninfinite number of different ways of making the cuts to solve the puzzlein four pieces. To this point I propose to return. 

[Illustration: Fig. 6]

[Illustration: Fig. 7]

[Illustration: Fig. 8]

[Illustration: Fig. 9]

[Illustration: Fig. 10]

[Illustration: Fig. 11]

It will be seen that every one of these puzzles has its reversepuzzle--to cut a square into pieces to form a Greek cross. But as asquare has not so many angles as the cross, it is not always equallyeasy to discover the true directions of the cuts. Yet in the case of theexamples given, I will leave the reader to determine their direction forhimself, as they are rather obvious from the diagrams. 

Cut a square into five pieces that will form two separate Greek crossesof _different sizes_. This is quite an easy puzzle. As will be seen inFig. 12, we have only to divide our square into 25 little squares andthen cut as shown. The cross A is cut out entire, and the pieces B, C, D, and E form the larger cross in Fig. 13. The reader may here like tocut the single piece, B, into four pieces all similar in shape toitself, and form a cross with them in the manner shown in Fig. 13. Ihardly need give the solution. 

[Illustration: FIG. 12. ]

[Illustration: FIG. 13. ]

Cut a square into five pieces that will form two separate Greek crossesof exactly the _same size_. This is more difficult. We make the cuts asin Fig. 14, where the cross A comes out entire and the other four piecesform the cross in Fig. 15. The direction of the cuts is pretty obvious. It will be seen that the sides of the square in Fig. 14 are marked offinto six equal parts. The sides of the cross are found by ruling linesfrom certain of these points to others. 

[Illustration: FIG. 14. ]

[Illustration: FIG. 15. ]

I will now explain, as I promised, why a Greek cross may be cut intofour pieces in an infinite number of different ways to make a square. Draw a cross, as in Fig. 16. Then draw on transparent paper the squareshown in Fig. 17, taking care that the distance c to d is exactlythe same as the distance a to b in the cross. Now place thetransparent paper over the cross and slide it about into differentpositions, only be very careful always to keep the square at the sameangle to the cross as shown, where a b is parallel to c d. Ifyou place the point c exactly over a the lines will indicate thesolution (Figs. 10 and 11). If you place c in the very centre of thedotted square, it will give the solution in Figs. 8 and 9. You will nowsee that by sliding the square about so that the point c is alwayswithin the dotted square you may get as many different solutions as youlike; because, since an infinite number of different points maytheoretically be placed within this square, there must be an infinitenumber of different solutions. But the point c need not necessarily beplaced within the dotted square. It may be placed, for example, at pointe to give a solution in four pieces. Here the joins at a and f maybe as slender as you like. Yet if you once get over the edge at a orf you no longer have a solution in four pieces. This proof will befound both entertaining and instructive. If you do not happen to haveany transparent paper at hand, any thin paper will of course do if youhold the two sheets against a pane of glass in the window. 

[Illustration: FIG. 16. ]

[Illustration: FIG. 17. ]

It may have been noticed from the solutions of the puzzles that I havegiven that the side of the square formed from the cross is always equalto the distance a to b in Fig. 16. This must necessarily be so, andI will presently try to make the point quite clear. 

We will now go one step further. I have already said that the idealsolution to a cutting-out puzzle is always that which requires thefewest possible pieces. We have just seen that two crosses of the samesize may be cut out of a square in five pieces. The reader whosucceeded in solving this perhaps asked himself: "Can it be done infewer pieces?" This is just the sort of question that the true puzzlelover is always asking, and it is the right attitude for him to adopt. The answer to the question is that the puzzle may be solved in fourpieces--the fewest possible. This, then, is a new puzzle. Cut a squareinto four pieces that will form two Greek crosses of the same size. 

[Illustration: FIG. 18. ]

[Illustration: FIG. 19. ]

[Illustration: FIG. 20. ]

The solution is very beautiful. If you divide by points the sides of thesquare into three equal parts, the directions of the lines in Fig. 18will be quite obvious. If you cut along these lines, the pieces A and Bwill form the cross in Fig. 19 and the pieces C and D the similar crossin Fig. 20. In this square we have another form of Swastika. 

The reader will here appreciate the truth of my remark to the effectthat it is easier to find the directions of the cuts when transforming across to a square than when converting a square into a cross. Thus, inFigs. 6, 8, and 10 the directions of the cuts are more obvious than inFig. 14, where we had first to divide the sides of the square into sixequal parts, and in Fig. 18, where we divide them into three equalparts. Then, supposing you were required to cut two equal Greek crosses, each into two pieces, to form a square, a glance at Figs. 19 and 20 willshow how absurdly more easy this is than the reverse puzzle of cuttingthe square to make two crosses. 

Referring to my remarks on "fallacies, " I will now give a little exampleof these "solutions" that are not solutions. Some years ago a youngcorrespondent sent me what he evidently thought was a brilliant newdiscovery--the transforming of a square into a Greek cross in fourpieces by cuts all parallel to the sides of the square. I give hisattempt in Figs. 21 and 22, where it will be seen that the four piecesdo not form a symmetrical Greek cross, because the four arms are notreally squares but oblongs. To make it a true Greek cross we shouldrequire the additions that I have indicated with dotted lines. Of coursehis solution produces a cross, but it is not the symmetrical Greekvariety required by the conditions of the puzzle. My young friendthought his attempt was "near enough" to be correct; but if he bought apenny apple with a sixpence he probably would not have thought it "nearenough" if he had been given only fourpence change. As the readeradvances he will realize the importance of this question of exactitude. 

[Illustration: FIG. 21. ]

[Illustration: FIG. 22. ]

In these cutting-out puzzles it is necessary not only to get thedirections of the cutting lines as correct as possible, but to rememberthat these lines have no width. If after cutting up one of the crossesin a manner indicated in these articles you find that the pieces do notexactly fit to form a square, you may be certain that the fault isentirely your own. Either your cross was not exactly drawn, or your cutswere not made quite in the right directions, or (if you used wood and afret-saw) your saw was not sufficiently fine. If you cut out the puzzlesin paper with scissors, or in cardboard with a penknife, no material islost; but with a saw, however fine, there is a certain loss. In the caseof most puzzles this slight loss is not sufficient to be appreciable, if the puzzle is cut out on a large scale, but there have beeninstances where I have found it desirable to draw and cut out each partseparately--not from one diagram--in order to produce a perfect result. 

[Illustration: FIG. 23. ]

[Illustration: FIG. 24. ]

Now for another puzzle. If you have cut out the five pieces indicated inFig. 14, you will find that these can be put together so as to form thecurious cross shown in Fig. 23. So if I asked you to cut Fig. 24 intofive pieces to form either a square or two equal Greek crosses you wouldknow how to do it. You would make the cuts as in Fig. 23, and place themtogether as in Figs. 14 and 15. But I want something better than that, and it is this. Cut Fig. 24 into only four pieces that will fit togetherand form a square. 

[Illustration: FIG. 25. ]

[Illustration: FIG. 26. ]

The solution to the puzzle is shown in Figs. 25 and 26. The direction ofthe cut dividing A and C in the first diagram is very obvious, and thesecond cut is made at right angles to it. That the four pieces shouldfit together and form a square will surprise the novice, who will dowell to study the puzzle with some care, as it is most instructive. 

I will now explain the beautiful rule by which we determine the size ofa square that shall have the same area as a Greek cross, for it isapplicable, and necessary, to the solution of almost every dissectionpuzzle that we meet with. It was first discovered by the philosopherPythagoras, who died 500 B. C. , and is the 47th proposition of Euclid. The young reader who knows nothing of the elements of geometry will getsome idea of the fascinating character of that science. The triangle ABCin Fig. 27 is what we call a right-angled triangle, because the side BCis at right angles to the side AB. Now if we build up a square on eachside of the triangle, the squares on AB and BC will together be exactlyequal to the square on the long side AC, which we call the hypotenuse. This is proved in the case I have given by subdividing the three squaresinto cells of equal dimensions. 

[Illustration: FIG. 27. ]

[Illustration: FIG. 28. ]

It will be seen that 9 added to 16 equals 25, the number of cells in thelarge square. If you make triangles with the sides 5, 12 and 13, or with8, 15 and 17, you will get similar arithmetical proofs, for these areall "rational" right-angled triangles, but the law is equally true forall cases. Supposing we cut off the lower arm of a Greek cross and placeit to the left of the upper arm, as in Fig. 28, then the square on EFadded to the square on DE exactly equals a square on DF. Therefore weknow that the square of DF will contain the same area as the cross. Thisfact we have proved practically by the solutions of the earlier puzzlesof this series. But whatever length we give to DE and EF, we can nevergive the exact length of DF in numbers, because the triangle is not a"rational" one. But the law is none the less geometrically true. 

[Illustration: FIG. 29. ]

[Illustration: FIG. 30. ]

Now look at Fig. 29, and you will see an elegant method for cutting apiece of wood of the shape of two squares (of any relative dimensions)into three pieces that will fit together and form a single square. Ifyou mark off the distance _ab_ equal to the side _cd_ the directions ofthe cuts are very evident. From what we have just been considering, youwill at once see why _bc_ must be the length of the side of the newsquare. Make the experiment as often as you like, taking differentrelative proportions for the two squares, and you will find the rulealways come true. If you make the two squares of exactly the same size, you will see that the diagonal of any square is always the side of asquare that is twice the size. All this, which is so simple that anybodycan understand it, is very essential to the solving of cutting-outpuzzles. It is in fact the key to most of them. And it is all sobeautiful that it seems a pity that it should not be familiar toeverybody. 

We will now go one step further and deal with the half-square. Take asquare and cut it in half diagonally. Now try to discover how to cutthis triangle into four pieces that will form a Greek cross. Thesolution is shown in Figs. 31 and 32. In this case it will be seen thatwe divide two of the sides of the triangle into three equal parts andthe long side into four equal parts. Then the direction of the cuts willbe easily found. It is a pretty puzzle, and a little more difficult thansome of the others that I have given. It should be noted again that itwould have been much easier to locate the cuts in the reverse puzzle ofcutting the cross to form a half-square triangle. 

[Illustration: FIG. 31. ]

[Illustration: FIG. 32. ]

[Illustration: FIG. 33. ]

[Illustration: FIG. 34. ]

Another ideal that the puzzle maker always keeps in mind is to contrivethat there shall, if possible, be only one correct solution. Thus, inthe case of the first puzzle, if we only require that a Greek crossshall be cut into four pieces to form a square, there is, as I haveshown, an infinite number of different solutions. It makes a betterpuzzle to add the condition that all the four pieces shall be of thesame size and shape, because it can then be solved in only one way, asin Figs. 8 and 9. In this way, too, a puzzle that is too easy to beinteresting may be improved by such an addition. Let us take an example. We have seen in Fig. 28 that Fig. 33 can be cut into two pieces to forma Greek cross. I suppose an intelligent child would do it in fiveminutes. But suppose we say that the puzzle has to be solved with apiece of wood that has a bad knot in the position shown in Fig. 33--aknot that we must not attempt to cut through--then a solution in twopieces is barred out, and it becomes a more interesting puzzle to solveit in three pieces. I have shown in Figs. 33 and 34 one way of doingthis, and it will be found entertaining to discover other ways of doingit. Of course I could bar out all these other ways by introducing moreknots, and so reduce the puzzle to a single solution, but it would thenbe overloaded with conditions. 

And this brings us to another point in seeking the ideal. Do notoverload your conditions, or you will make your puzzle too complex to beinteresting. The simpler the conditions of a puzzle are, the better. Thesolution may be as complex and difficult as you like, or as happens, butthe conditions ought to be easily understood, or people will not attempta solution. 

If the reader were now asked "to cut a half-square into as few pieces aspossible to form a Greek cross, " he would probably produce our solution, Figs. 31-32, and confidently claim that he had solved the puzzlecorrectly. In this way he would be wrong, because it is not now statedthat the square is to be divided diagonally. Although we should alwaysobserve the exact conditions of a puzzle we must not read into itconditions that are not there. Many puzzles are based entirely on thetendency that people have to do this. 

The very first essential in solving a puzzle is to be sure that youunderstand the exact conditions. Now, if you divided your square in halfso as to produce Fig. 35 it is possible to cut it into as few as threepieces to form a Greek cross. We thus save a piece. 

I give another puzzle in Fig. 36. The dotted lines are added merely toshow the correct proportions of the figure--a square of 25 cells withthe four corner cells cut out. The puzzle is to cut this figure intofive pieces that will form a Greek cross (entire) and a square. 

[Illustration: FIG. 35. ]

[Illustration: FIG. 36. ]

The solution to the first of the two puzzles last given--to cut arectangle of the shape of a half-square into three pieces that will forma Greek cross--is shown in Figs. 37 and 38. It will be seen that wedivide the long sides of the oblong into six equal parts and the shortsides into three equal parts, in order to get the points that willindicate the direction of the cuts. The reader should compare thissolution with some of the previous illustrations. He will see, forexample, that if we continue the cut that divides B and C in the cross, we get Fig. 15. 

[Illustration: FIG. 37. ]

[Illustration: FIG. 38. ]

The other puzzle, like the one illustrated in Figs. 12 and 13, will showhow useful a little arithmetic may sometimes prove to be in the solutionof dissection puzzles. There are twenty-one of those little square cellsinto which our figure is subdivided, from which we have to form both asquare and a Greek cross. Now, as the cross is built up of five squares, and 5 from 21 leaves 16--a square number--we ought easily to be led tothe solution shown in Fig. 39. It will be seen that the cross is cut outentire, while the four remaining pieces form the square in Fig. 40. 

[Illustration: FIG. 39]

[Illustration: FIG. 40]

Of course a half-square rectangle is the same as a double square, or twoequal squares joined together. Therefore, if you want to solve thepuzzle of cutting a Greek cross into four pieces to form two separatesquares of the same size, all you have to do is to continue the shortcut in Fig. 38 right across the cross, and you will have four pieces ofthe same size and shape. Now divide Fig. 37 into two equal squares by ahorizontal cut midway and you will see the four pieces forming the twosquares. 

[Illustration: FIG. 41]

Cut a Greek cross into five pieces that will form two separate squares, one of which shall contain half the area of one of the arms of thecross. In further illustration of what I have already written, if thetwo squares of the same size A B C D and B C F E, in Fig. 41, are cut inthe manner indicated by the dotted lines, the four pieces will form thelarge square A G E C. We thus see that the diagonal A C is the side of asquare twice the size of A B C D. It is also clear that half thediagonal of any square is equal to the side of a square of half thearea. Therefore, if the large square in the diagram is one of the armsof your cross, the small square is the size of one of the squaresrequired in the puzzle. 

The solution is shown in Figs. 42 and 43. It will be seen that the smallsquare is cut out whole and the large square composed of the four piecesB, C, D, and E. After what I have written, the reader will have nodifficulty in seeing that the square A is half the size of one of thearms of the cross, because the length of the diagonal of the former isclearly the same as the side of the latter. The thing is nowself-evident. I have thus tried to show that some of these puzzles thatmany people are apt to regard as quite wonderful and bewildering, arereally not difficult if only we use a little thought and judgment. Inconclusion of this particular subject I will give four Greek crosspuzzles, with detached solutions. 

142. --THE SILK PATCHWORK. 

The lady members of the Wilkinson family had made a simple patchworkquilt, as a small Christmas present, all composed of square pieces ofthe same size, as shown in the illustration. It only lacked the fourcorner pieces to make it complete. Somebody pointed out to them that ifyou unpicked the Greek cross in the middle and then cut the stitchesalong the dark joins, the four pieces all of the same size and shapewould fit together and form a square. This the reader knows, from thesolution in Fig. 39, is quite easily done. But George Wilkinson suddenlysuggested to them this poser. He said, "Instead of picking out the crossentire, and forming the square from four equal pieces, can you cut out asquare entire and four equal pieces that will form a perfect Greekcross?" The puzzle is, of course, now quite easy. 

143. --TWO CROSSES FROM ONE. 

Cut a Greek cross into five pieces that will form two such crosses, bothof the same size. The solution of this puzzle is very beautiful. 

144. --THE CROSS AND THE TRIANGLE. 

Cut a Greek cross into six pieces that will form an equilateraltriangle. This is another hard problem, and I will state here that asolution is practically impossible without a previous knowledge of mymethod of transforming an equilateral triangle into a square (see No. 26, "Canterbury Puzzles"). 

145. --THE FOLDED CROSS. 

Cut out of paper a Greek cross; then so fold it that with a singlestraight cut of the scissors the four pieces produced will form asquare. 

VARIOUS DISSECTION PUZZLES. 

We will now consider a small miscellaneous selection of cutting-outpuzzles, varying in degrees of difficulty. 

146. --AN EASY DISSECTION PUZZLE. 

First, cut out a piece of paper or cardboard of the shape shown in theillustration. It will be seen at once that the proportions are simplythose of a square attached to half of another similar square, divideddiagonally. The puzzle is to cut it into four pieces all of preciselythe same size and shape. 

147. --AN EASY SQUARE PUZZLE. 

If you take a rectangular piece of cardboard, twice as long as it isbroad, and cut it in half diagonally, you will get two of the piecesshown in the illustration. The puzzle is with five such pieces of equalsize to form a square. One of the pieces may be cut in two, but theothers must be used intact. 

148. --THE BUN PUZZLE. 

THE three circles represent three buns, and it is simply required toshow how these may be equally divided among four boys. The buns must beregarded as of equal thickness throughout and of equal thickness to eachother. Of course, they must be cut into as few pieces as possible. Tosimplify it I will state the rather surprising fact that only fivepieces are necessary, from which it will be seen that one boy gets hisshare in two pieces and the other three receive theirs in a singlepiece. I am aware that this statement "gives away" the puzzle, but itshould not destroy its interest to those who like to discover the"reason why. "

149. --THE CHOCOLATE SQUARES. 

Here is a slab of chocolate, indented at the dotted lines so that thetwenty squares can be easily separated. Make a copy of the slab in paperor cardboard and then try to cut it into nine pieces so that they willform four perfect squares all of exactly the same size. 

150. --DISSECTING A MITRE. 

The figure that is perplexing the carpenter in the illustrationrepresents a mitre. It will be seen that its proportions are those of asquare with one quarter removed. The puzzle is to cut it into fivepieces that will fit together and form a perfect square. I show anattempt, published in America, to perform the feat in four pieces, basedon what is known as the "step principle, " but it is a fallacy. 

[Illustration]

We are told first to cut oft the pieces 1 and 2 and pack them into thetriangular space marked off by the dotted line, and so form a rectangle. 

So far, so good. Now, we are directed to apply the old step principle, as shown, and, by moving down the piece 4 one step, form the requiredsquare. But, unfortunately, it does _not_ produce a square: only anoblong. Call the three long sides of the mitre 84 in. Each. Then, beforecutting the steps, our rectangle in three pieces will be 84 × 63. Thesteps must be 10½ in. In height and 12 in. In breadth. Therefore, bymoving down a step we reduce by 12 in. The side 84 in. And increase by10½ in. The side 63 in. Hence our final rectangle must be 72 in. × 73½in. , which certainly is not a square! The fact is, the step principlecan only be applied to rectangles with sides of particular relativelengths. For example, if the shorter side in this case were 61+5/7(instead of 63), then the step method would apply. For the steps wouldthen be 10+2/7 in. In height and 12 in. In breadth. Note that 61+5/7 ×84 = the square of 72. At present no solution has been found in fourpieces, and I do not believe one possible. 

151. --THE JOINER'S PROBLEM. 

I have often had occasion to remark on the practical utility of puzzles, arising out of an application to the ordinary affairs of life of thelittle tricks and "wrinkles" that we learn while solving recreationproblems. 

[Illustration]

The joiner, in the illustration, wants to cut the piece of wood into asfew pieces as possible to form a square table-top, without any waste ofmaterial. How should he go to work? How many pieces would you require?

152. --ANOTHER JOINER'S PROBLEM. 

[Illustration]

A joiner had two pieces of wood of the shapes and relative proportionsshown in the diagram. He wished to cut them into as few pieces aspossible so that they could be fitted together, without waste, to form aperfectly square table-top. How should he have done it? There is nonecessity to give measurements, for if the smaller piece (which is halfa square) be made a little too large or a little too small it will notaffect the method of solution. 

153--A CUTTING-OUT PUZZLE. 

Here is a little cutting-out poser. I take a strip of paper, measuringfive inches by one inch, and, by cutting it into five pieces, the partsfit together and form a square, as shown in the illustration. Now, it isquite an interesting puzzle to discover how we can do this in only fourpieces. 

[Illustration]

154. --MRS. HOBSON'S HEARTHRUG. 

[Illustration]

Mrs. Hobson's boy had an accident when playing with the fire, and burnttwo of the corners of a pretty hearthrug. The damaged corners have beencut away, and it now has the appearance and proportions shown in mydiagram. How is Mrs. Hobson to cut the rug into the fewest possiblepieces that will fit together and form a perfectly square rug? It willbe seen that the rug is in the proportions 36 × 27 (it does not matterwhether we say inches or yards), and each piece cut away measured 12 and6 on the outside. 

155. --THE PENTAGON AND SQUARE. 

I wonder how many of my readers, amongst those who have not given anyclose attention to the elements of geometry, could draw a regularpentagon, or five-sided figure, if they suddenly required to do so. Aregular hexagon, or six-sided figure, is easy enough, for everybodyknows that all you have to do is to describe a circle and then, takingthe radius as the length of one of the sides, mark off the six pointsround the circumference. But a pentagon is quite another matter. So, asmy puzzle has to do with the cutting up of a regular pentagon, it willperhaps be well if I first show my less experienced readers how thisfigure is to be correctly drawn. Describe a circle and draw the twolines H B and D G, in the diagram, through the centre at right angles. Now find the point A, midway between C and B. Next place the point ofyour compasses at A and with the distance A D describe the arc cutting HB at E. Then place the point of your compasses at D and with thedistance D E describe the arc cutting the circumference at F. Now, D Fis one of the sides of your pentagon, and you have simply to mark offthe other sides round the circle. Quite simple when you know how, butotherwise somewhat of a poser. 

[Illustration]

Having formed your pentagon, the puzzle is to cut it into the fewestpossible pieces that will fit together and form a perfect square. 

[Illustration]

156. --THE DISSECTED TRIANGLE. 

A good puzzle is that which the gentleman in the illustration is showingto his friends. He has simply cut out of paper an equilateraltriangle--that is, a triangle with all its three sides of the samelength. He proposes that it shall be cut into five pieces in such a waythat they will fit together and form either two or three smallerequilateral triangles, using all the material in each case. Can youdiscover how the cuts should be made?

Remember that when you have made your five pieces, you must be able, asdesired, to put them together to form either the single originaltriangle or to form two triangles or to form three triangles--allequilateral. 

157. --THE TABLE-TOP AND STOOLS. 

I have frequently had occasion to show that the published answers to agreat many of the oldest and most widely known puzzles are either quiteincorrect or capable of improvement. I propose to consider the old poserof the table-top and stools that most of my readers have probably seenin some form or another in books compiled for the recreation ofchildhood. 

The story is told that an economical and ingenious schoolmaster oncewished to convert a circular table-top, for which he had no use, intoseats for two oval stools, each with a hand-hole in the centre. Heinstructed the carpenter to make the cuts as in the illustration andthen join the eight pieces together in the manner shown. So impressedwas he with the ingenuity of his performance that he set the puzzle tohis geometry class as a little study in dissection. But the remainder ofthe story has never been published, because, so it is said, it was acharacteristic of the principals of academies that they would neveradmit that they could err. I get my information from a descendant of theoriginal boy who had most reason to be interested in the matter. 

The clever youth suggested modestly to the master that the hand-holeswere too big, and that a small boy might perhaps fall through them. Hetherefore proposed another way of making the cuts that would get overthis objection. For his impertinence he received such severechastisement that he became convinced that the larger the hand-hole inthe stools the more comfortable might they be. 

[Illustration]

Now what was the method the boy proposed?

Can you show how the circular table-top may be cut into eight piecesthat will fit together and form two oval seats for stools (each ofexactly the same size and shape) and each having similar hand-holes ofsmaller dimensions than in the case shown above? Of course, all the woodmust be used. 

158. --THE GREAT MONAD. 

[Illustration]

Here is a symbol of tremendous antiquity which is worthy of notice. Itis borne on the Korean ensign and merchant flag, and has been adopted asa trade sign by the Northern Pacific Railroad Company, though probablyfew are aware that it is the Great Monad, as shown in the sketch below. This sign is to the Chinaman what the cross is to the Christian. It isthe sign of Deity and eternity, while the two parts into which thecircle is divided are called the Yin and the Yan--the male and femaleforces of nature. A writer on the subject more than three thousand yearsago is reported to have said in reference to it: "The illimitableproduces the great extreme. The great extreme produces the twoprinciples. The two principles produce the four quarters, and from thefour quarters we develop the quadrature of the eight diagrams ofFeuh-hi. " I hope readers will not ask me to explain this, for I have notthe slightest idea what it means. Yet I am persuaded that for ages thesymbol has had occult and probably mathematical meanings for theesoteric student. 

I will introduce the Monad in its elementary form. Here are three easyquestions respecting this great symbol:--

(I. ) Which has the greater area, the inner circle containing the Yin andthe Yan, or the outer ring?

(II. ) Divide the Yin and the Yan into four pieces of the same size andshape by one cut. 

(III. ) Divide the Yin and the Yan into four pieces of the same size, butdifferent shape, by one straight cut. 

159. --THE SQUARE OF VENEER. 

The following represents a piece of wood in my possession, 5 in. Square. By markings on the surface it is divided into twenty-five square inches. I want to discover a way of cutting this piece of wood into the fewestpossible pieces that will fit together and form two perfect squares ofdifferent sizes and of known dimensions. But, unfortunately, at everyone of the sixteen intersections of the cross lines a small nail hasbeen driven in at some time or other, and my fret-saw will be injured ifit comes in contact with any of these. I have therefore to find a methodof doing the work that will not necessitate my cutting through any ofthose sixteen points. How is it to be done? Remember, the exactdimensions of the two squares must be given. 

[Illustration]

160. --THE TWO HORSESHOES. 

[Illustration]

Why horseshoes should be considered "lucky" is one of those thingswhich no man can understand. It is a very old superstition, and JohnAubrey (1626-1700) says, "Most houses at the West End of London have ahorseshoe on the threshold. " In Monmouth Street there were seventeen in1813 and seven so late as 1855. Even Lord Nelson had one nailed to themast of the ship _Victory_. To-day we find it more conducive to "goodluck" to see that they are securely nailed on the feet of the horse weare about to drive. 

Nevertheless, so far as the horseshoe, like the Swastika and otheremblems that I have had occasion at times to deal with, has served tosymbolize health, prosperity, and goodwill towards men, we may welltreat it with a certain amount of respectful interest. May there not, moreover, be some esoteric or lost mathematical mystery concealed in theform of a horseshoe? I have been looking into this matter, and I wish todraw my readers' attention to the very remarkable fact that the pair ofhorseshoes shown in my illustration are related in a striking andbeautiful manner to the circle, which is the symbol of eternity. Ipresent this fact in the form of a simple problem, so that it may beseen how subtly this relation has been concealed for ages and ages. Myreaders will, I know, be pleased when they find the key to the mystery. 

Cut out the two horseshoes carefully round the outline and then cut theminto four pieces, all different in shape, that will fit together andform a perfect circle. Each shoe must be cut into two pieces and all thepart of the horse's hoof contained within the outline is to be used andregarded as part of the area. 

161. --THE BETSY ROSS PUZZLE. 

A correspondent asked me to supply him with the solution to an oldpuzzle that is attributed to a certain Betsy Ross, of Philadelphia, whoshowed it to George Washington. It consists in so folding a piece ofpaper that with one clip of the scissors a five-pointed star of Freedommay be produced. Whether the story of the puzzle's origin is a true oneor not I cannot say, but I have a print of the old house in Philadelphiawhere the lady is said to have lived, and I believe it still standsthere. But my readers will doubtless be interested in the little poser. 

Take a circular piece of paper and so fold it that with one cut of thescissors you can produce a perfect five-pointed star. 

162. --THE CARDBOARD CHAIN. 

[Illustration]

Can you cut this chain out of a piece of cardboard without any joinwhatever? Every link is solid; without its having been split andafterwards joined at any place. It is an interesting old puzzle that Ilearnt as a child, but I have no knowledge as to its inventor. 

163. --THE PAPER BOX. 

It may be interesting to introduce here, though it is not strictly apuzzle, an ingenious method for making a paper box. 

Take a square of stout paper and by successive foldings make all thecreases indicated by the dotted lines in the illustration. Then cut awaythe eight little triangular pieces that are shaded, and cut through thepaper along the dark lines. The second illustration shows the box halffolded up, and the reader will have no difficulty in effecting itscompletion. Before folding up, the reader might cut out the circularpiece indicated in the diagram, for a purpose I will now explain. 

This box will be found to serve excellently for the production of vortexrings. These rings, which were discussed by Von Helmholtz in 1858, aremost interesting, and the box (with the hole cut out) will produce themto perfection. Fill the box with tobacco smoke by blowing it gentlythrough the hole. Now, if you hold it horizontally, and softly tap theside that is opposite to the hole, an immense number of perfect ringscan be produced from one mouthful of smoke. It is best that there shouldbe no currents of air in the room. People often do not realise thatthese rings are formed in the air when no smoke is used. The smoke onlymakes them visible. Now, one of these rings, if properly directed on itscourse, will travel across the room and put out the flame of a candle, and this feat is much more striking if you can manage to do it withoutthe smoke. Of course, with a little practice, the rings may be blownfrom the mouth, but the box produces them in much greater perfection, and no skill whatever is required. Lord Kelvin propounded the theorythat matter may consist of vortex rings in a fluid that fills all space, and by a development of the hypothesis he was able to explain chemicalcombination. 

[Illustration:

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]

[Illustration]

164. --THE POTATO PUZZLE. 

Take a circular slice of potato, place it on the table, and see into howlarge a number of pieces you can divide it with six cuts of a knife. Ofcourse you must not readjust the pieces or pile them after a cut. Whatis the greatest number of pieces you can make?

[Illustration:

 -------- / \ 1/ \ / \ 2 \/ 3 / \ / \ /\ / \ / \ 4 \/ 5\/ 6 / \ | \ /\ /\ / | \ 7\/ 8\/ 9\/10 / \ /\ /\ /\ / \/11\/12\/13\/ \ /\ /\ / \/14\/15\/ \ /\ / \/16\/ -----

]

The illustration shows how to make sixteen pieces. This can, of course, be easily beaten. 

165. --THE SEVEN PIGS. 

[Illustration]

 +------------------------------+ | | | P | | | | P | | P | | P | | P | | P | | P | | | +------------------------------+

Here is a little puzzle that was put to one of the sons of Erin theother day and perplexed him unduly, for it is really quite easy. It willbe seen from the illustration that he was shown a sketch of a square pencontaining seven pigs. He was asked how he would intersect the pen withthree straight fences so as to enclose every pig in a separate sty. Inother words, all you have to do is to take your pencil and, with threestraight strokes across the square, enclose each pig separately. Nothingcould be simpler. 

[Illustration]

The Irishman complained that the pigs would not keep still while he wasputting up the fences. He said that they would all flock together, orone obstinate beast would go into a corner and flock all by himself. Itwas pointed out to him that for the purposes of the puzzle the pigs werestationary. He answered that Irish pigs are not stationery--they arepork. Being persuaded to make the attempt, he drew three lines, one ofwhich cut through a pig. When it was explained that this is not allowed, he protested that a pig was no use until you cut its throat. "Begorra, if it's bacon ye want without cutting your pig, it will be all gammon. "We will not do the Irishman the injustice of suggesting that themiserable pun was intentional. However, he failed to solve the puzzle. Can you do it?

166. --THE LANDOWNER'S FENCES. 

The landowner in the illustration is consulting with his bailiff over arather puzzling little question. He has a large plan of one of hisfields, in which there are eleven trees. Now, he wants to divide thefield into just eleven enclosures by means of straight fences, so thatevery enclosure shall contain one tree as a shelter for his cattle. Howis he to do it with as few fences as possible? Take your pencil and drawstraight lines across the field until you have marked off the elevenenclosures (and no more), and then see how many fences you require. Ofcourse the fences may cross one another. 

167. --THE WIZARD'S CATS. 

[Illustration]

A wizard placed ten cats inside a magic circle as shown in ourillustration, and hypnotized them so that they should remain stationaryduring his pleasure. He then proposed to draw three circles inside thelarge one, so that no cat could approach another cat without crossing amagic circle. Try to draw the three circles so that every cat has itsown enclosure and cannot reach another cat without crossing a line. 

168. --THE CHRISTMAS PUDDING. 

[Illustration]

"Speaking of Christmas puddings, " said the host, as he glanced at theimposing delicacy at the other end of the table. "I am reminded of thefact that a friend gave me a new puzzle the other day respecting one. Here it is, " he added, diving into his breast pocket. 

"'Problem: To find the contents, ' I suppose, " said the Eton boy. 

"No; the proof of that is in the eating. I will read you theconditions. "

"'Cut the pudding into two parts, each of exactly the same size andshape, without touching any of the plums. The pudding is to be regardedas a flat disc, not as a sphere. '"

"Why should you regard a Christmas pudding as a disc? And why should anyreasonable person ever wish to make such an accurate division?" askedthe cynic. 

"It is just a puzzle--a problem in dissection. " All in turn had a lookat the puzzle, but nobody succeeded in solving it. It is a littledifficult unless you are acquainted with the principle involved in themaking of such puddings, but easy enough when you know how it is done. 

169. --A TANGRAM PARADOX. 

Many pastimes of great antiquity, such as chess, have so developed andchanged down the centuries that their original inventors would scarcelyrecognize them. This is not the case with Tangrams, a recreation thatappears to be at least four thousand years old, that has apparentlynever been dormant, and that has not been altered or "improved upon"since the legendary Chinaman Tan first cut out the seven pieces shown inDiagram I. If you mark the point B, midway between A and C, on one sideof a square of any size, and D, midway between C and E, on an adjoiningside, the direction of the cuts is too obvious to need furtherexplanation. Every design in this article is built up from the sevenpieces of blackened cardboard. It will at once be understood that thepossible combinations are infinite. 

[Illustration]

The late Mr. Sam Loyd, of New York, who published a small book of veryingenious designs, possessed the manuscripts of the late Mr. Challenor, who made a long and close study of Tangrams. This gentleman, it is said, records that there were originally seven books of Tangrams, compiled inChina two thousand years before the Christian era. These books are sorare that, after forty years' residence in the country, he onlysucceeded in seeing perfect copies of the first and seventh volumes withfragments of the second. Portions of one of the books, printed in goldleaf upon parchment, were found in Peking by an English soldier and soldfor three hundred pounds. 

A few years ago a little book came into my possession, from the libraryof the late Lewis Carroll, entitled _The Fashionable Chinese Puzzle_. Itcontains three hundred and twenty-three Tangram designs, mostlynondescript geometrical figures, to be constructed from the sevenpieces. It was "Published by J. And E. Wallis, 42 Skinner Street, and J. Wallis, Jun. , Marine Library, Sidmouth" (South Devon). There is no date, but the following note fixes the time of publication pretty closely:"This ingenious contrivance has for some time past been the favouriteamusement of the ex-Emperor Napoleon, who, being now in a debilitatedstate and living very retired, passes many hours a day in thusexercising his patience and ingenuity. " The reader will find, as did thegreat exile, that much amusement, not wholly uninstructive, may bederived from forming the designs of others. He will find many of theillustrations to this article quite easy to build up, and some ratherdifficult. Every picture may thus be regarded as a puzzle. 

But it is another pastime altogether to create new and original designsof a pictorial character, and it is surprising what extraordinary scopethe Tangrams afford for producing pictures of real life--angular andoften grotesque, it is true, but full of character. I give an example ofa recumbent figure (2) that is particularly graceful, and only needssome slight reduction of its angularities to produce an entirelysatisfactory outline. 

As I have referred to the author of _Alice in Wonderland_, I give alsomy designs of the March Hare (3) and the Hatter (4). I also give anattempt at Napoleon (5), and a very excellent Red Indian with his Squawby Mr. Loyd (6 and 7). A large number of other designs will be found inan article by me in _The Strand Magazine_ for November, 1908. 

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

On the appearance of this magazine article, the late Sir James Murray, the eminent philologist, tried, with that amazing industry thatcharacterized all his work, to trace the word "tangram" to its source. At length he wrote as follows:--"One of my sons is a professor in theAnglo-Chinese college at Tientsin. Through him, his colleagues, and hisstudents, I was able to make inquiries as to the alleged Tan amongChinese scholars. Our Chinese professor here (Oxford) also took aninterest in the matter and obtained information from the secretary ofthe Chinese Legation in London, who is a very eminent representative ofthe Chinese literati. "

[Illustration: 5]

"The result has been to show that the man Tan, the god Tan, and the'Book of Tan' are entirely unknown to Chinese literature, history, ortradition. By most of the learned men the name, or allegation of theexistence, of these had never been heard of. The puzzle is, of course, well known. It is called in Chinese _ch'i ch'iao t'u_; literally, 'seven-ingenious-plan' or 'ingenious-puzzle figure of seven pieces. ' Noname approaching 'tangram, ' or even 'tan, ' occurs in Chinese, and theonly suggestions for the latter were the Chinese _t'an_, 'to extend'; or_t'ang_, Cantonese dialect for 'Chinese. ' It was suggested that probablysome American or Englishman who knew a little Chinese or Cantonese, wanting a name for the puzzle, might concoct one out of one of thesewords and the European ending 'gram. ' I should say the name 'tangram'was probably invented by an American some little time before 1864 andafter 1847, but I cannot find it in print before the 1864 edition ofWebster. I have therefore had to deal very shortly with the word in thedictionary, telling what it is applied to and what conjectures orguesses have been made at the name, and giving a few quotations, onefrom your own article, which has enabled me to make more of the subjectthan I could otherwise have done. "

[Illustration: 6]

[Illustration: 7]

Several correspondents have informed me that they possess, or hadpossessed, specimens of the old Chinese books. An American gentlemanwrites to me as follows:--"I have in my possession a book made of tissuepaper, printed in black (with a Chinese inscription on the front page), containing over three hundred designs, which belongs to the box of'tangrams, ' which I also own. The blocks are seven in number, made ofmother-of-pearl, highly polished and finely engraved on either side. These are contained in a rosewood box 2+1/8 in. Square. My great uncle, ----, was one of the first missionaries to visit China. This box andbook, along with quite a collection of other relics, were sent to mygrandfather and descended to myself. "

My correspondent kindly supplied me with rubbings of the Tangrams, fromwhich it is clear that they are cut in the exact proportions that I haveindicated. I reproduce the Chinese inscription (8) for this reason. Theowner of the book informs me that he has submitted it to a number ofChinamen in the United States and offered as much as a dollar for atranslation. But they all steadfastly refused to read the words, offering the lame excuse that the inscription is Japanese. Natives ofJapan, however, insist that it is Chinese. Is there something occult andesoteric about Tangrams, that it is so difficult to lift the veil?Perhaps this page will come under the eye of some reader acquainted withthe Chinese language, who will supply the required translation, whichmay, or may not, throw a little light on this curious question. 

[Illustration: 8]

By using several sets of Tangrams at the same time we may construct moreambitious pictures. I was advised by a friend not to send my picture, "AGame of Billiards" (9), to the Academy. He assured me that it would notbe accepted because the "judges are so hide-bound by convention. "Perhaps he was right, and it will be more appreciated byPost-impressionists and Cubists. The players are considering a verydelicate stroke at the top of the table. Of course, the two men, thetable, and the clock are formed from four sets of Tangrams. My secondpicture is named "The Orchestra" (10), and it was designed for thedecoration of a large hall of music. Here we have the conductor, thepianist, the fat little cornet-player, the left-handed player of thedouble-bass, whose attitude is life-like, though he does stand at anunusual distance from his instrument, and the drummer-boy, with hisimposing music-stand. The dog at the back of the pianoforte is nothowling: he is an appreciative listener. 

[Illustration: 9]

[Illustration: 10]

One remarkable thing about these Tangram pictures is that they suggestto the imagination such a lot that is not really there. Who, forexample, can look for a few minutes at Lady Belinda (11) and the Dutchgirl (12) without soon feeling the haughty expression in the one caseand the arch look in the other? Then look again at the stork (13), andsee how it is suggested to the mind that the leg is actually much moreslender than any one of the pieces employed. It is really an opticalillusion. Again, notice in the case of the yacht (14) how, by leavingthat little angular point at the top, a complete mast is suggested. Ifyou place your Tangrams together on white paper so that they do notquite touch one another, in some cases the effect is improved by thewhite lines; in other cases it is almost destroyed. 

[Illustration: 11]

[Illustration: 12]

Finally, I give an example from the many curious paradoxes that onehappens upon in manipulating Tangrams. I show designs of two dignifiedindividuals (15 and 16) who appear to be exactly alike, except for thefact that one has a foot and the other has not. Now, both of thesefigures are made from the same seven Tangrams. Where does the second manget his foot from?

[Illustration: 13]

[Illustration: 14]

[Illustration: 15]

[Illustration: 16]

PATCHWORK PUZZLES. 

"Of shreds and patches. "--_Hamlet_, iii. 4. 

170. --THE CUSHION COVERS. 

[Illustration]

The above represents a square of brocade. A lady wishes to cut it infour pieces so that two pieces will form one perfectly square cushiontop, and the remaining two pieces another square cushion top. How is sheto do it? Of course, she can only cut along the lines that divide thetwenty-five squares, and the pattern must "match" properly without anyirregularity whatever in the design of the material. There is only oneway of doing it. Can you find it?

171. --THE BANNER PUZZLE. 

[Illustration]

A Lady had a square piece of bunting with two lions on it, of which theillustration is an exactly reproduced reduction. She wished to cut thestuff into pieces that would fit together and form two square bannerswith a lion on each banner. She discovered that this could be done in asfew as four pieces. How did she manage it? Of course, to cut the BritishLion would be an unpardonable offence, so you must be careful that nocut passes through any portion of either of them. Ladies are informedthat no allowance whatever has to be made for "turnings, " and no part ofthe material may be wasted. It is quite a simple little dissectionpuzzle if rightly attacked. Remember that the banners have to be perfectsquares, though they need not be both of the same size. 

172. --MRS. SMILEY'S CHRISTMAS PRESENT. 

Mrs. Smiley's expression of pleasure was sincere when her sixgranddaughters sent to her, as a Christmas present, a very prettypatchwork quilt, which they had made with their own hands. It wasconstructed of square pieces of silk material, all of one size, and asthey made a large quilt with fourteen of these little squares on eachside, it is obvious that just 196 pieces had been stitched into it. Now, the six granddaughters each contributed a part of the work in the formof a perfect square (all six portions being different in size), but inorder to join them up to form the square quilt it was necessary that thework of one girl should be unpicked into three separate pieces. Can youshow how the joins might have been made? Of course, no portion can beturned over. 

[Illustration]

173. --MRS. PERKINS'S QUILT. 

[Illustration]

It will be seen that in this case the square patchwork quilt is built upof 169 pieces. The puzzle is to find the smallest possible number ofsquare portions of which the quilt could be composed and show how theymight be joined together. Or, to put it the reverse way, divide thequilt into as few square portions as possible by merely cutting thestitches. 

174. --THE SQUARES OF BROCADE. 

[Illustration]

I happened to be paying a call at the house of a lady, when I took upfrom a table two lovely squares of brocade. They were beautifulspecimens of Eastern workmanship--both of the same design, a delicatechequered pattern. 

"Are they not exquisite?" said my friend. "They were brought to me by acousin who has just returned from India. Now, I want you to give me alittle assistance. You see, I have decided to join them together so asto make one large square cushion-cover. How should I do this so as tomutilate the material as little as possible? Of course I propose to makemy cuts only along the lines that divide the little chequers. "

[Illustration]

I cut the two squares in the manner desired into four pieces that wouldfit together and form another larger square, taking care that thepattern should match properly, and when I had finished I noticed thattwo of the pieces were of exactly the same area; that is, each of thetwo contained the same number of chequers. Can you show how the cutswere made in accordance with these conditions?

175--ANOTHER PATCHWORK PUZZLE. 

[Illustration]

A lady was presented, by two of her girl friends, with the pretty piecesof silk patchwork shown in our illustration. It will be seen that bothpieces are made up of squares all of the same size--one 12 × 12 and theother 5 × 5. She proposes to join them together and make one squarepatchwork quilt, 13 × 13, but, of course, she will not cut any of thematerial--merely cut the stitches where necessary and join togetheragain. What perplexes her is this. A friend assures her that there needbe no more than four pieces in all to join up for the new quilt. Couldyou show her how this little needlework puzzle is to be solved in so fewpieces?

176. --LINOLEUM CUTTING. 

[Illustration]

The diagram herewith represents two separate pieces of linoleum. Thechequered pattern is not repeated at the back, so that the pieces cannotbe turned over. The puzzle is to cut the two squares into four pieces sothat they shall fit together and form one perfect square 10 × 10, sothat the pattern shall properly match, and so that the larger pieceshall have as small a portion as possible cut from it. 

177. --ANOTHER LINOLEUM PUZZLE. 

[Illustration]

Can you cut this piece of linoleum into four pieces that will fittogether and form a perfect square? Of course the cuts may only be madealong the lines. 

VARIOUS GEOMETRICAL PUZZLES. 

 "So various are the tastes of men. " MARK AKENSIDE. 

178. --THE CARDBOARD BOX. 

This puzzle is not difficult, but it will be found entertaining todiscover the simple rule for its solution. I have a rectangularcardboard box. The top has an area of 120 square inches, the side 96square inches, and the end 80 square inches. What are the exactdimensions of the box?

179. --STEALING THE BELL-ROPES. 

Two men broke into a church tower one night to steal the bell-ropes. Thetwo ropes passed through holes in the wooden ceiling high above them, and they lost no time in climbing to the top. Then one man drew hisknife and cut the rope above his head, in consequence of which he fellto the floor and was badly injured. His fellow-thief called out that itserved him right for being such a fool. He said that he should have doneas he was doing, upon which he cut the rope below the place at which heheld on. Then, to his dismay, he found that he was in no better plight, for, after hanging on as long as his strength lasted, he was compelledto let go and fall beside his comrade. Here they were both found thenext morning with their limbs broken. How far did they fall? One of theropes when they found it was just touching the floor, and when youpulled the end to the wall, keeping the rope taut, it touched a pointjust three inches above the floor, and the wall was four feet from therope when it hung at rest. How long was the rope from floor to ceiling?

180. --THE FOUR SONS. 

Readers will recognize the diagram as a familiar friend of their youth. A man possessed a square-shaped estate. He bequeathed to his widow thequarter of it that is shaded off. The remainder was to be dividedequitably amongst his four sons, so that each should receive land ofexactly the same area and exactly similar in shape. We are shown howthis was done. But the remainder of the story is not so generally known. In the centre of the estate was a well, indicated by the dark spot, andBenjamin, Charles, and David complained that the division was not"equitable, " since Alfred had access to this well, while they could notreach it without trespassing on somebody else's land. The puzzle is toshow how the estate is to be apportioned so that each son shall haveland of the same shape and area, and each have access to the wellwithout going off his own land. 

[Illustration]

181. --THE THREE RAILWAY STATIONS. 

As I sat in a railway carriage I noticed at the other end of thecompartment a worthy squire, whom I knew by sight, engaged inconversation with another passenger, who was evidently a friend of his. 

"How far have you to drive to your place from the railway station?"asked the stranger. 

"Well, " replied the squire, "if I get out at Appleford, it is just thesame distance as if I go to Bridgefield, another fifteen miles fartheron; and if I changed at Appleford and went thirteen miles from there toCarterton, it would still be the same distance. You see, I amequidistant from the three stations, so I get a good choice of trains. "

Now I happened to know that Bridgefield is just fourteen miles fromCarterton, so I amused myself in working out the exact distance that thesquire had to drive home whichever station he got out at. What was thedistance?

182. --THE GARDEN PUZZLE. 

Professor Rackbrain tells me that he was recently smoking a friendlypipe under a tree in the garden of a country acquaintance. The gardenwas enclosed by four straight walls, and his friend informed him that hehad measured these and found the lengths to be 80, 45, 100, and 63 yardsrespectively. "Then, " said the professor, "we can calculate the exactarea of the garden. " "Impossible, " his host replied, "because you canget an infinite number of different shapes with those four sides. " "Butyou forget, " Rackbrane said, with a twinkle in his eye, "that you toldme once you had planted this tree equidistant from all the four cornersof the garden. " Can you work out the garden's area?

183. --DRAWING A SPIRAL. 

If you hold the page horizontally and give it a quick rotary motionwhile looking at the centre of the spiral, it will appear to revolve. Perhaps a good many readers are acquainted with this little opticalillusion. But the puzzle is to show how I was able to draw this spiralwith so much exactitude without using anything but a pair of compassesand the sheet of paper on which the diagram was made. How would youproceed in such circumstances?

[Illustration]

184. --HOW TO DRAW AN OVAL. 

Can you draw a perfect oval on a sheet of paper with one sweep of thecompasses? It is one of the easiest things in the world when you knowhow. 

185. --ST. GEORGE'S BANNER. 

At a celebration of the national festival of St. George's Day I wascontemplating the familiar banner of the patron saint of our country. Weall know the red cross on a white ground, shown in our illustration. This is the banner of St. George. The banner of St. Andrew (Scotland) isa white "St. Andrew's Cross" on a blue ground. That of St. Patrick(Ireland) is a similar cross in red on a white ground. These three areunited in one to form our Union Jack. 

Now on looking at St. George's banner it occurred to me that thefollowing question would make a simple but pretty little puzzle. Supposing the flag measures four feet by three feet, how wide must thearm of the cross be if it is required that there shall be used just thesame quantity of red and of white bunting?

[Illustration]

186. --THE CLOTHES LINE PUZZLE. 

A boy tied a clothes line from the top of each of two poles to the baseof the other. He then proposed to his father the following question. Asone pole was exactly seven feet above the ground and the other exactlyfive feet, what was the height from the ground where the two cordscrossed one another?

187. --THE MILKMAID PUZZLE. 

[Illustration]

Here is a little pastoral puzzle that the reader may, at first sight, beled into supposing is very profound, involving deep calculations. He mayeven say that it is quite impossible to give any answer unless we aretold something definite as to the distances. And yet it is really quite"childlike and bland. "

In the corner of a field is seen a milkmaid milking a cow, and on theother side of the field is the dairy where the extract has to bedeposited. But it has been noticed that the young woman always goes downto the river with her pail before returning to the dairy. Here thesuspicious reader will perhaps ask why she pays these visits to theriver. I can only reply that it is no business of ours. The alleged milkis entirely for local consumption. 

 "Where are you going to, my pretty maid?" "Down to the river, sir, " she said. "I'll _not_ choose your dairy, my pretty maid. " "Nobody axed you, sir, " she said. 

If one had any curiosity in the matter, such an independent spirit wouldentirely disarm one. So we will pass from the point of commercialmorality to the subject of the puzzle. 

Draw a line from the milking-stool down to the river and thence to thedoor of the dairy, which shall indicate the shortest possible route forthe milkmaid. That is all. It is quite easy to indicate the exact spoton the bank of the river to which she should direct her steps if shewants as short a walk as possible. Can you find that spot?

188. --THE BALL PROBLEM. 

[Illustration]

A stonemason was engaged the other day in cutting out a round ball forthe purpose of some architectural decoration, when a smart schoolboycame upon the scene. 

"Look here, " said the mason, "you seem to be a sharp youngster, can youtell me this? If I placed this ball on the level ground, how many otherballs of the same size could I lay around it (also on the ground) sothat every ball should touch this one?"

The boy at once gave the correct answer, and then put this littlequestion to the mason:--

"If the surface of that ball contained just as many square feet as itsvolume contained cubic feet, what would be the length of its diameter?"

The stonemason could not give an answer. Could you have repliedcorrectly to the mason's and the boy's questions?

189. --THE YORKSHIRE ESTATES. 

[Illustration]

I was on a visit to one of the large towns of Yorkshire. While walkingto the railway station on the day of my departure a man thrust ahand-bill upon me, and I took this into the railway carriage and read itat my leisure. It informed me that three Yorkshire neighbouring estateswere to be offered for sale. Each estate was square in shape, and theyjoined one another at their corners, just as shown in the diagram. Estate A contains exactly 370 acres, B contains 116 acres, and C 74acres. 

Now, the little triangular bit of land enclosed by the three squareestates was not offered for sale, and, for no reason in particular, Ibecame curious as to the area of that piece. How many acres did itcontain?

190. --FARMER WURZEL'S ESTATE. 

[Illustration]

I will now present another land problem. The demonstration of the answerthat I shall give will, I think, be found both interesting and easy ofcomprehension. 

Farmer Wurzel owned the three square fields shown in the annexed plan, containing respectively 18, 20, and 26 acres. In order to get aring-fence round his property he bought the four intervening triangularfields. The puzzle is to discover what was then the whole area of hisestate. 

191. --THE CRESCENT PUZZLE. 

[Illustration]

Here is an easy geometrical puzzle. The crescent is formed by twocircles, and C is the centre of the larger circle. The width of thecrescent between B and D is 9 inches, and between E and F 5 inches. Whatare the diameters of the two circles?

192. --THE PUZZLE WALL. 

[Illustration]

There was a small lake, around which four poor men built their cottages. Four rich men afterwards built their mansions, as shown in theillustration, and they wished to have the lake to themselves, so theyinstructed a builder to put up the shortest possible wall that wouldexclude the cottagers, but give themselves free access to the lake. Howwas the wall to be built?

193. --THE SHEEPFOLD. 

It is a curious fact that the answers always given to some of thebest-known puzzles that appear in every little book of firesiderecreations that has been published for the last fifty or a hundredyears are either quite unsatisfactory or clearly wrong. Yet nobody everseems to detect their faults. Here is an example:--A farmer had a penmade of fifty hurdles, capable of holding a hundred sheep only. Supposing he wanted to make it sufficiently large to hold double thatnumber, how many additional hurdles must he have?

194. --THE GARDEN WALLS. 

[Illustration]

A speculative country builder has a circular field, on which he haserected four cottages, as shown in the illustration. The field issurrounded by a brick wall, and the owner undertook to put up threeother brick walls, so that the neighbours should not be overlooked byeach other, but the four tenants insist that there shall be nofavouritism, and that each shall have exactly the same length of wallspace for his wall fruit trees. The puzzle is to show how the threewalls may be built so that each tenant shall have the same area ofground, and precisely the same length of wall. 

Of course, each garden must be entirely enclosed by its walls, and itmust be possible to prove that each garden has exactly the same lengthof wall. If the puzzle is properly solved no figures are necessary. 

195. --LADY BELINDA'S GARDEN. 

Lady Belinda is an enthusiastic gardener. In the illustration she isdepicted in the act of worrying out a pleasant little problem which Iwill relate. One of her gardens is oblong in shape, enclosed by a highholly hedge, and she is turning it into a rosary for the cultivation ofsome of her choicest roses. She wants to devote exactly half of the areaof the garden to the flowers, in one large bed, and the other half to bea path going all round it of equal breadth throughout. Such a garden isshown in the diagram at the foot of the picture. How is she to mark outthe garden under these simple conditions? She has only a tape, thelength of the garden, to do it with, and, as the holly hedge is so thickand dense, she must make all her measurements inside. Lady Belinda didnot know the exact dimensions of the garden, and, as it was notnecessary for her to know, I also give no dimensions. It is quite asimple task no matter what the size or proportions of the garden may be. Yet how many lady gardeners would know just how to proceed? The tape maybe quite plain--that is, it need not be a graduated measure. 

[Illustration]

196. --THE TETHERED GOAT. 

[Illustration]

Here is a little problem that everybody should know how to solve. Thegoat is placed in a half-acre meadow, that is in shape an equilateraltriangle. It is tethered to a post at one corner of the field. Whatshould be the length of the tether (to the nearest inch) in order thatthe goat shall be able to eat just half the grass in the field? It isassumed that the goat can feed to the end of the tether. 

197. --THE COMPASSES PUZZLE. 

It is curious how an added condition or restriction will sometimesconvert an absurdly easy puzzle into an interesting and perhapsdifficult one. I remember buying in the street many years ago a littlemechanical puzzle that had a tremendous sale at the time. It consistedof a medal with holes in it, and the puzzle was to work a ring with agap in it from hole to hole until it was finally detached. As I waswalking along the street I very soon acquired the trick of taking offthe ring with one hand while holding the puzzle in my pocket. A friendto whom I showed the little feat set about accomplishing it himself, andwhen I met him some days afterwards he exhibited his proficiency in theart. But he was a little taken aback when I then took the puzzle fromhim and, while simply holding the medal between the finger and thumb ofone hand, by a series of little shakes and jerks caused the ring, without my even touching it, to fall off upon the floor. The followinglittle poser will probably prove a rather tough nut for a great manyreaders, simply on account of the restricted conditions:--

Show how to find exactly the middle of any straight line by means of thecompasses only. You are not allowed to use any ruler, pencil, or otherarticle--only the compasses; and no trick or dodge, such as folding thepaper, will be permitted. You must simply use the compasses in theordinary legitimate way. 

198. --THE EIGHT STICKS. 

I have eight sticks, four of them being exactly half the length of theothers. I lay every one of these on the table, so that they enclosethree squares, all of the same size. How do I do it? There must be noloose ends hanging over. 

199. --PAPA'S PUZZLE. 

Here is a puzzle by Pappus, who lived at Alexandria about the end of thethird century. It is the fifth proposition in the eighth book of his_Mathematical Collections_. I give it in the form that I presented itsome years ago under the title "Papa's Puzzle, " just to see how manyreaders would discover that it was by Pappus himself. "The little maid'spapa has taken two different-sized rectangular pieces of cardboard, andhas clipped off a triangular piece from one of them, so that when it issuspended by a thread from the point A it hangs with the long sideperfectly horizontal, as shown in the illustration. He has perplexed thechild by asking her to find the point A on the other card, so as toproduce a similar result when cut and suspended by a thread. " Of course, the point must not be found by trial clippings. A curious and prettypoint is involved in this setting of the puzzle. Can the reader discoverit?

[Illustration]

200. --A KITE-FLYING PUZZLE. 

While accompanying my friend Professor Highflite during a scientifickite-flying competition on the South Downs of Sussex I was led into alittle calculation that ought to interest my readers. The Professor waspaying out the wire to which his kite was attached from a winch on whichit had been rolled into a perfectly spherical form. This ball of wirewas just two feet in diameter, and the wire had a diameter ofone-hundredth of an inch. What was the length of the wire?

Now, a simple little question like this that everybody can perfectlyunderstand will puzzle many people to answer in any way. Let us seewhether, without going into any profound mathematical calculations, wecan get the answer roughly--say, within a mile of what is correct! Wewill assume that when the wire is all wound up the ball is perfectlysolid throughout, and that no allowance has to be made for the axle thatpasses through it. With that simplification, I wonder how many readerscan state within even a mile of the correct answer the length of thatwire. 

201. --HOW TO MAKE CISTERNS. 

[Illustration]

Our friend in the illustration has a large sheet of zinc, measuring(before cutting) eight feet by three feet, and he has cut out squarepieces (all of the same size) from the four corners and now proposes tofold up the sides, solder the edges, and make a cistern. But the pointthat puzzles him is this: Has he cut out those square pieces of thecorrect size in order that the cistern may hold the greatest possiblequantity of water? You see, if you cut them very small you get a veryshallow cistern; if you cut them large you get a tall and slender one. It is all a question of finding a way of cutting put these four squarepieces exactly the right size. How are we to avoid making them too smallor too large?

202. --THE CONE PUZZLE. 

[Illustration]

I have a wooden cone, as shown in Fig. 1. How am I to cut out of it thegreatest possible cylinder? It will be seen that I can cut out one thatis long and slender, like Fig. 2, or short and thick, like Fig. 3. Butneither is the largest possible. A child could tell you where to cut, ifhe knew the rule. Can you find this simple rule?

203. --CONCERNING WHEELS. 

[Illustration]

There are some curious facts concerning the movements of wheels that areapt to perplex the novice. For example: when a railway train istravelling from London to Crewe certain parts of the train at any givenmoment are actually moving from Crewe towards London. Can you indicatethose parts? It seems absurd that parts of the same train can at anytime travel in opposite directions, but such is the case. 

In the accompanying illustration we have two wheels. The lower one issupposed to be fixed and the upper one running round it in the directionof the arrows. Now, how many times does the upper wheel turn on its ownaxis in making a complete revolution of the other wheel? Do not be in ahurry with your answer, or you are almost certain to be wrong. Experiment with two pennies on the table and the correct answer willsurprise you, when you succeed in seeing it. 

204. --A NEW MATCH PUZZLE. 

[Illustration]

In the illustration eighteen matches are shown arranged so that theyenclose two spaces, one just twice as large as the other. Can yourearrange them (1) so as to enclose two four-sided spaces, one exactlythree times as large as the other, and (2) so as to enclose twofive-sided spaces, one exactly three times as large as the other? Allthe eighteen matches must be fairly used in each case; the two spacesmust be quite detached, and there must be no loose ends or duplicatedmatches. 

205. --THE SIX SHEEP-PENS. 

[Illustration]

Here is a new little puzzle with matches. It will be seen in theillustration that thirteen matches, representing a farmer's hurdles, have been so placed that they enclose six sheep-pens all of the samesize. Now, one of these hurdles was stolen, and the farmer wanted stillto enclose six pens of equal size with the remaining twelve. How was heto do it? All the twelve matches must be fairly used, and there must beno duplicated matches or loose ends. 

POINTS AND LINES PROBLEMS. 

"Line upon line, line upon line; here a little and there alittle. "--_Isa_. Xxviii. 10. 

What are known as "Points and Lines" puzzles are found very interestingby many people. The most familiar example, here given, to plant ninetrees so that they shall form ten straight rows with three trees inevery row, is attributed to Sir Isaac Newton, but the earliestcollection of such puzzles is, I believe, in a rare little book that Ipossess--published in 1821--_Rational Amusement for Winter Evenings_, byJohn Jackson. The author gives ten examples of "Trees planted in Rows. "

These tree-planting puzzles have always been a matter of greatperplexity. They are real "puzzles, " in the truest sense of the word, because nobody has yet succeeded in finding a direct and certain way ofsolving them. They demand the exercise of sagacity, ingenuity, andpatience, and what we call "luck" is also sometimes of service. Perhapssome day a genius will discover the key to the whole mystery. Rememberthat the trees must be regarded as mere points, for if we were allowedto make our trees big enough we might easily "fudge" our diagrams andget in a few extra straight rows that were more apparent than real. 

[Illustration]

206. --THE KING AND THE CASTLES. 

There was once, in ancient times, a powerful king, who had eccentricideas on the subject of military architecture. He held that there wasgreat strength and economy in symmetrical forms, and always cited theexample of the bees, who construct their combs in perfect hexagonalcells, to prove that he had nature to support him. He resolved to buildten new castles in his country all to be connected by fortified walls, which should form five lines with four castles in every line. The royalarchitect presented his preliminary plan in the form I have shown. Butthe monarch pointed out that every castle could be approached from theoutside, and commanded that the plan should be so modified that as manycastles as possible should be free from attack from the outside, andcould only be reached by crossing the fortified walls. The architectreplied that he thought it impossible so to arrange them that even onecastle, which the king proposed to use as a royal residence, could be soprotected, but his majesty soon enlightened him by pointing out how itmight be done. How would you have built the ten castles andfortifications so as best to fulfil the king's requirements? Rememberthat they must form five straight lines with four castles in every line. 

[Illustration]

207. --CHERRIES AND PLUMS. 

[Illustration]

The illustration is a plan of a cottage as it stands surrounded by anorchard of fifty-five trees. Ten of these trees are cherries, ten areplums, and the remainder apples. The cherries are so planted as to formfive straight lines, with four cherry trees in every line. The plumtrees are also planted so as to form five straight lines with four plumtrees in every line. The puzzle is to show which are the ten cherrytrees and which are the ten plums. In order that the cherries and plumsshould have the most favourable aspect, as few as possible (under theconditions) are planted on the north and east sides of the orchard. Ofcourse in picking out a group of ten trees (cherry or plum, as the casemay be) you ignore all intervening trees. That is to say, four trees maybe in a straight line irrespective of other trees (or the house) beingin between. After the last puzzle this will be quite easy. 

208. --A PLANTATION PUZZLE. 

[Illustration]

A man had a square plantation of forty-nine trees, but, as will be seenby the omissions in the illustration, four trees were blown down andremoved. He now wants to cut down all the remainder except ten trees, which are to be so left that they shall form five straight rows withfour trees in every row. Which are the ten trees that he must leave?

209. --THE TWENTY-ONE TREES. 

A gentleman wished to plant twenty-one trees in his park so that theyshould form twelve straight rows with five trees in every row. Could youhave supplied him with a pretty symmetrical arrangement that wouldsatisfy these conditions?

210. --THE TEN COINS. 

Place ten pennies on a large sheet of paper or cardboard, as shown inthe diagram, five on each edge. Now remove four of the coins, withoutdisturbing the others, and replace them on the paper so that the tenshall form five straight lines with four coins in every line. This initself is not difficult, but you should try to discover in how manydifferent ways the puzzle may be solved, assuming that in every case thetwo rows at starting are exactly the same. 

[Illustration]

211. --THE TWELVE MINCE-PIES. 

It will be seen in our illustration how twelve mince-pies may be placedon the table so as to form six straight rows with four pies in everyrow. The puzzle is to remove only four of them to new positions so thatthere shall be _seven_ straight rows with four in every row. Which fourwould you remove, and where would you replace them?

[Illustration]

212. --THE BURMESE PLANTATION. 

[Illustration]

A short time ago I received an interesting communication from theBritish chaplain at Meiktila, Upper Burma, in which my correspondentinformed me that he had found some amusement on board ship on his wayout in trying to solve this little poser. 

If he has a plantation of forty-nine trees, planted in the form of asquare as shown in the accompanying illustration, he wishes to know howhe may cut down twenty-seven of the trees so that the twenty-two leftstanding shall form as many rows as possible with four trees in everyrow. 

Of course there may not be more than four trees in any row. 

213. --TURKS AND RUSSIANS. 

This puzzle is on the lines of the Afridi problem published by me in_Tit-Bits_ some years ago. 

On an open level tract of country a party of Russian infantry, no two ofwhom were stationed at the same spot, were suddenly surprised bythirty-two Turks, who opened fire on the Russians from all directions. Each of the Turks simultaneously fired a bullet, and each bullet passedimmediately over the heads of three Russian soldiers. As each of thesebullets when fired killed a different man, the puzzle is to discoverwhat is the smallest possible number of soldiers of which the Russianparty could have consisted and what were the casualties on each side. 

MOVING COUNTER PROBLEMS. 

 "I cannot do't without counters. "

 _Winter's Tale_, iv. 3. 

Puzzles of this class, except so far as they occur in connection withactual games, such as chess, seem to be a comparatively modernintroduction. Mathematicians in recent times, notably Vandermonde andReiss, have devoted some attention to them, but they do not appear tohave been considered by the old writers. So far as games with countersare concerned, perhaps the most ancient and widely known in old times is"Nine Men's Morris" (known also, as I shall show, under a great manyother names), unless the simpler game, distinctly mentioned in the worksof Ovid (No. 110, "Ovid's Game, " in _The Canterbury Puzzles_), fromwhich "Noughts and Crosses" seems to be derived, is still more ancient. 

In France the game is called Marelle, in Poland Siegen Wulf Myll(She-goat Wolf Mill, or Fight), in Germany and Austria it is calledMuhle (the Mill), in Iceland it goes by the name of Mylla, while theBogas (or native bargees) of South America are said to play it, and onthe Amazon it is called Trique, and held to be of Indian origin. In ourown country it has different names in different districts, such as MegMerrylegs, Peg Meryll, Nine Peg o'Merryal, Nine-Pin Miracle, Merry Peg, and Merry Hole. Shakespeare refers to it in "Midsummer Night's Dream"(Act ii. , scene 1):--

 "The nine-men's morris is filled up with mud; And the quaint mazes in the wanton green, For lack of tread, are undistinguishable. "

It was played by the shepherds with stones in holes cut in the turf. John Clare, the peasant poet of Northamptonshire, in "The Shepherd Boy"(1835) says:--"Oft we track his haunts . . . . By nine-peg-morris nickedupon the green. " It is also mentioned by Drayton in his "Polyolbion. "

It was found on an old Roman tile discovered during the excavations atSilchester, and cut upon the steps of the Acropolis at Athens. Whenvisiting the Christiania Museum a few years ago I was shown the greatViking ship that was discovered at Gokstad in 1880. On the oak planksforming the deck of the vessel were found boles and lines marking outthe game, the holes being made to receive pegs. While inspecting theancient oak furniture in the Rijks Museum at Amsterdam I becameinterested in an old catechumen's settle, and was surprised to find thegame diagram cut in the centre of the seat--quite conveniently forsurreptitious play. It has been discovered cut in the choir stalls ofseveral of our English cathedrals. In the early eighties it was foundscratched upon a stone built into a wall (probably about the date 1200), during the restoration of Hargrave church in Northamptonshire. Thisstone is now in the Northampton Museum. A similar stone has since beenfound at Sempringham, Lincolnshire. It is to be seen on an ancienttombstone in the Isle of Man, and painted on old Dutch tiles. And in1901 a stone was dug out of a gravel pit near Oswestry bearing anundoubted diagram of the game. 

The game has been played with different rules at different periods andplaces. I give a copy of the board. Sometimes the diagonal lines areomitted, but this evidently was not intended to affect the play: itsimply meant that the angles alone were thought sufficient to indicatethe points. This is how Strutt, in _Sports and Pastimes_, describes thegame, and it agrees with the way I played it as a boy:--"Two persons, having each of them nine pieces, or men, lay them down alternately, oneby one, upon the spots; and the business of either party is to preventhis antagonist from placing three of his pieces so as to form a row ofthree, without the intervention of an opponent piece. If a row beformed, he that made it is at liberty to take up one of his competitor'spieces from any part he thinks most to his advantage; excepting he hasmade a row, which must not be touched if he have another piece upon theboard that is not a component part of that row. When all the pieces arelaid down, they are played backwards and forwards, in any direction thatthe lines run, but only can move from one spot to another (next to it)at one time. He that takes off all his antagonist's pieces is theconqueror. "

[Illustration]

214. --THE SIX FROGS. 

[Illustration]

The six educated frogs in the illustration are trained to reverse theirorder, so that their numbers shall read 6, 5, 4, 3, 2, 1, with the blanksquare in its present position. They can jump to the next square (ifvacant) or leap over one frog to the next square beyond (if vacant), just as we move in the game of draughts, and can go backwards orforwards at pleasure. Can you show how they perform their feat in thefewest possible moves? It is quite easy, so when you have done it add aseventh frog to the right and try again. Then add more frogs until youare able to give the shortest solution for any number. For it can alwaysbe done, with that single vacant square, no matter how many frogs thereare. 

215. --THE GRASSHOPPER PUZZLE. 

It has been suggested that this puzzle was a great favourite among theyoung apprentices of the City of London in the sixteenth and seventeenthcenturies. Readers will have noticed the curious brass grasshopper onthe Royal Exchange. This long-lived creature escaped the fires of 1666and 1838. The grasshopper, after his kind, was the crest of Sir ThomasGresham, merchant grocer, who died in 1579, and from this cause it hasbeen used as a sign by grocers in general. Unfortunately for the legendas to its origin, the puzzle was only produced by myself so late as theyear 1900. On twelve of the thirteen black discs are placed numberedcounters or grasshoppers. The puzzle is to reverse their order, so thatthey shall read, 1, 2, 3, 4, etc. , in the opposite direction, with thevacant disc left in the same position as at present. Move one at a timein any order, either to the adjoining vacant disc or by jumping over onegrasshopper, like the moves in draughts. The moves or leaps may be madein either direction that is at any time possible. What are the fewestpossible moves in which it can be done?

[Illustration]

216. --THE EDUCATED FROGS. 

[Illustration]

Our six educated frogs have learnt a new and pretty feat. When placed onglass tumblers, as shown in the illustration, they change sides so thatthe three black ones are to the left and the white frogs to the right, with the unoccupied tumbler at the opposite end--No. 7. They can jump tothe next tumbler (if unoccupied), or over one, or two, frogs to anunoccupied tumbler. The jumps can be made in either direction, and afrog may jump over his own or the opposite colour, or both colours. Foursuccessive specimen jumps will make everything quite plain: 4 to 1, 5 to4, 3 to 5, 6 to 3. Can you show how they do it in ten jumps?

217. --THE TWICKENHAM PUZZLE. 

[Illustration:

 ( I ) ((N))

 ( M ) ((A))

 ( H ) ((T))

 ( E ) ((W))

 ( C ) ((K)) ( )

]

In the illustration we have eleven discs in a circle. On five of thediscs we place white counters with black letters--as shown--and on fiveother discs the black counters with white letters. The bottom disc isleft vacant. Starting thus, it is required to get the counters intoorder so that they spell the word "Twickenham" in a clockwise direction, leaving the vacant disc in the original position. The black countersmove in the direction that a clock-hand revolves, and the white countersgo the opposite way. A counter may jump over one of the opposite colourif the vacant disc is next beyond. Thus, if your first move is with K, then C can jump over K. If then K moves towards E, you may next jump Wover C, and so on. The puzzle may be solved in twenty-six moves. Remember a counter cannot jump over one of its own colour. 

218. --THE VICTORIA CROSS PUZZLE. 

[Illustration:

 +---------------------+ | \. . . A . . . / | | (I) |. . . . . . . | (V) | |\_____|_______|_____/| |. . . . . . | |------| |. . R . | |. I . . | |. . . . . . | |. . . . . . | | _____|_______|_____ | |/ |. . . . . . . | \| | (O) |. . T . . | (C) | | /. . . . . . . . . \ | +---------------------+

]

The puzzle-maker is peculiarly a "snapper-up of unconsidered trifles, "and his productions are often built up with the slenderest materials. Trivialities that might entirely escape the observation of others, or, if they were observed, would be regarded as of no possible moment, oftensupply the man who is in quest of posers with a pretty theme or an ideathat he thinks possesses some "basal value. "

When seated opposite to a lady in a railway carriage at the time ofQueen Victoria's Diamond Jubilee, my attention was attracted to a broochthat she was wearing. It was in the form of a Maltese or Victoria Cross, and bore the letters of the word VICTORIA. The number and arrangement ofthe letters immediately gave me the suggestion for the puzzle which Inow present. 

The diagram, it will be seen, is composed of nine divisions. The puzzleis to place eight counters, bearing the letters of the word VICTORIA, exactly in the manner shown, and then slide one letter at a time fromblack to white and white to black alternately, until the word readsround in the same direction, only with the initial letter V on one ofthe black arms of the cross. At no time may two letters be in the samedivision. It is required to find the shortest method. 

Leaping moves are, of course, not permitted. The first move mustobviously be made with A, I, T, or R. Supposing you move T to thecentre, the next counter played will be O or C, since I or R cannot bemoved. There is something a little remarkable in the solution of thispuzzle which I will explain. 

219. --THE LETTER BLOCK PUZZLE. 

[Illustration:

 +-----+-----+-----+\ | | | | | | G | E | F | | | | | | | +-----+-----+-----+\| | | | | | | H | C | B | | | | | | | +-----+-----+-----+\| | |\____| | | | D || | A | | | || | | | +-----+-----+-----+ | \_________________\|

]

Here is a little reminiscence of our old friend the Fifteen BlockPuzzle. Eight wooden blocks are lettered, and are placed in a box, asshown in the illustration. It will be seen that you can only move oneblock at a time to the place vacant for the time being, as no block maybe lifted out of the box. The puzzle is to shift them about until youget them in the order--

 A B C D E F G H

This you will find by no means difficult if you are allowed as manymoves as you like. But the puzzle is to do it in the fewest possiblemoves. I will not say what this smallest number of moves is, because thereader may like to discover it for himself. In writing down your movesyou will find it necessary to record no more than the letters in theorder that they are shifted. Thus, your first five moves might be C, H, G, E, F; and this notation can have no possible ambiguity. In practiceyou only need eight counters and a simple diagram on a sheet of paper. 

220. --A LODGING-HOUSE DIFFICULTY. 

[Illustration]

The Dobsons secured apartments at Slocomb-on-Sea. There were six roomson the same floor, all communicating, as shown in the diagram. The roomsthey took were numbers 4, 5, and 6, all facing the sea. But a littledifficulty arose. Mr. Dobson insisted that the piano and the bookcaseshould change rooms. This was wily, for the Dobsons were not musical, but they wanted to prevent any one else playing the instrument. Now, therooms were very small and the pieces of furniture indicated were verybig, so that no two of these articles could be got into any room at thesame time. How was the exchange to be made with the least possiblelabour? Suppose, for example, you first move the wardrobe into No. 2;then you can move the bookcase to No. 5 and the piano to No. 6, and soon. It is a fascinating puzzle, but the landlady had reasons for notappreciating it. Try to solve her difficulty in the fewest possibleremovals with counters on a sheet of paper. 

221. --THE EIGHT ENGINES. 

The diagram represents the engine-yard of a railway company undereccentric management. The engines are allowed to be stationary only atthe nine points indicated, one of which is at present vacant. It isrequired to move the engines, one at a time, from point to point, inseventeen moves, so that their numbers shall be in numerical order roundthe circle, with the central point left vacant. But one of the engineshas had its fire drawn, and therefore cannot move. How is the thing tobe done? And which engine remains stationary throughout?

[Illustration]

222. --A RAILWAY PUZZLE. 

[Illustration]

Make a diagram, on a large sheet of paper, like the illustration, andhave three counters marked A, three marked B, and three marked C. Itwill be seen that at the intersection of lines there are ninestopping-places, and a tenth stopping-place is attached to the outercircle like the tail of a Q. Place the three counters or engines markedA, the three marked B, and the three marked C at the places indicated. The puzzle is to move the engines, one at a time, along the lines, fromstopping-place to stopping-place, until you succeed in getting an A, aB, and a C on each circle, and also A, B, and C on each straight line. You are required to do this in as few moves as possible. How many movesdo you need?

223. --A RAILWAY MUDDLE. 

The plan represents a portion of the line of the London, Clodville, andMudford Railway Company. It is a single line with a loop. There is onlyroom for eight wagons, or seven wagons and an engine, between B and C oneither the left line or the right line of the loop. It happened that twogoods trains (each consisting of an engine and sixteen wagons) got intothe position shown in the illustration. It looked like a hopelessdeadlock, and each engine-driver wanted the other to go back to the nextstation and take off nine wagons. But an ingenious stoker undertook topass the trains and send them on their respective journeys with theirengines properly in front. He also contrived to reverse the engines thefewest times possible. Could you have performed the feat? And how manytimes would you require to reverse the engines? A "reversal" means achange of direction, backward or forward. No rope-shunting, fly-shunting, or other trick is allowed. All the work must be donelegitimately by the two engines. It is a simple but interesting puzzleif attempted with counters. 

[Illustration]

224. --THE MOTOR-GARAGE PUZZLE. 

[Illustration]

The difficulties of the proprietor of a motor garage are converted intoa little pastime of a kind that has a peculiar fascination. All you needis to make a simple plan or diagram on a sheet of paper or cardboard andnumber eight counters, 1 to 8. Then a whole family can enter into anamusing competition to find the best possible solution of thedifficulty. 

The illustration represents the plan of a motor garage, withaccommodation for twelve cars. But the premises are so inconvenientlyrestricted that the proprietor is often caused considerable perplexity. Suppose, for example, that the eight cars numbered 1 to 8 are in thepositions shown, how are they to be shifted in the quickest possible wayso that 1, 2, 3, and 4 shall change places with 5, 6, 7, and 8--that is, with the numbers still running from left to right, as at present, butthe top row exchanged with the bottom row? What are the fewest possiblemoves?

One car moves at a time, and any distance counts as one move. To preventmisunderstanding, the stopping-places are marked in squares, and onlyone car can be in a square at the same time. 

225. --THE TEN PRISONERS. 

If prisons had no other use, they might still be preserved for thespecial benefit of puzzle-makers. They appear to be an inexhaustiblemine of perplexing ideas. Here is a little poser that will perhapsinterest the reader for a short period. We have in the illustration aprison of sixteen cells. The locations of the ten prisoners will beseen. The jailer has queer superstitions about odd and even numbers, andhe wants to rearrange the ten prisoners so that there shall be as manyeven rows of men, vertically, horizontally, and diagonally, aspossible. At present it will be seen, as indicated by the arrows, thatthere are only twelve such rows of 2 and 4. I will state at once thatthe greatest number of such rows that is possible is sixteen. But thejailer only allows four men to be removed to other cells, and informs methat, as the man who is seated in the bottom right-hand corner isinfirm, he must not be moved. Now, how are we to get those sixteen rowsof even numbers under such conditions?

[Illustration]

226. --ROUND THE COAST. 

[Illustration]

Here is a puzzle that will, I think, be found as amusing as instructive. We are given a ring of eight circles. Leaving circle 8 blank, we arerequired to write in the name of a seven-lettered port in the UnitedKingdom in this manner. Touch a blank circle with your pencil, then jumpover two circles in either direction round the ring, and write down thefirst letter. Then touch another vacant circle, jump over two circles, and write down your second letter. Proceed similarly with the otherletters in their proper order until you have completed the word. Thus, suppose we select "Glasgow, " and proceed as follows: 6--1, 7--2, 8--3, 7--4, 8--5, which means that we touch 6, jump over 7 and and write down"G" on 1; then touch 7, jump over 8 and 1, and write down "l" on 2; andso on. It will be found that after we have written down the first fiveletters--"Glasg"--as above, we cannot go any further. Either there issomething wrong with "Glasgow, " or we have not managed our jumpsproperly. Can you get to the bottom of the mystery?

227. --CENTRAL SOLITAIRE. 

[Illustration]

This ancient puzzle was a great favourite with our grandmothers, andmost of us, I imagine, have on occasions come across a "Solitaire"board--a round polished board with holes cut in it in a geometricalpattern, and a glass marble in every hole. Sometimes I have noticed oneon a side table in a suburban front parlour, or found one on a shelf ina country cottage, or had one brought under my notice at a wayside inn. Sometimes they are of the form shown above, but it is equally common forthe board to have four more holes, at the points indicated by dots. Iselect the simpler form. 

Though "Solitaire" boards are still sold at the toy shops, it will besufficient if the reader will make an enlarged copy of the above on asheet of cardboard or paper, number the "holes, " and provide himselfwith 33 counters, buttons, or beans. Now place a counter in every holeexcept the central one, No. 17, and the puzzle is to take off all thecounters in a series of jumps, except the last counter, which must beleft in that central hole. You are allowed to jump one counter over thenext one to a vacant hole beyond, just as in the game of draughts, andthe counter jumped over is immediately taken off the board. Onlyremember every move must be a jump; consequently you will take off acounter at each move, and thirty-one single jumps will of course removeall the thirty-one counters. But compound moves are allowed (as indraughts, again), for so long as one counter continues to jump, thejumps all count as one move. 

Here is the beginning of an imaginary solution which will serve to makethe manner of moving perfectly plain, and show how the solver shouldwrite out his attempts: 5-17, 12-10, 26-12, 24-26 (13-11, 11-25), 9-11(26-24, 24-10, 10-12), etc. , etc. The jumps contained within bracketscount as one move, because they are made with the same counter. Find thefewest possible moves. Of course, no diagonal jumps are permitted; youcan only jump in the direction of the lines. 

228. --THE TEN APPLES. 

[Illustration]

The family represented in the illustration are amusing themselves withthis little puzzle, which is not very difficult but quite interesting. They have, it will be seen, placed sixteen plates on the table in theform of a square, and put an apple in each of ten plates. They want tofind a way of removing all the apples except one by jumping over one ata time to the next vacant square, as in draughts; or, better, as insolitaire, for you are not allowed to make any diagonal moves--onlymoves parallel to the sides of the square. It is obvious that as theapples stand no move can be made, but you are permitted to transfer anysingle apple you like to a vacant plate before starting. Then the movesmust be all leaps, taking off the apples leaped over. 

229. --THE NINE ALMONDS. 

"Here is a little puzzle, " said a Parson, "that I have found peculiarlyfascinating. It is so simple, and yet it keeps you interestedindefinitely. "

The reverend gentleman took a sheet of paper and divided it off intotwenty-five squares, like a square portion of a chessboard. Then heplaced nine almonds on the central squares, as shown in theillustration, where we have represented numbered counters forconvenience in giving the solution. 

"Now, the puzzle is, " continued the Parson, "to remove eight of thealmonds and leave the ninth in the central square. You make the removalsby jumping one almond over another to the vacant square beyond andtaking off the one jumped over--just as in draughts, only here you canjump in any direction, and not diagonally only. The point is to do thething in the fewest possible moves. "

The following specimen attempt will make everything clear. Jump 4 over1, 5 over 9, 3 over 6, 5 over 3, 7 over 5 and 2, 4 over 7, 8 over 4. But8 is not left in the central square, as it should be. Remember to removethose you jump over. Any number of jumps in succession with the samealmond count as one move. 

[Illustration]

230. --THE TWELVE PENNIES. 

Here is a pretty little puzzle that only requires twelve pennies orcounters. Arrange them in a circle, as shown in the illustration. Nowtake up one penny at a time and, passing it over two pennies, place iton the third penny. Then take up another single penny and do the samething, and so on, until, in six such moves, you have the coins in sixpairs in the positions 1, 2, 3, 4, 5, 6. You can move in eitherdirection round the circle at every play, and it does not matterwhether the two jumped over are separate or a pair. This is quite easyif you use just a little thought. 

[Illustration]

231. --PLATES AND COINS. 

Place twelve plates, as shown, on a round table, with a penny or orangein every plate. Start from any plate you like and, always going in onedirection round the table, take up one penny, pass it over two otherpennies, and place it in the next plate. Go on again; take up anotherpenny and, having passed it over two pennies, place it in a plate; andso continue your journey. Six coins only are to be removed, and whenthese have been placed there should be two coins in each of six platesand six plates empty. An important point of the puzzle is to go roundthe table as few times as possible. It does not matter whether the twocoins passed over are in one or two plates, nor how many empty platesyou pass a coin over. But you must always go in one direction round thetable and end at the point from which you set out. Your hand, that is tosay, goes steadily forward in one direction, without ever movingbackwards. 

[Illustration]

232. --CATCHING THE MICE. 

[Illustration]

"Play fair!" said the mice. "You know the rules of the game. "

"Yes, I know the rules, " said the cat. "I've got to go round and roundthe circle, in the direction that you are looking, and eat everythirteenth mouse, but I must keep the white mouse for a tit-bit at thefinish. Thirteen is an unlucky number, but I will do my best to obligeyou. "

"Hurry up, then!" shouted the mice. 

"Give a fellow time to think, " said the cat. "I don't know which of youto start at. I must figure it out. "

While the cat was working out the puzzle he fell asleep, and, the spellbeing thus broken, the mice returned home in safety. At which mouseshould the cat have started the count in order that the white mouseshould be the last eaten?

When the reader has solved that little puzzle, here is a second one forhim. What is the smallest number that the cat can count round and roundthe circle, if he must start at the white mouse (calling that "one" inthe count) and still eat the white mouse last of all?

And as a third puzzle try to discover what is the smallest number thatthe cat can count round and round if she must start at the white mouse(calling that "one") and make the white mouse the third eaten. 

233. --THE ECCENTRIC CHEESEMONGER. 

[Illustration]

The cheesemonger depicted in the illustration is an inveterate puzzlelover. One of his favourite puzzles is the piling of cheeses in hiswarehouse, an amusement that he finds good exercise for the body as wellas for the mind. He places sixteen cheeses on the floor in a straightrow and then makes them into four piles, with four cheeses in everypile, by always passing a cheese over four others. If you use sixteencounters and number them in order from 1 to 16, then you may place 1 on6, 11 on 1, 7 on 4, and so on, until there are four in every pile. Itwill be seen that it does not matter whether the four passed over arestanding alone or piled; they count just the same, and you can alwayscarry a cheese in either direction. There are a great many differentways of doing it in twelve moves, so it makes a good game of "patience"to try to solve it so that the four piles shall be left in differentstipulated places. For example, try to leave the piles at the extremeends of the row, on Nos. 1, 2, 15 and 16; this is quite easy. Then tryto leave three piles together, on Nos. 13, 14, and 15. Then again playso that they shall be left on Nos. 3, 5, 12, and 14. 

234. --THE EXCHANGE PUZZLE. 

Here is a rather entertaining little puzzle with moving counters. Youonly need twelve counters--six of one colour, marked A, C, E, G, I, andK, and the other six marked B, D, F, H, J, and L. You first place themon the diagram, as shown in the illustration, and the puzzle is to getthem into regular alphabetical order, as follows:--

 A B C D E F G H I J K L

The moves are made by exchanges of opposite colours standing on the sameline. Thus, G and J may exchange places, or F and A, but you cannotexchange G and C, or F and D, because in one case they are both whiteand in the other case both black. Can you bring about the requiredarrangement in seventeen exchanges?

[Illustration]

It cannot be done in fewer moves. The puzzle is really much easier thanit looks, if properly attacked. 

235. --TORPEDO PRACTICE. 

[Illustration]

If a fleet of sixteen men-of-war were lying at anchor and surrounded bythe enemy, how many ships might be sunk if every torpedo, projected in astraight line, passed under three vessels and sank the fourth? In thediagram we have arranged the fleet in square formation, where it will beseen that as many as seven ships may be sunk (those in the top row andfirst column) by firing the torpedoes indicated by arrows. Anchoring thefleet as we like, to what extent can we increase this number? Rememberthat each successive ship is sunk before another torpedo is launched, and that every torpedo proceeds in a different direction; otherwise, byplacing the ships in a straight line, we might sink as many as thirteen!It is an interesting little study in naval warfare, and eminentlypractical--provided the enemy will allow you to arrange his fleet foryour convenience and promise to lie still and do nothing!

236. --THE HAT PUZZLE. 

Ten hats were hung on pegs as shown in the illustration--five silk hatsand five felt "bowlers, " alternately silk and felt. The two pegs at theend of the row were empty. 

[Illustration]

The puzzle is to remove two contiguous hats to the vacant pegs, then twoother adjoining hats to the pegs now unoccupied, and so on until fivepairs have been moved and the hats again hang in an unbroken row, butwith all the silk ones together and all the felt hats together. 

Remember, the two hats removed must always be contiguous ones, and youmust take one in each hand and place them on their new pegs withoutreversing their relative position. You are not allowed to cross yourhands, nor to hang up one at a time. 

Can you solve this old puzzle, which I give as introductory to the next?Try it with counters of two colours or with coins, and remember that thetwo empty pegs must be left at one end of the row. 

237. --BOYS AND GIRLS. 

If you mark off ten divisions on a sheet of paper to represent thechairs, and use eight numbered counters for the children, you will havea fascinating pastime. Let the odd numbers represent boys and evennumbers girls, or you can use counters of two colours, or coins. 

The puzzle is to remove two children who are occupying adjoining chairsand place them in two empty chairs, _making them first change sides_;then remove a second pair of children from adjoining chairs and placethem in the two now vacant, making them change sides; and so on, untilall the boys are together and all the girls together, with the twovacant chairs at one end as at present. To solve the puzzle you must dothis in five moves. The two children must always be taken from chairsthat are next to one another; and remember the important point of makingthe two children change sides, as this latter is the distinctive featureof the puzzle. By "change sides" I simply mean that if, for example, youfirst move 1 and 2 to the vacant chairs, then the first (the outside)chair will be occupied by 2 and the second one by 1. 

[Illustration]

238. --ARRANGING THE JAMPOTS. 

I happened to see a little girl sorting out some jam in a cupboard forher mother. She was putting each different kind of preserve apart on theshelves. I noticed that she took a pot of damson in one hand and a potof gooseberry in the other and made them change places; then she changeda strawberry with a raspberry, and so on. It was interesting to observewhat a lot of unnecessary trouble she gave herself by making moreinterchanges than there was any need for, and I thought it would workinto a good puzzle. 

It will be seen in the illustration that little Dorothy has tomanipulate twenty-four large jampots in as many pigeon-holes. She wantsto get them in correct numerical order--that is, 1, 2, 3, 4, 5, 6 on thetop shelf, 7, 8, 9, 10, 11, 12 on the next shelf, and so on. Now, if shealways takes one pot in the right hand and another in the left and makesthem change places, how many of these interchanges will be necessary toget all the jampots in proper order? She would naturally first changethe 1 and the 3, then the 2 and the 3, when she would have the firstthree pots in their places. How would you advise her to go on then?Place some numbered counters on a sheet of paper divided into squaresfor the pigeon-holes, and you will find it an amusing puzzle. 

[Illustration]

UNICURSAL AND ROUTE PROBLEMS. 

 "I see them on their winding way. " REGINALD HEBER. 

It is reasonable to suppose that from the earliest ages one man hasasked another such questions as these: "Which is the nearest way home?""Which is the easiest or pleasantest way?" "How can we find a way thatwill enable us to dodge the mastodon and the plesiosaurus?" "How can weget there without ever crossing the track of the enemy?" All these areelementary route problems, and they can be turned into good puzzles bythe introduction of some conditions that complicate matters. A varietyof such complications will be found in the following examples. I havealso included some enumerations of more or less difficulty. These affordexcellent practice for the reasoning faculties, and enable one togeneralize in the case of symmetrical forms in a manner that is mostinstructive. 

239. --A JUVENILE PUZZLE. 

For years I have been perpetually consulted by my juvenile friends aboutthis little puzzle. Most children seem to know it, and yet, curiouslyenough, they are invariably unacquainted with the answer. The questionthey always ask is, "Do, please, tell me whether it is really possible. "I believe Houdin the conjurer used to be very fond of giving it to hischild friends, but I cannot say whether he invented the little puzzle ornot. No doubt a large number of my readers will be glad to have themystery of the solution cleared up, so I make no apology for introducingthis old "teaser. "

The puzzle is to draw with three strokes of the pencil the diagram thatthe little girl is exhibiting in the illustration. Of course, you mustnot remove your pencil from the paper during a stroke or go over thesame line a second time. You will find that you can get in a good dealof the figure with one continuous stroke, but it will always appear asif four strokes are necessary. 

[Illustration]

Another form of the puzzle is to draw the diagram on a slate and thenrub it out in three rubs. 

240. --THE UNION JACK. 

[Illustration]

The illustration is a rough sketch somewhat resembling the British flag, the Union Jack. It is not possible to draw the whole of it withoutlifting the pencil from the paper or going over the same line twice. Thepuzzle is to find out just _how much_ of the drawing it is possible tomake without lifting your pencil or going twice over the same line. Takeyour pencil and see what is the best you can do. 

241. --THE DISSECTED CIRCLE. 

How many continuous strokes, without lifting your pencil from the paper, do you require to draw the design shown in our illustration? Directlyyou change the direction of your pencil it begins a new stroke. You maygo over the same line more than once if you like. It requires just alittle care, or you may find yourself beaten by one stroke. 

[Illustration]

242. --THE TUBE INSPECTOR'S PUZZLE. 

The man in our illustration is in a little dilemma. He has just beenappointed inspector of a certain system of tube railways, and it is hisduty to inspect regularly, within a stated period, all the company'sseventeen lines connecting twelve stations, as shown on the big posterplan that he is contemplating. Now he wants to arrange his route so thatit shall take him over all the lines with as little travelling aspossible. He may begin where he likes and end where he likes. What ishis shortest route?

Could anything be simpler? But the reader will soon find that, howeverhe decides to proceed, the inspector must go over some of the lines morethan once. In other words, if we say that the stations are a mile apart, he will have to travel more than seventeen miles to inspect every line. There is the little difficulty. How far is he compelled to travel, andwhich route do you recommend?

[Illustration]

243. --VISITING THE TOWNS. 

[Illustration]

A traveller, starting from town No. 1, wishes to visit every one of thetowns once, and once only, going only by roads indicated by straightlines. How many different routes are there from which he can select? Ofcourse, he must end his journey at No. 1, from which he started, andmust take no notice of cross roads, but go straight from town to town. This is an absurdly easy puzzle, if you go the right way to work. 

244. --THE FIFTEEN TURNINGS. 

Here is another queer travelling puzzle, the solution of which calls foringenuity. In this case the traveller starts from the black town andwishes to go as far as possible while making only fifteen turnings andnever going along the same road twice. The towns are supposed to be amile apart. Supposing, for example, that he went straight to A, thenstraight to B, then to C, D, E, and F, you will then find that he hastravelled thirty-seven miles in five turnings. Now, how far can he go infifteen turnings?

[Illustration]

245. --THE FLY ON THE OCTAHEDRON. 

"Look here, " said the professor to his colleague, "I have been watchingthat fly on the octahedron, and it confines its walks entirely to theedges. What can be its reason for avoiding the sides?"

"Perhaps it is trying to solve some route problem, " suggested the other. "Supposing it to start from the top point, how many different routes arethere by which it may walk over all the edges, without ever going twicealong the same edge in any route?"

[Illustration]

The problem was a harder one than they expected, and after working at itduring leisure moments for several days their results did not agree--infact, they were both wrong. If the reader is surprised at their failure, let him attempt the little puzzle himself. I will just explain that theoctahedron is one of the five regular, or Platonic, bodies, and iscontained under eight equal and equilateral triangles. If you cut outthe two pieces of cardboard of the shape shown in the margin of theillustration, cut half through along the dotted lines and then bend themand put them together, you will have a perfect octahedron. In any routeover all the edges it will be found that the fly must end at the pointof departure at the top. 

246. --THE ICOSAHEDRON PUZZLE. 

The icosahedron is another of the five regular, or Platonic, bodieshaving all their sides, angles, and planes similar and equal. It isbounded by twenty similar equilateral triangles. If you cut out a pieceof cardboard of the form shown in the smaller diagram, and cut halfthrough along the dotted lines, it will fold up and form a perfecticosahedron. 

Now, a Platonic body does not mean a heavenly body; but it will suit thepurpose of our puzzle if we suppose there to be a habitable planet ofthis shape. We will also suppose that, owing to a superfluity of water, the only dry land is along the edges, and that the inhabitants have noknowledge of navigation. If every one of those edges is 10, 000 mileslong and a solitary traveller is placed at the North Pole (the highestpoint shown), how far will he have to travel before he will have visitedevery habitable part of the planet--that is, have traversed every one ofthe edges?

[Illustration]

247. --INSPECTING A MINE. 

The diagram is supposed to represent the passages or galleries in amine. We will assume that every passage, A to B, B to C, C to H, H to I, and so on, is one furlong in length. It will be seen that there arethirty-one of these passages. Now, an official has to inspect all ofthem, and he descends by the shaft to the point A. How far must hetravel, and what route do you recommend? The reader may at first say, "As there are thirty-one passages, each a furlong in length, he willhave to travel just thirty-one furlongs. " But this is assuming that heneed never go along a passage more than once, which is not the case. Take your pencil and try to find the shortest route. You will soondiscover that there is room for considerable judgment. In fact, it is aperplexing puzzle. 

[Illustration]

248. --THE CYCLISTS' TOUR. 

Two cyclists were consulting a road map in preparation for a little tourtogether. The circles represent towns, and all the good roads arerepresented by lines. They are starting from the town with a star, andmust complete their tour at E. But before arriving there they want tovisit every other town once, and only once. That is the difficulty. Mr. Spicer said, "I am certain we can find a way of doing it;" but Mr. Maggsreplied, "No way, I'm sure. " Now, which of them was correct? Take yourpencil and see if you can find any way of doing it. Of course you mustkeep to the roads indicated. 

[Illustration]

249. --THE SAILOR'S PUZZLE. 

The sailor depicted in the illustration stated that he had since hisboyhood been engaged in trading with a small vessel among some twentylittle islands in the Pacific. He supplied the rough chart of which Ihave given a copy, and explained that the lines from island to islandrepresented the only routes that he ever adopted. He always started fromisland A at the beginning of the season, and then visited every islandonce, and once only, finishing up his tour at the starting-point A. Buthe always put off his visit to C as long as possible, for trade reasonsthat I need not enter into. The puzzle is to discover his exact route, and this can be done with certainty. Take your pencil and, starting atA, try to trace it out. If you write down the islands in the order inwhich you visit them--thus, for example, A, I, O, L, G, etc. --you can atonce see if you have visited an island twice or omitted any. Of course, the crossings of the lines must be ignored--that is, you must continueyour route direct, and you are not allowed to switch off at a crossingand proceed in another direction. There is no trick of this kind in thepuzzle. The sailor knew the best route. Can you find it?

[Illustration]

250. --THE GRAND TOUR. 

One of the everyday puzzles of life is the working out of routes. If youare taking a holiday on your bicycle, or a motor tour, there alwaysarises the question of how you are to make the best of your time andother resources. You have determined to get as far as some particularplace, to include visits to such-and-such a town, to try to seesomething of special interest elsewhere, and perhaps to try to look upan old friend at a spot that will not take you much out of your way. Then you have to plan your route so as to avoid bad roads, uninterestingcountry, and, if possible, the necessity of a return by the same waythat you went. With a map before you, the interesting puzzle is attackedand solved. I will present a little poser based on these lines. 

I give a rough map of a country--it is not necessary to say whatparticular country--the circles representing towns and the dotted linesthe railways connecting them. Now there lived in the town marked A a manwho was born there, and during the whole of his life had never once lefthis native place. From his youth upwards he had been very industrious, sticking incessantly to his trade, and had no desire whatever to roamabroad. However, on attaining his fiftieth birthday he decided to seesomething of his country, and especially to pay a visit to a very oldfriend living at the town marked Z. What he proposed was this: that hewould start from his home, enter every town once and only once, andfinish his journey at Z. As he made up his mind to perform this grandtour by rail only, he found it rather a puzzle to work out his route, but he at length succeeded in doing so. How did he manage it? Do notforget that every town has to be visited once, and not more than once. 

[Illustration]

251. --WATER, GAS, AND ELECTRICITY. 

There are some half-dozen puzzles, as old as the hills, that areperpetually cropping up, and there is hardly a month in the year thatdoes not bring inquiries as to their solution. Occasionally one ofthese, that one had thought was an extinct volcano, bursts into eruptionin a surprising manner. I have received an extraordinary number ofletters respecting the ancient puzzle that I have called "Water, Gas, and Electricity. " It is much older than electric lighting, or even gas, but the new dress brings it up to date. The puzzle is to lay on water, gas, and electricity, from W, G, and E, to each of the three houses, A, B, and C, without any pipe crossing another. Take your pencil and drawlines showing how this should be done. You will soon find yourselflanded in difficulties. 

[Illustration]

252. --A PUZZLE FOR MOTORISTS. 

[Illustration]

Eight motorists drove to church one morning. Their respective housesand churches, together with the only roads available (the dotted lines), are shown. One went from his house A to his church A, another from hishouse B to his church B, another from C to C, and so on, but it wasafterwards found that no driver ever crossed the track of another car. Take your pencil and try to trace out their various routes. 

253. --A BANK HOLIDAY PUZZLE. 

Two friends were spending their bank holiday on a cycling trip. Stoppingfor a rest at a village inn, they consulted a route map, which isrepresented in our illustration in an exceedingly simplified form, forthe puzzle is interesting enough without all the original complexities. They started from the town in the top left-hand corner marked A. It willbe seen that there are one hundred and twenty such towns, all connectedby straight roads. Now they discovered that there are exactly 1, 365different routes by which they may reach their destination, alwaystravelling either due south or due east. The puzzle is to discover whichtown is their destination. 

[Illustration]

Of course, if you find that there are more than 1, 365 different routesto a town it cannot be the right one. 

254. --THE MOTOR-CAR TOUR. 

[Illustration]

In the above diagram the circles represent towns and the lines goodroads. In just how many different ways can a motorist, starting fromLondon (marked with an L), make a tour of all these towns, visitingevery town once, and only once, on a tour, and always coming back toLondon on the last ride? The exact reverse of any route is not countedas different. 

255. --THE LEVEL PUZZLE. 

[Illustration]

This is a simple counting puzzle. In how many different ways can youspell out the word LEVEL by placing the point of your pencil on an L andthen passing along the lines from letter to letter. You may go in anydirection, backwards or forwards. Of course you are not allowed to missletters--that is to say, if you come to a letter you must use it. 

256. --THE DIAMOND PUZZLE. 

IN how many different ways may the word DIAMOND be read in thearrangement shown? You may start wherever you like at a D and go up ordown, backwards or forwards, in and out, in any direction you like, solong as you always pass from one letter to another that adjoins it. Howmany ways are there?

[Illustration]

257. --THE DEIFIED PUZZLE. 

In how many different ways may the word DEIFIED be read in thisarrangement under the same conditions as in the last puzzle, with theaddition that you can use any letters twice in the same reading?

[Illustration]

258. --THE VOTERS' PUZZLE. 

[Illustration]

Here we have, perhaps, the most interesting form of the puzzle. In howmany different ways can you read the political injunction, "RISE TOVOTE, SIR, " under the same conditions as before? In this case everyreading of the palindrome requires the use of the central V as themiddle letter. 

259. --HANNAH'S PUZZLE. 

A man was in love with a young lady whose Christian name was Hannah. When he asked her to be his wife she wrote down the letters of her namein this manner:--

 H H H H H H H A A A A H H A N N A H H A N N A H H A A A A H H H H H H H

and promised that she would be his if he could tell her correctly in howmany different ways it was possible to spell out her name, alwayspassing from one letter to another that was adjacent. Diagonal steps arehere allowed. Whether she did this merely to tease him or to test hiscleverness is not recorded, but it is satisfactory to know that hesucceeded. Would you have been equally successful? Take your pencil andtry. You may start from any of the H's and go backwards or forwards andin any direction, so long as all the letters in a spelling are adjoiningone another. How many ways are there, no two exactly alike?

260. --THE HONEYCOMB PUZZLE. 

[Illustration]

Here is a little puzzle with the simplest possible conditions. Place thepoint of your pencil on a letter in one of the cells of the honeycomb, and trace out a very familiar proverb by passing always from a cell toone that is contiguous to it. If you take the right route you will havevisited every cell once, and only once. The puzzle is much easier thanit looks. 

261. --THE MONK AND THE BRIDGES. 

In this case I give a rough plan of a river with an island and fivebridges. On one side of the river is a monastery, and on the other sideis seen a monk in the foreground. Now, the monk has decided that he willcross every bridge once, and only once, on his return to the monastery. This is, of course, quite easy to do, but on the way he thought tohimself, "I wonder how many different routes there are from which Imight have selected. " Could you have told him? That is the puzzle. Takeyour pencil and trace out a route that will take you once over all thefive bridges. Then trace out a second route, then a third, and see ifyou can count all the variations. You will find that the difficulty istwofold: you have to avoid dropping routes on the one hand and countingthe same routes more than once on the other. 

[Illustration]

COMBINATION AND GROUP PROBLEMS. 

 "A combination and a form indeed. " _Hamlet_, iii. 4. 

Various puzzles in this class might be termed problems in the "geometryof situation, " but their solution really depends on the theory ofcombinations which, in its turn, is derived directly from the theory ofpermutations. It has seemed convenient to include here certain grouppuzzles and enumerations that might, perhaps, with equal reason havebeen placed elsewhere; but readers are again asked not to be toocritical about the classification, which is very difficult andarbitrary. As I have included my problem of "The Round Table" (No. 273), perhaps a few remarks on another well-known problem of the same class, known by the French as La Problême des Ménages, may be interesting. Ifn married ladies are seated at a round table in any determined order, in how many different ways may their n husbands be placed so thatevery man is between two ladies but never next to his own wife?

This difficult problem was first solved by Laisant, and the method shownin the following table is due to Moreau:--

 4 0 2 5 3 13 6 13 80 7 83 579 8 592 4738 9 4821 43387 10 43979 439792

The first column shows the number of married couples. The numbers in thesecond column are obtained in this way: 5 × 3 + 0 - 2 = 13; 6 × 13 + 3 +2 = 83; 7 × 83 + 13 - 2 = 592; 8 × 592 + 83 + 2 = 4821; and so on. Findall the numbers, except 2, in the table, and the method will be evident. It will be noted that the 2 is subtracted when the first number (thenumber of couples) is odd, and added when that number is even. Thenumbers in the third column are obtained thus: 13 - 0 = 13; 83 - 3 = 80;592 - 13 = 579; 4821 - 83 = 4738; and so on. The numbers in this lastcolumn give the required solutions. Thus, four husbands may be seated intwo ways, five husbands may be placed in thirteen ways, and six husbandsin eighty ways. 

The following method, by Lucas, will show the remarkable way in whichchessboard analysis may be applied to the solution of a circular problemof this kind. Divide a square into thirty-six cells, six by six, andstrike out all the cells in the long diagonal from the bottom left-handcorner to the top right-hand corner, also the five cells in the diagonalnext above it and the cell in the bottom right-hand corner. The answerfor six couples will be the same as the number of ways in which you canplace six rooks (not using the cancelled cells) so that no rook shallever attack another rook. It will be found that the six rooks may beplaced in eighty different ways, which agrees with the above table. 

262. --THOSE FIFTEEN SHEEP. 

A certain cyclopædia has the following curious problem, I am told:"Place fifteen sheep in four pens so that there shall be the same numberof sheep in each pen. " No answer whatever is vouchsafed, so I thought Iwould investigate the matter. I saw that in dealing with apples orbricks the thing would appear to be quite impossible, since four timesany number must be an even number, while fifteen is an odd number. Ithought, therefore, that there must be some quality peculiar to thesheep that was not generally known. So I decided to interview somefarmers on the subject. The first one pointed out that if we put one peninside another, like the rings of a target, and placed all sheep in thesmallest pen, it would be all right. But I objected to this, because youadmittedly place all the sheep in one pen, not in four pens. The secondman said that if I placed four sheep in each of three pens and threesheep in the last pen (that is fifteen sheep in all), and one of theewes in the last pen had a lamb during the night, there would be thesame number in each pen in the morning. This also failed to satisfy me. 

[Illustration]

The third farmer said, "I've got four hurdle pens down in one of myfields, and a small flock of wethers, so if you will just step down withme I will show you how it is done. " The illustration depicts my friendas he is about to demonstrate the matter to me. His lucid explanationwas evidently that which was in the mind of the writer of the article inthe cyclopædia. What was it? Can you place those fifteen sheep?

263. --KING ARTHUR'S KNIGHTS. 

King Arthur sat at the Round Table on three successive evenings with hisknights--Beleobus, Caradoc, Driam, Eric, Floll, and Galahad--but on nooccasion did any person have as his neighbour one who had before satnext to him. On the first evening they sat in alphabetical order roundthe table. But afterwards King Arthur arranged the two next sittings sothat he might have Beleobus as near to him as possible and Galahad asfar away from him as could be managed. How did he seat the knights tothe best advantage, remembering that rule that no knight may have thesame neighbour twice?

264. --THE CITY LUNCHEONS. 

Twelve men connected with a large firm in the City of London sit down toluncheon together every day in the same room. The tables are small onesthat only accommodate two persons at the same time. Can you show howthese twelve men may lunch together on eleven days in pairs, so that notwo of them shall ever sit twice together? We will represent the men bythe first twelve letters of the alphabet, and suppose the first day'spairing to be as follows--

 (A B) (C D) (E F) (G H) (I J) (K L). 

Then give any pairing you like for the next day, say--

 (A C) (B D) (E G) (F H) (I K) (J L), and so on, until you have completed your eleven lines, with no pair everoccurring twice. There are a good many different arrangements possible. Try to find one of them. 

265. --A PUZZLE FOR CARD-PLAYERS. 

Twelve members of a club arranged to play bridge together on elevenevenings, but no player was ever to have the same partner more thanonce, or the same opponent more than twice. Can you draw up a schemeshowing how they may all sit down at three tables every evening? Callthe twelve players by the first twelve letters of the alphabet and tryto group them. 

266. --A TENNIS TOURNAMENT. 

Four married couples played a "mixed double" tennis tournament, a manand a lady always playing against a man and a lady. But no person everplayed with or against any other person more than once. Can you show howthey all could have played together in the two courts on threesuccessive days? This is a little puzzle of a quite practical kind, andit is just perplexing enough to be interesting. 

267. --THE WRONG HATS. 

"One of the most perplexing things I have come across lately, " said Mr. Wilson, "is this. Eight men had been dining not wisely but too well at acertain London restaurant. They were the last to leave, but not one manwas in a condition to identify his own hat. Now, considering that theytook their hats at random, what are the chances that every man took ahat that did not belong to him?"

"The first thing, " said Mr. Waterson, "is to see in how many differentways the eight hats could be taken. "

"That is quite easy, " Mr. Stubbs explained. "Multiply together thenumbers, 1, 2, 3, 4, 5, 6, 7, and 8. Let me see--half a minute--yes;there are 40, 320 different ways. "

"Now all you've got to do is to see in how many of these cases no manhas his own hat, " said Mr. Waterson. 

"Thank you, I'm not taking any, " said Mr. Packhurst. "I don't envy theman who attempts the task of writing out all those forty-thousand-oddcases and then picking out the ones he wants. "

They all agreed that life is not long enough for that sort of amusement;and as nobody saw any other way of getting at the answer, the matter waspostponed indefinitely. Can you solve the puzzle?

268. --THE PEAL OF BELLS. 

A correspondent, who is apparently much interested in campanology, asksme how he is to construct what he calls a "true and correct" peal forfour bells. He says that every possible permutation of the four bellsmust be rung once, and once only. He adds that no bell must move morethan one place at a time, that no bell must make more than twosuccessive strokes in either the first or the last place, and that thelast change must be able to pass into the first. These fantasticconditions will be found to be observed in the little peal for threebells, as follows:--

 1 2 3 2 1 3 2 3 1 3 2 1 3 1 2 1 3 2

How are we to give him a correct solution for his four bells?

269. --THREE MEN IN A BOAT. 

A certain generous London manufacturer gives his workmen every year aweek's holiday at the seaside at his own expense. One year fifteen ofhis men paid a visit to Herne Bay. On the morning of their departurefrom London they were addressed by their employer, who expressed thehope that they would have a very pleasant time. 

"I have been given to understand, " he added, "that some of you fellowsare very fond of rowing, so I propose on this occasion to provide youwith this recreation, and at the same time give you an amusing littlepuzzle to solve. During the seven days that you are at Herne Bay everyone of you will go out every day at the same time for a row, but theremust always be three men in a boat and no more. No two men may ever goout in a boat together more than once, and no man is allowed to go outtwice in the same boat. If you can manage to do this, and use as fewdifferent boats as possible, you may charge the firm with the expense. "

One of the men tells me that the experience he has gained in suchmatters soon enabled him to work out the answer to the entiresatisfaction of themselves and their employer. But the amusing part ofthe thing is that they never really solved the little mystery. I findtheir method to have been quite incorrect, and I think it will amuse myreaders to discover how the men should have been placed in the boats. Astheir names happen to have been Andrews, Baker, Carter, Danby, Edwards, Frith, Gay, Hart, Isaacs, Jackson, Kent, Lang, Mason, Napper, andOnslow, we can call them by their initials and write out the five groupsfor each of the seven days in the following simple way:

 1 2 3 4 5 First Day: (ABC) (DEF) (GHI) (JKL) (MNO). 

The men within each pair of brackets are here seen to be in the sameboat, and therefore A can never go out with B or with C again, and C cannever go out again with B. The same applies to the other four boats. Thefigures show the number on the boat, so that A, B, or C, for example, can never go out in boat No. 1 again. 

270. --THE GLASS BALLS. 

A number of clever marksmen were staying at a country house, and thehost, to provide a little amusement, suspended strings of glass balls, as shown in the illustration, to be fired at. After they had all puttheir skill to a sufficient test, somebody asked the following question:"What is the total number of different ways in which these sixteen ballsmay be broken, if we must always break the lowest ball that remains onany string?" Thus, one way would be to break all the four balls on eachstring in succession, taking the strings from left to right. Anotherwould be to break all the fourth balls on the four strings first, thenbreak the three remaining on the first string, then take the balls onthe three other strings alternately from right to left, and so on. Thereis such a vast number of different ways (since every little variation oforder makes a different way) that one is apt to be at first impressed bythe great difficulty of the problem. Yet it is really quite simple whenonce you have hit on the proper method of attacking it. How manydifferent ways are there?

[Illustration]

271. --FIFTEEN LETTER PUZZLE. 

 ALE FOE HOD BGN CAB HEN JOG KFM HAG GEM MOB BFH FAN KIN JEK DFL JAM HIM GCL LJH AID JIB FCJ NJD OAK FIG HCK MLN BED OIL MCD BLK ICE CON DGK

The above is the solution of a puzzle I gave in _Tit-bits_ in the summerof 1896. It was required to take the letters, A, B, C, D, E, F, G, H, I, J, K, L, M, N, and O, and with them form thirty-five groups of threeletters so that the combinations should include the greatest numberpossible of common English words. No two letters may appear together ina group more than once. Thus, A and L having been together in ALE, mustnever be found together again; nor may A appear again in a group with E, nor L with E. These conditions will be found complied with in the abovesolution, and the number of words formed is twenty-one. Many personshave since tried hard to beat this number, but so far have notsucceeded. 

More than thirty-five combinations of the fifteen letters cannot beformed within the conditions. Theoretically, there cannot possibly bemore than twenty-three words formed, because only this number ofcombinations is possible with a vowel or vowels in each. And as noEnglish word can be formed from three of the given vowels (A, E, I, andO), we must reduce the number of possible words to twenty-two. This iscorrect theoretically, but practically that twenty-second word cannot begot in. If JEK, shown above, were a word it would be all right; but itis not, and no amount of juggling with the other letters has resulted ina better answer than the one shown. I should, say that proper nouns andabbreviations, such as Joe, Jim, Alf, Hal, Flo, Ike, etc. , aredisallowed. 

Now, the present puzzle is a variation of the above. It is simply this:Instead of using the fifteen letters given, the reader is allowed toselect any fifteen different letters of the alphabet that he may prefer. Then construct thirty-five groups in accordance with the conditions, andshow as many good English words as possible. 

272. --THE NINE SCHOOLBOYS. 

This is a new and interesting companion puzzle to the "FifteenSchoolgirls" (see solution of No. 269), and even in the simplestpossible form in which I present it there are unquestionabledifficulties. Nine schoolboys walk out in triplets on the six week daysso that no boy ever walks _side by side_ with any other boy more thanonce. How would you arrange them?

If we represent them by the first nine letters of the alphabet, theymight be grouped on the first day as follows:--

 A B C D E F G H I

Then A can never walk again side by side with B, or B with C, or D withE, and so on. But A can, of course, walk side by side with C. It is herenot a question of being together in the same triplet, but of walkingside by side in a triplet. Under these conditions they can walk out onsix days; under the "Schoolgirls" conditions they can only walk on fourdays. 

273. --THE ROUND TABLE. 

Seat the same n persons at a round table on

 (n - 1)(n - 2) -------------- 2

occasions so that no person shall ever have the same two neighbourstwice. This is, of course, equivalent to saying that every person mustsit once, and once only, between every possible pair. 

274. --THE MOUSE-TRAP PUZZLE. 

[Illustration

 6 20 2 19 13 21 7 5 3 18 17 8 15 11 14 16 1 9 10 4 12

]

This is a modern version, with a difference, of an old puzzle of thesame name. Number twenty-one cards, 1, 2, 3, etc. , up to 21, and placethem in a circle in the particular order shown in the illustration. These cards represent mice. You start from any card, calling that card"one, " and count, "one, two, three, " etc. , in a clockwise direction, andwhen your count agrees with the number on the card, you have made a"catch, " and you remove the card. Then start at the next card, callingthat "one, " and try again to make another "catch. " And so on. Supposingyou start at 18, calling that card "one, " your first "catch" will be 19. Remove 19 and your next "catch" is 10. Remove 10 and your next "catch"is 1. Remove the 1, and if you count up to 21 (you must never gobeyond), you cannot make another "catch. " Now, the ideal is to "catch"all the twenty-one mice, but this is not here possible, and if it wereit would merely require twenty-one different trials, at the most, tosucceed. But the reader may make any two cards change places before hebegins. Thus, you can change the 6 with the 2, or the 7 with the 11, orany other pair. This can be done in several ways so as to enable you to"catch" all the twenty-one mice, if you then start at the right place. You may never pass over a "catch"; you must always remove the card andstart afresh. 

275. --THE SIXTEEN SHEEP. 

[Illustration:

 +========================+ || | | | || || 0 | 0 | 0 | 0 || +-----+-----+-----+------+ || | | | || || 0 | 0 | 0 | 0 || +========================+ || || | || || || 0 || 0 | 0 || 0 || +-----+=====+=====+------+ || | || | || || 0 | 0 || 0 | 0 || +========================+

]

Here is a new puzzle with matches and counters or coins. In theillustration the matches represent hurdles and the counters sheep. Thesixteen hurdles on the outside, and the sheep, must be regarded asimmovable; the puzzle has to do entirely with the nine hurdles on theinside. It will be seen that at present these nine hurdles enclose fourgroups of 8, 3, 3, and 2 sheep. The farmer requires to readjust some ofthe hurdles so as to enclose 6, 6, and 4 sheep. Can you do it by onlyreplacing two hurdles? When you have succeeded, then try to do it byreplacing three hurdles; then four, five, six, and seven in succession. Of course, the hurdles must be legitimately laid on the dotted lines, and no such tricks are allowed as leaving unconnected ends of hurdles, or two hurdles placed side by side, or merely making hurdles changeplaces. In fact, the conditions are so simple that any farm labourerwill understand it directly. 

276. --THE EIGHT VILLAS. 

In one of the outlying suburbs of London a man had a square plot ofground on which he decided to build eight villas, as shown in theillustration, with a common recreation ground in the middle. After thehouses were completed, and all or some of them let, he discovered thatthe number of occupants in the three houses forming a side of the squarewas in every case nine. He did not state how the occupants weredistributed, but I have shown by the numbers on the sides of the housesone way in which it might have happened. The puzzle is to discover thetotal number of ways in which all or any of the houses might beoccupied, so that there should be nine persons on each side. In orderthat there may be no misunderstanding, I will explain that although B iswhat we call a reflection of A, these would count as two differentarrangements, while C, if it is turned round, will give fourarrangements; and if turned round in front of a mirror, four otherarrangements. All eight must be counted. 

[Illustration:

 /\ /\ /\ |2 | |5 | |2 |

 /\ /\ |5 | |5 |

 /\ /\ /\ |2 | |5 | |2 |

 +---+---+---+ +---+---+---+ +---+---+---+ | 1 | 6 | 2 | | 2 | 6 | 1 | | 1 | 6 | 2 | +---+---+---+ +---+---+---+ +---+---+---+ | 6 | | 6 | | 6 | | 6 | | 4 | | 4 | +---+---+---+ +---+---+---+ +---+---+---+ | 2 | 6 | 1 | | 1 | 6 | 2 | | 4 | 2 | 3 | +---+---+---+ +---+---+---+ +---+---+---+ A B C

]

277. --COUNTER CROSSES. 

All that we need for this puzzle is nine counters, numbered 1, 2, 3, 4, 5, 6, 7, 8, and 9. It will be seen that in the illustration A these arearranged so as to form a Greek cross, while in the case of B they form aLatin cross. In both cases the reader will find that the sum of thenumbers in the upright of the cross is the same as the sum of thenumbers in the horizontal arm. It is quite easy to hit on such anarrangement by trial, but the problem is to discover in exactly how manydifferent ways it may be done in each case. Remember that reversals andreflections do not count as different. That is to say, if you turn thispage round you get four arrangements of the Greek cross, and if you turnit round again in front of a mirror you will get four more. But theseeight are all regarded as one and the same. Now, how many different waysare there in each case?

[Illustration:

 (1) (2)

 (2) (4) (5) (1) (6) (7)

 (3) (4) (9) (5) (6) (3)

 (7) (8)

 A (8) B (9)

]

278. --A DORMITORY PUZZLE. 

In a certain convent there were eight large dormitories on one floor, approached by a spiral staircase in the centre, as shown in our plan. Onan inspection one Monday by the abbess it was found that the southaspect was so much preferred that six times as many nuns slept on thesouth side as on each of the other three sides. She objected to thisovercrowding, and ordered that it should be reduced. On Tuesday shefound that five times as many slept on the south side as on each of theother sides. Again she complained. On Wednesday she found four times asmany on the south side, on Thursday three times as many, and on Fridaytwice as many. Urging the nuns to further efforts, she was pleased tofind on Saturday that an equal number slept on each of the four sides ofthe house. What is the smallest number of nuns there could have been, and how might they have arranged themselves on each of the six nights?No room may ever be unoccupied. 

[Illustration

 +---+---+---+ | | | | | | | | | | | | +---+---+---+ | |\|/| | | |-*-| | | |/|\| | +---+---+---+ | | | | | | | | | | | | +---+---+---+

]

279. --THE BARRELS OF BALSAM. 

A merchant of Bagdad had ten barrels of precious balsam for sale. Theywere numbered, and were arranged in two rows, one on top of the other, as shown in the picture. The smaller the number on the barrel, thegreater was its value. So that the best quality was numbered "1" and theworst numbered "10, " and all the other numbers of graduating values. Now, the rule of Ahmed Assan, the merchant, was that he never put abarrel either beneath or to the right of one of less value. Thearrangement shown is, of course, the simplest way of complying with thiscondition. But there are many other ways--such, for example, as this:--

 1 2 5 7 8 3 4 6 9 10

Here, again, no barrel has a smaller number than itself on its right orbeneath it. The puzzle is to discover in how many different ways themerchant of Bagdad might have arranged his barrels in the two rowswithout breaking his rule. Can you count the number of ways?

280. --BUILDING THE TETRAHEDRON. 

I possess a tetrahedron, or triangular pyramid, formed of six sticksglued together, as shown in the illustration. Can you count correctlythe number of different ways in which these six sticks might have beenstuck together so as to form the pyramid?

Some friends worked at it together one evening, each person providinghimself with six lucifer matches to aid his thoughts; but it was foundthat no two results were the same. You see, if we remove one of thesticks and turn it round the other way, that will be a differentpyramid. If we make two of the sticks change places the result willagain be different. But remember that every pyramid may be made to standon either of its four sides without being a different one. How many waysare there altogether?

[Illustration]

281. --PAINTING A PYRAMID. 

This puzzle concerns the painting of the four sides of a tetrahedron, ortriangular pyramid. If you cut out a piece of cardboard of thetriangular shape shown in Fig. 1, and then cut half through along thedotted lines, it will fold up and form a perfect triangular pyramid. AndI would first remind my readers that the primary colours of the solarspectrum are seven--violet, indigo, blue, green, yellow, orange, andred. When I was a child I was taught to remember these by the ungainlyword formed by the initials of the colours, "Vibgyor. "

[Illustration]

In how many different ways may the triangular pyramid be coloured, usingin every case one, two, three, or four colours of the solar spectrum? Ofcourse a side can only receive a single colour, and no side can be leftuncoloured. But there is one point that I must make quite clear. Thefour sides are not to be regarded as individually distinct. That is tosay, if you paint your pyramid as shown in Fig. 2 (where the bottom sideis green and the other side that is out of view is yellow), and thenpaint another in the order shown in Fig. 3, these are really both thesame and count as one way. For if you tilt over No. 2 to the right itwill so fall as to represent No. 3. The avoidance of repetitions of thiskind is the real puzzle of the thing. If a coloured pyramid cannot beplaced so that it exactly resembles in its colours and their relativeorder another pyramid, then they are different. Remember that one waywould be to colour all the four sides red, another to colour two sidesgreen, and the remaining sides yellow and blue; and so on. 

282. --THE ANTIQUARY'S CHAIN. 

An antiquary possessed a number of curious old links, which he took to ablacksmith, and told him to join together to form one straight piece ofchain, with the sole condition that the two circular links were not tobe together. The following illustration shows the appearance of thechain and the form of each link. Now, supposing the owner shouldseparate the links again, and then take them to another smith and repeathis former instructions exactly, what are the chances against the linksbeing put together exactly as they were by the first man? Remember thatevery successive link can be joined on to another in one of two ways, just as you can put a ring on your finger in two ways, or link yourforefingers and thumbs in two ways. 

[Illustration]

283. --THE FIFTEEN DOMINOES. 

In this case we do not use the complete set of twenty-eight dominoes tobe found in the ordinary box. We dispense with all those dominoes thathave a five or a six on them and limit ourselves to the fifteen thatremain, where the double-four is the highest. 

In how many different ways may the fifteen dominoes be arranged in astraight line in accordance with the simple rule of the game that anumber must always be placed against a similar number--that is, a fouragainst a four, a blank against a blank, and so on? Left to right andright to left of the same arrangement are to be counted as two differentways. 

384. --THE CROSS TARGET. 

 +-+-+ |*|*| +-+-+ |*|*| +-+-+-+-+-+-+ | | | |*| | | +-+-+-+-+-+-+ | | |*| |*| | +-+-+-+-+-+-+ | |*| +-+-+ | | | +-+-+

In the illustration we have a somewhat curious target designed by aneccentric sharpshooter. His idea was that in order to score you must hitfour circles in as many shots so that those four shots shall form asquare. It will be seen by the results recorded on the target that twoattempts have been successful. The first man hit the four circles at thetop of the cross, and thus formed his square. The second man intended tohit the four in the bottom arm, but his second shot, on the left, wenttoo high. This compelled him to complete his four in a different waythan he intended. It will thus be seen that though it is immaterialwhich circle you hit at the first shot, the second shot may commit youto a definite procedure if you are to get your square. Now, the puzzleis to say in just how many different ways it is possible to form asquare on the target with four shots. 

285. --THE FOUR POSTAGE STAMPS. 

 +---+----+----+----+ | 1 | 2 | 3 | 4 | +---+----+----+----+ | 5 | 6 | 7 | 8 | +---+----+----+----+ | 9 | 10 | 11 | 12 | +---+----+----+----+

"It is as easy as counting, " is an expression one sometimes hears. Butmere counting may be puzzling at times. Take the following simpleexample. Suppose you have just bought twelve postage stamps, in thisform--three by four--and a friend asks you to oblige him with fourstamps, all joined together--no stamp hanging on by a mere corner. Inhow many different ways is it possible for you to tear off those fourstamps? You see, you can give him 1, 2, 3, 4, or 2, 3, 6, 7, or 1, 2, 3, 6, or 1, 2, 3, 7, or 2, 3, 4, 8, and so on. Can you count the number ofdifferent ways in which those four stamps might be delivered? There arenot many more than fifty ways, so it is not a big count. Can you get theexact number?

286. --PAINTING THE DIE. 

In how many different ways may the numbers on a single die be marked, with the only condition that the 1 and 6, the 2 and 5, and the 3 and 4must be on opposite sides? It is a simple enough question, and yet itwill puzzle a good many people. 

287. --AN ACROSTIC PUZZLE. 

In the making or solving of double acrostics, has it ever occurred toyou to consider the variety and limitation of the pair of initial andfinal letters available for cross words? You may have to find a wordbeginning with A and ending with B, or A and C, or A and D, and so on. Some combinations are obviously impossible--such, for example, as thosewith Q at the end. But let us assume that a good English word can befound for every case. Then how many possible pairs of letters areavailable?

CHESSBOARD PROBLEMS. 

 "You and I will goe to the chesse. "

 GREENE'S _Groatsworth of Wit. _

During a heavy gale a chimney-pot was hurled through the air, andcrashed upon the pavement just in front of a pedestrian. He quite calmlysaid, "I have no use for it: I do not smoke. " Some readers, when theyhappen to see a puzzle represented on a chessboard with chess pieces, are apt to make the equally inconsequent remark, "I have no use for it:I do not play chess. " This is largely a result of the common, buterroneous, notion that the ordinary chess puzzle with which we arefamiliar in the press (dignified, for some reason, with the name"problem") has a vital connection with the game of chess itself. Butthere is no condition in the game that you shall checkmate your opponentin two moves, in three moves, or in four moves, while the majority ofthe positions given in these puzzles are such that one player would haveso great a superiority in pieces that the other would have resignedbefore the situations were reached. And the solving of them helps youbut little, and that quite indirectly, in playing the game, it beingwell known that, as a rule, the best "chess problemists" are indifferentplayers, and _vice versa_. Occasionally a man will be found strong onboth subjects, but he is the exception to the rule. 

Yet the simple chequered board and the characteristic moves of thepieces lend themselves in a very remarkable manner to the devising ofthe most entertaining puzzles. There is room for such infinite varietythat the true puzzle lover cannot afford to neglect them. It was with aview to securing the interest of readers who are frightened off by themere presentation of a chessboard that so many puzzles of this classwere originally published by me in various fanciful dresses. Some ofthese posers I still retain in their disguised form; others I havetranslated into terms of the chessboard. In the majority of cases thereader will not need any knowledge whatever of chess, but I have thoughtit best to assume throughout that he is acquainted with the terminology, the moves, and the notation of the game. 

I first deal with a few questions affecting the chessboard itself; thenwith certain statical puzzles relating to the Rook, the Bishop, theQueen, and the Knight in turn; then dynamical puzzles with the pieces inthe same order; and, finally, with some miscellaneous puzzles on thechessboard. It is hoped that the formulæ and tables given at the end ofthe statical puzzles will be of interest, as they are, for the mostpart, published for the first time. 

THE CHESSBOARD. 

 "Good company's a chessboard. " BYRON'S _Don Juan_, xiii. 89. 

A chessboard is essentially a square plane divided into sixty-foursmaller squares by straight lines at right angles. Originally it was notchequered (that is, made with its rows and columns alternately black andwhite, or of any other two colours), and this improvement was introducedmerely to help the eye in actual play. The utility of the chequers isunquestionable. For example, it facilitates the operation of thebishops, enabling us to see at the merest glance that our king or pawnson black squares are not open to attack from an opponent's bishoprunning on the white diagonals. Yet the chequering of the board is notessential to the game of chess. Also, when we are propounding puzzles onthe chessboard, it is often well to remember that additional interestmay result from "generalizing" for boards containing any number ofsquares, or from limiting ourselves to some particular chequeredarrangement, not necessarily a square. We will give a few puzzlesdealing with chequered boards in this general way. 

288. --CHEQUERED BOARD DIVISIONS. 

I recently asked myself the question: In how many different ways may achessboard be divided into two parts of the same size and shape by cutsalong the lines dividing the squares? The problem soon proved to be bothfascinating and bristling with difficulties. I present it in asimplified form, taking a board of smaller dimensions. 

[Illustration:

 +---+---*---+---+ +---+---+---*---+ +---+---+---*---+ | | H | | | | | H | | | | H | +---+---*---+---+ +---+---*===*---+ +---*===*---*---+ | | H | | | | H | | | H H H | +---+---*---+---+ +---+---*---+---+ +---*---*---*---+ | | H | | | | H | | | H H H | +---+---*---+---+ +---*===*---+---+ +---*---*===*---+ | | H | | | H | | | | H | | | +---+---*---+---+ +---*---+---+---+ +---*---+---+---+

 +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+ | | | | | | | +---+---+---+---+---+---+

 +---+---*---+---+ +---+---+---*---+ +---+---+---*---+ | | H | | | | | H | | | | H | +---*===*---+---+ +---*===*===*---+ +---+---*===*---+ | H | | | | H | | | | | H | | +---*===*===*---+ +---*===*===*---+ +---+---*---+---+ | | | H | | | | H | | | H | | +---+---*===*---+ +---*===*===*---+ +---*===*---+---+ | | H | | | H | | | | H | | | +---+---*---+---+ +---*---+---+---+ +---*---+---+---+

]

It is obvious that a board of four squares can only be so divided in oneway--by a straight cut down the centre--because we shall not countreversals and reflections as different. In the case of a board ofsixteen squares--four by four--there are just six different ways. I havegiven all these in the diagram, and the reader will not find any others. Now, take the larger board of thirty-six squares, and try to discover inhow many ways it may be cut into two parts of the same size and shape. 

289. --LIONS AND CROWNS. 

The young lady in the illustration is confronted with a littlecutting-out difficulty in which the reader may be glad to assist her. She wishes, for some reason that she has not communicated to me, to cutthat square piece of valuable material into four parts, all of exactlythe same size and shape, but it is important that every piece shallcontain a lion and a crown. As she insists that the cuts can only bemade along the lines dividing the squares, she is considerably perplexedto find out how it is to be done. Can you show her the way? There isonly one possible method of cutting the stuff. 

[Illustration:

 +-+-+-+-+-+-+ | | | | | | | +-+-+-+-+-+-+ | |L|L|L| | | +-+-+-+-+-+-+ | | |C|C| | | +-+-+-+-+-+-+ | | |C|C| | | +-+-+-+-+-+-+ |L| | | | | | +-+-+-+-+-+-+ | | | | | | | +-+-+-+-+-+-+

]

290. --BOARDS WITH AN ODD NUMBER OF SQUARES. 

We will here consider the question of those boards that contain an oddnumber of squares. We will suppose that the central square is first cutout, so as to leave an even number of squares for division. Now, it isobvious that a square three by three can only be divided in one way, asshown in Fig. 1. It will be seen that the pieces A and B are of the samesize and shape, and that any other way of cutting would only produce thesame shaped pieces, so remember that these variations are not counted asdifferent ways. The puzzle I propose is to cut the board five by five(Fig. 2) into two pieces of the same size and shape in as many differentways as possible. I have shown in the illustration one way of doing it. How many different ways are there altogether? A piece which when turnedover resembles another piece is not considered to be of a differentshape. 

[Illustration:

 +---*---+---+ | H | | +---*===*---+ | HHHHH | +---*===*---+ | | H | +---+---*---+

Fig 1]

[Illustration:

 +---+---+---+---+---+ | | | | | | *===*===*===*---+---+ | | | H | | +---+---*===*---+---+ | | HHHHH | | +---+---*===*---+---+ | | H | | | +---+---*===*===*===* | H | | | | +---*---+---+---+---+

Fig 2]

291. --THE GRAND LAMA'S PROBLEM. 

Once upon a time there was a Grand Lama who had a chessboard made ofpure gold, magnificently engraved, and, of course, of great value. Everyyear a tournament was held at Lhassa among the priests, and whenever anyone beat the Grand Lama it was considered a great honour, and his namewas inscribed on the back of the board, and a costly jewel set in theparticular square on which the checkmate had been given. After thissovereign pontiff had been defeated on four occasions he died--possiblyof chagrin. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | * | | | | | | | | +---+---+---+---+---+---+---+---+ | | * | | | | | | | +---+---+---+---+---+---+---+---+ | | | * | | | | | | +---+---+---+---+---+---+---+---+ | | | | * | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+

]

Now the new Grand Lama was an inferior chess-player, and preferred otherforms of innocent amusement, such as cutting off people's heads. So hediscouraged chess as a degrading game, that did not improve either themind or the morals, and abolished the tournament summarily. Then he sentfor the four priests who had had the effrontery to play better than aGrand Lama, and addressed them as follows: "Miserable and heathenishmen, calling yourselves priests! Know ye not that to lay claim to acapacity to do anything better than my predecessor is a capital offence?Take that chessboard and, before day dawns upon the torture chamber, cutit into four equal parts of the same shape, each containing sixteenperfect squares, with one of the gems in each part! If in this you fail, then shall other sports be devised for your special delectation. Go!"The four priests succeeded in their apparently hopeless task. Can youshow how the board may be divided into four equal parts, each ofexactly the same shape, by cuts along the lines dividing the squares, each part to contain one of the gems?

292. --THE ABBOT'S WINDOW. 

[Illustration]

Once upon a time the Lord Abbot of St. Edmondsbury, in consequence of"devotions too strong for his head, " fell sick and was unable to leavehis bed. As he lay awake, tossing his head restlessly from side to side, the attentive monks noticed that something was disturbing his mind; butnobody dared ask what it might be, for the abbot was of a sterndisposition, and never would brook inquisitiveness. Suddenly he calledfor Father John, and that venerable monk was soon at the bedside. 

"Father John, " said the Abbot, "dost thou know that I came into thiswicked world on a Christmas Even?"

The monk nodded assent. 

"And have I not often told thee that, having been born on ChristmasEven, I have no love for the things that are odd? Look there!"

The Abbot pointed to the large dormitory window, of which I give asketch. The monk looked, and was perplexed. 

"Dost thou not see that the sixty-four lights add up an even numbervertically and horizontally, but that all the _diagonal_ lines, exceptfourteen are of a number that is odd? Why is this?"

"Of a truth, my Lord Abbot, it is of the very nature of things, andcannot be changed. "

"Nay, but it _shall_ be changed. I command thee that certain of thelights be closed this day, so that every line shall have an even numberof lights. See thou that this be done without delay, lest the cellars belocked up for a month and other grievous troubles befall thee. "

Father John was at his wits' end, but after consultation with one whowas learned in strange mysteries, a way was found to satisfy the whim ofthe Lord Abbot. Which lights were blocked up, so that those whichremained added up an even number in every line horizontally, vertically, and diagonally, while the least possible obstruction of light wascaused?

293. --THE CHINESE CHESSBOARD. 

Into how large a number of different pieces may the chessboard be cut(by cuts along the lines only), no two pieces being exactly alike?Remember that the arrangement of black and white constitutes adifference. Thus, a single black square will be different from a singlewhite square, a row of three containing two white squares will differfrom a row of three containing two black, and so on. If two piecescannot be placed on the table so as to be exactly alike, they count asdifferent. And as the back of the board is plain, the pieces cannot beturned over. 

294. --THE CHESSBOARD SENTENCE. 

[Illustration]

I once set myself the amusing task of so dissecting an ordinarychessboard into letters of the alphabet that they would form a completesentence. It will be seen from the illustration that the piecesassembled give the sentence, "CUT THY LIFE, " with the stops between. Theideal sentence would, of course, have only one full stop, but that I didnot succeed in obtaining. 

The sentence is an appeal to the transgressor to cut himself adrift fromthe evil life he is living. Can you fit these pieces together to form aperfect chessboard?

STATICAL CHESS PUZZLES. 

 "They also serve who only stand and wait. " MILTON. 

295. --THE EIGHT ROOKS. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | R | R | R | R | R | R | R | R | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+

FIG. 1. ]

[Illustration:

 +---+---+---+---+---+---+---+---+ | R | | | | | | | | +---+---+---+---+---+---+---+---+ | | R | | | | | | | +---+---+---+---+---+---+---+---+ | | | R | | | | | | +---+---+---+---+---+---+---+---+ | | | | R | | | | | +---+---+---+---+---+---+---+---+ | | | | | R | | | | +---+---+---+---+---+---+---+---+ | | | | | | R | | | +---+---+---+---+---+---+---+---+ | | | | | | | R | | +---+---+---+---+---+---+---+---+ | | | | | | | | R | +---+---+---+---+---+---+---+---+

FIG. 2. ]

It will be seen in the first diagram that every square on the board iseither occupied or attacked by a rook, and that every rook is "guarded"(if they were alternately black and white rooks we should say"attacked") by another rook. Placing the eight rooks on any row or fileobviously will have the same effect. In diagram 2 every square is againeither occupied or attacked, but in this case every rook is unguarded. Now, in how many different ways can you so place the eight rooks on theboard that every square shall be occupied or attacked and no rook everguarded by another? I do not wish to go into the question of reversalsand reflections on this occasion, so that placing the rooks on the otherdiagonal will count as different, and similarly with other repetitionsobtained by turning the board round. 

296. --THE FOUR LIONS. 

The puzzle is to find in how many different ways the four lions may beplaced so that there shall never be more than one lion in any row orcolumn. Mere reversals and reflections will not count as different. Thus, regarding the example given, if we place the lions in the otherdiagonal, it will be considered the same arrangement. For if you holdthe second arrangement in front of a mirror or give it a quarter turn, you merely get the first arrangement. It is a simple little puzzle, butrequires a certain amount of careful consideration. 

[Illustration

 +---+---+---+---+ | L | | | | +---+---+---+---+ | | L | | | +---+---+---+---+ | | | L | | +---+---+---+---+ | | | | L | +---+---+---+---+

]

297. --BISHOPS--UNGUARDED. 

Place as few bishops as possible on an ordinary chessboard so that everysquare of the board shall be either occupied or attacked. It will beseen that the rook has more scope than the bishop: for wherever youplace the former, it will always attack fourteen other squares; whereasthe latter will attack seven, nine, eleven, or thirteen squares, according to the position of the diagonal on which it is placed. And itis well here to state that when we speak of "diagonals" in connectionwith the chessboard, we do not limit ourselves to the two long diagonalsfrom corner to corner, but include all the shorter lines that areparallel to these. To prevent misunderstanding on future occasions, itwill be well for the reader to note carefully this fact. 

298. --BISHOPS--GUARDED. 

Now, how many bishops are necessary in order that every square shall beeither occupied or attacked, and every bishop guarded by another bishop?And how may they be placed?

299. --BISHOPS IN CONVOCATION. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | B | B | B | B | B | B | B | B | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | B | B | B | B | B | B | | +---+---+---+---+---+---+---+---+

]

The greatest number of bishops that can be placed at the same time onthe chessboard, without any bishop attacking another, is fourteen. Ishow, in diagram, the simplest way of doing this. In fact, on a squarechequered board of any number of squares the greatest number of bishopsthat can be placed without attack is always two less than twice thenumber of squares on the side. It is an interesting puzzle to discoverin just how many different ways the fourteen bishops may be so placedwithout mutual attack. I shall give an exceedingly simple rule fordetermining the number of ways for a square chequered board of anynumber of squares. 

300. --THE EIGHT QUEENS. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | | | | . . Q | | | | +---+---+---+. . . +---+---+---+---+ | | . . Q. . | | | | | +---+. . . +---+---+---+---+---+---+ | Q. . | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | Q | | +---+---+---+---+---+---+---+---+ | | Q | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | . . Q | +---+---+---+---+---+---+. . . +---+ | | | | | . . Q. . | | +---+---+---+---+. . . +---+---+---+ | | | | Q. . | | | | +---+---+---+---+---+---+---+---+

]

The queen is by far the strongest piece on the chessboard. If you placeher on one of the four squares in the centre of the board, she attacksno fewer than twenty-seven other squares; and if you try to hide her ina corner, she still attacks twenty-one squares. Eight queens may beplaced on the board so that no queen attacks another, and it is an oldpuzzle (first proposed by Nauck in 1850, and it has quite a littleliterature of its own) to discover in just how many different ways thismay be done. I show one way in the diagram, and there are in all twelveof these fundamentally different ways. These twelve produce ninety-twoways if we regard reversals and reflections as different. The diagram isin a way a symmetrical arrangement. If you turn the page upside down, itwill reproduce itself exactly; but if you look at it with one of theother sides at the bottom, you get another way that is not identical. Then if you reflect these two ways in a mirror you get two more ways. Now, all the other eleven solutions are non-symmetrical, and thereforeeach of them may be presented in eight ways by these reversals andreflections. It will thus be seen why the twelve fundamentally differentsolutions produce only ninety-two arrangements, as I have said, and notninety-six, as would happen if all twelve were non-symmetrical. It iswell to have a clear understanding on the matter of reversals andreflections when dealing with puzzles on the chessboard. 

Can the reader place the eight queens on the board so that no queenshall attack another and so that no three queens shall be in a straightline in any oblique direction? Another glance at the diagram will showthat this arrangement will not answer the conditions, for in the twodirections indicated by the dotted lines there are three queens in astraight line. There is only one of the twelve fundamental ways thatwill solve the puzzle. Can you find it?

301. --THE EIGHT STARS. 

[Illustration:

 +---+---+---+---+---+---+---+---+ |///| | | | | | |///| +---+---+---+---+---+---+---+---+ | |///| | | | |///| * | +---+---+---+---+---+---+---+---+ | | |///| | |///| | | +---+---+---+---+---+---+---+---+ | | | |///|///| | | | +---+---+---+---+---+---+---+---+ | | | |///|///| | | | +---+---+---+---+---+---+---+---+ | | |///| | |///| | | +---+---+---+---+---+---+---+---+ | |///| | | | |///| | +---+---+---+---+---+---+---+---+ |///| | | | | | |///| +---+---+---+---+---+---+---+---+

]

The puzzle in this case is to place eight stars in the diagram so thatno star shall be in line with another star horizontally, vertically, ordiagonally. One star is already placed, and that must not be moved, sothere are only seven for the reader now to place. But you must not placea star on any one of the shaded squares. There is only one way ofsolving this little puzzle. 

302. --A PROBLEM IN MOSAICS. 

The art of producing pictures or designs by means of joining togetherpieces of hard substances, either naturally or artificially coloured, isof very great antiquity. It was certainly known in the time of thePharaohs, and we find a reference in the Book of Esther to "a pavementof red, and blue, and white, and black marble. " Some of this ancientwork that has come down to us, especially some of the Roman mosaics, would seem to show clearly, even where design is not at first evident, that much thought was bestowed upon apparently disorderly arrangements. Where, for example, the work has been produced with a very limitednumber of colours, there are evidences of great ingenuity in preventingthe same tints coming in close proximity. Lady readers who are familiarwith the construction of patchwork quilts will know how desirable it issometimes, when they are limited in the choice of material, to preventpieces of the same stuff coming too near together. Now, this puzzle willapply equally to patchwork quilts or tesselated pavements. 

It will be seen from the diagram how a square piece of flooring may bepaved with sixty-two square tiles of the eight colours violet, red, yellow, green, orange, purple, white, and blue (indicated by the initialletters), so that no tile is in line with a similarly coloured tile, vertically, horizontally, or diagonally. Sixty-four such tiles could notpossibly be placed under these conditions, but the two shaded squareshappen to be occupied by iron ventilators. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | V | R | Y | G | O | P | W | B | +---+---+---+---+---+---+---+---+ | W | B | O | P | Y | G | V | R | +---+---*===*---+---*===*---+---+ | G | P H W H V | B H R H Y | O | +---+---*===*---+---*===*---+---+ | R | Y | B | O | G | V | P | W | +---+---+---+---+---+---+---+---+ | B | G | R | Y | P | W | O | V | +---+---+---+---+---+---+---+---+ | O | V | P | W | R | Y | B | G | +---+---+---+---+---+---+---+---+ | P | W | G | B | V | O | R | Y | +---+---+---+---+---+---+---+---+ |///| O | V | R | W | B | G |///| +---+---+---+---+---+---+---+---+

]

The puzzle is this. These two ventilators have to be removed to thepositions indicated by the darkly bordered tiles, and two tiles placedin those bottom corner squares. Can you readjust the thirty-two tiles sothat no two of the same colour shall still be in line?

303. --UNDER THE VEIL. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | | | V | E | I | L | | | +---+---+---+---+---+---+---+---+ | | | I | L | V | E | | | +---+---+---+---+---+---+---+---+ | I | V | | | | | L | E | +---+---+---+---+---+---+---+---+ | L | E | | | | | I | V | +---+---+---+---+---+---+---+---+ | V | I | | | | | E | L | +---+---+---+---+---+---+---+---+ | E | L | | | | | V | I | +---+---+---+---+---+---+---+---+ | | | E | V | L | I | | | +---+---+---+---+---+---+---+---+ | | | L | I | E | V | | | +---+---+---+---+---+---+---+---+

]

If the reader will examine the above diagram, he will see that I have soplaced eight V's, eight E's, eight I's, and eight L's in the diagramthat no letter is in line with a similar one horizontally, vertically, or diagonally. Thus, no V is in line with another V, no E with anotherE, and so on. There are a great many different ways of arranging theletters under this condition. The puzzle is to find an arrangement thatproduces the greatest possible number of four-letter words, readingupwards and downwards, backwards and forwards, or diagonally. Allrepetitions count as different words, and the five variations that maybe used are: VEIL, VILE, LEVI, LIVE, and EVIL. 

This will be made perfectly clear when I say that the above arrangementscores eight, because the top and bottom row both give VEIL; the secondand seventh columns both give VEIL; and the two diagonals, starting fromthe L in the 5th row and E in the 8th row, both give LIVE and EVIL. There are therefore eight different readings of the words in all. 

This difficult word puzzle is given as an example of the use ofchessboard analysis in solving such things. Only a person who isfamiliar with the "Eight Queens" problem could hope to solve it. 

304. --BACHET'S SQUARE. 

One of the oldest card puzzles is by Claude Caspar Bachet de Méziriac, first published, I believe, in the 1624 edition of his work. Rearrangethe sixteen court cards (including the aces) in a square so that in norow of four cards, horizontal, vertical, or diagonal, shall be found twocards of the same suit or the same value. This in itself is easy enough, but a point of the puzzle is to find in how many different ways this maybe done. The eminent French mathematician A. Labosne, in his modernedition of Bachet, gives the answer incorrectly. And yet the puzzle isreally quite easy. Any arrangement produces seven more by turning thesquare round and reflecting it in a mirror. These are counted asdifferent by Bachet. 

Note "row of four cards, " so that the only diagonals we have here toconsider are the two long ones. 

305. --THE THIRTY-SIX LETTER-BLOCKS. 

[Illustration]

The illustration represents a box containing thirty-six letter-blocks. The puzzle is to rearrange these blocks so that no A shall be in a linevertically, horizontally, or diagonally with another A, no B withanother B, no C with another C, and so on. You will find it impossibleto get all the letters into the box under these conditions, but thepoint is to place as many as possible. Of course no letters other thanthose shown may be used. 

306. --THE CROWDED CHESSBOARD. 

[Illustration]

The puzzle is to rearrange the fifty-one pieces on the chessboard sothat no queen shall attack another queen, no rook attack another rook, no bishop attack another bishop, and no knight attack another knight. Nonotice is to be taken of the intervention of pieces of another type fromthat under consideration--that is, two queens will be considered toattack one another although there may be, say, a rook, a bishop, and aknight between them. And so with the rooks and bishops. It is notdifficult to dispose of each type of piece separately; the difficultycomes in when you have to find room for all the arrangements on theboard simultaneously. 

307. --THE COLOURED COUNTERS. 

[Illustration]

The diagram represents twenty-five coloured counters, Red, Blue, Yellow, Orange, and Green (indicated by their initials), and there are five ofeach colour, numbered 1, 2, 3, 4, and 5. The problem is so to place themin a square that neither colour nor number shall be found repeated inany one of the five rows, five columns, and two diagonals. Can you sorearrange them?

308. --THE GENTLE ART OF STAMP-LICKING. 

The Insurance Act is a most prolific source of entertaining puzzles, particularly entertaining if you happen to be among the exempt. One'sinitiation into the gentle art of stamp-licking suggests the followinglittle poser: If you have a card divided into sixteen spaces (4 × 4), and are provided with plenty of stamps of the values 1d. , 2d. , 3d. , 4d. , and 5d. , what is the greatest value that you can stick on the card ifthe Chancellor of the Exchequer forbids you to place any stamp in astraight line (that is, horizontally, vertically, or diagonally) withanother stamp of similar value? Of course, only one stamp can be affixedin a space. The reader will probably find, when he sees the solution, that, like the stamps themselves, he is licked He will most likely betwopence short of the maximum. A friend asked the Post Office how it wasto be done; but they sent him to the Customs and Excise officer, whosent him to the Insurance Commissioners, who sent him to an approvedsociety, who profanely sent him--but no matter. 

309. --THE FORTY-NINE COUNTERS. 

[Illustration]

Can you rearrange the above forty-nine counters in a square so that noletter, and also no number, shall be in line with a similar one, vertically, horizontally, or diagonally? Here I, of course, mean in thelines parallel with the diagonals, in the chessboard sense. 

310. --THE THREE SHEEP. 

[Illustration]

A farmer had three sheep and an arrangement of sixteen pens, divided offby hurdles in the manner indicated in the illustration. In how manydifferent ways could he place those sheep, each in a separate pen, sothat every pen should be either occupied or in line (horizontally, vertically, or diagonally) with at least one sheep? I have given onearrangement that fulfils the conditions. How many others can you find?Mere reversals and reflections must not be counted as different. Thereader may regard the sheep as queens. The problem is then to place thethree queens so that every square shall be either occupied or attackedby at least one queen--in the maximum number of different ways. 

311. --THE FIVE DOGS PUZZLE. 

In 1863, C. F. De Jaenisch first discussed the "Five Queens Puzzle"--toplace five queens on the chessboard so that every square shall beattacked or occupied--which was propounded by his friend, a "Mr. De R. "Jaenisch showed that if no queen may attack another there are ninety-onedifferent ways of placing the five queens, reversals and reflections notcounting as different. If the queens may attack one another, I haverecorded hundreds of ways, but it is not practicable to enumerate themexactly. 

[Illustration]

The illustration is supposed to represent an arrangement of sixty-fourkennels. It will be seen that five kennels each contain a dog, and onfurther examination it will be seen that every one of the sixty-fourkennels is in a straight line with at least one dog--eitherhorizontally, vertically, or diagonally. Take any kennel you like, andyou will find that you can draw a straight line to a dog in one or otherof the three ways mentioned. The puzzle is to replace the five dogs anddiscover in just how many different ways they may be placed in fivekennels _in a straight row_, so that every kennel shall always be inline with at least one dog. Reversals and reflections are here countedas different. 

312. --THE FIVE CRESCENTS OF BYZANTIUM. 

When Philip of Macedon, the father of Alexander the Great, found himselfconfronted with great difficulties in the siege of Byzantium, he set hismen to undermine the walls. His desires, however, miscarried, for nosooner had the operations been begun than a crescent moon suddenlyappeared in the heavens and discovered his plans to his adversaries. TheByzantines were naturally elated, and in order to show their gratitudethey erected a statue to Diana, and the crescent became thenceforward asymbol of the state. In the temple that contained the statue was asquare pavement composed of sixty-four large and costly tiles. Thesewere all plain, with the exception of five, which bore the symbol of thecrescent. These five were for occult reasons so placed that every tileshould be watched over by (that is, in a straight line, vertically, horizontally, or diagonally with) at least one of the crescents. Thearrangement adopted by the Byzantine architect was as follows:--

[Illustration]

Now, to cover up one of these five crescents was a capital offence, thedeath being something very painful and lingering. But on a certainoccasion of festivity it was necessary to lay down on this pavement asquare carpet of the largest dimensions possible, and I have shown inthe illustration by dark shading the largest dimensions that would beavailable. 

The puzzle is to show how the architect, if he had foreseen thisquestion of the carpet, might have so arranged his five crescent tilesin accordance with the required conditions, and yet have allowed for thelargest possible square carpet to be laid down without any one of thefive crescent tiles being covered, or any portion of them. 

313. --QUEENS AND BISHOP PUZZLE. 

It will be seen that every square of the board is either occupied orattacked. The puzzle is to substitute a bishop for the rook on the samesquare, and then place the four queens on other squares so that everysquare shall again be either occupied or attacked. 

[Illustration]

314. --THE SOUTHERN CROSS. 

[Illustration]

In the above illustration we have five Planets and eighty-one FixedStars, five of the latter being hidden by the Planets. It will be foundthat every Star, with the exception of the ten that have a black spot intheir centres, is in a straight line, vertically, horizontally, ordiagonally, with at least one of the Planets. The puzzle is so torearrange the Planets that all the Stars shall be in line with one ormore of them. 

In rearranging the Planets, each of the five may be moved once in astraight line, in either of the three directions mentioned. They will, of course, obscure five other Stars in place of those at presentcovered. 

315. --THE HAT-PEG PUZZLE. 

Here is a five-queen puzzle that I gave in a fanciful dress in 1897. Asthe queens were there represented as hats on sixty-four pegs, I willkeep to the title, "The Hat-Peg Puzzle. " It will be seen that everysquare is occupied or attacked. The puzzle is to remove one queen to adifferent square so that still every square is occupied or attacked, then move a second queen under a similar condition, then a third queen, and finally a fourth queen. After the fourth move every square must beattacked or occupied, but no queen must then attack another. Of course, the moves need not be "queen moves;" you can move a queen to any part ofthe board. 

[Illustration]

316. --THE AMAZONS. 

[Illustration]

This puzzle is based on one by Captain Turton. Remove three of thequeens to other squares so that there shall be eleven squares on theboard that are not attacked. The removal of the three queens need not beby "queen moves. " You may take them up and place them anywhere. There isonly one solution. 

317. --A PUZZLE WITH PAWNS. 

Place two pawns in the middle of the chessboard, one at Q 4 and theother at K 5. Now, place the remaining fourteen pawns (sixteen in all)so that no three shall be in a straight line in any possible direction. 

Note that I purposely do not say queens, because by the words "anypossible direction" I go beyond attacks on diagonals. The pawns must beregarded as mere points in space--at the centres of the squares. Seedotted lines in the case of No. 300, "The Eight Queens. "

318. --LION-HUNTING. 

[Illustration]

My friend Captain Potham Hall, the renowned hunter of big game, saysthere is nothing more exhilarating than a brush with a herd--a pack--ateam--a flock--a swarm (it has taken me a full quarter of an hour torecall the right word, but I have it at last)--a _pride_ of lions. Why anumber of lions are called a "pride, " a number of whales a "school, " anda number of foxes a "skulk" are mysteries of philology into which I willnot enter. 

Well, the captain says that if a spirited lion crosses your path in thedesert it becomes lively, for the lion has generally been looking forthe man just as much as the man has sought the king of the forest. Andyet when they meet they always quarrel and fight it out. A littlecontemplation of this unfortunate and long-standing feud between twoestimable families has led me to figure out a few calculations as to theprobability of the man and the lion crossing one another's path in thejungle. In all these cases one has to start on certain more or lessarbitrary assumptions. That is why in the above illustration I havethought it necessary to represent the paths in the desert with suchrigid regularity. Though the captain assures me that the tracks of thelions usually run much in this way, I have doubts. 

The puzzle is simply to find out in how many different ways the man andthe lion may be placed on two different spots that are not on the samepath. By "paths" it must be understood that I only refer to the ruledlines. Thus, with the exception of the four corner spots, each combatantis always on two paths and no more. It will be seen that there is a lotof scope for evading one another in the desert, which is just what onehas always understood. 

319. --THE KNIGHT-GUARDS. 

[Illustration]

The knight is the irresponsible low comedian of the chessboard. "He is avery uncertain, sneaking, and demoralizing rascal, " says an Americanwriter. "He can only move two squares, but makes up in the quality ofhis locomotion for its quantity, for he can spring one square sidewaysand one forward simultaneously, like a cat; can stand on one leg in themiddle of the board and jump to any one of eight squares he chooses; canget on one side of a fence and blackguard three or four men on theother; has an objectionable way of inserting himself in safe placeswhere he can scare the king and compel him to move, and then gobble aqueen. For pure cussedness the knight has no equal, and when you chasehim out of one hole he skips into another. " Attempts have been made overand over again to obtain a short, simple, and exact definition of themove of the knight--without success. It really consists in moving onesquare like a rook, and then another square like a bishop--the twooperations being done in one leap, so that it does not matter whetherthe first square passed over is occupied by another piece or not. It is, in fact, the only leaping move in chess. But difficult as it is todefine, a child can learn it by inspection in a few minutes. 

I have shown in the diagram how twelve knights (the fewest possible thatwill perform the feat) may be placed on the chessboard so that everysquare is either occupied or attacked by a knight. Examine every squarein turn, and you will find that this is so. Now, the puzzle in this caseis to discover what is the smallest possible number of knights that isrequired in order that every square shall be either occupied orattacked, and every knight protected by another knight. And how wouldyou arrange them? It will be found that of the twelve shown in thediagram only four are thus protected by being a knight's move fromanother knight. 

THE GUARDED CHESSBOARD. 

On an ordinary chessboard, 8 by 8, every square can be guarded--that is, either occupied or attacked--by 5 queens, the fewest possible. There areexactly 91 fundamentally different arrangements in which no queenattacks another queen. If every queen must attack (or be protected by)another queen, there are at fewest 41 arrangements, and I have recordedsome 150 ways in which some of the queens are attacked and some not, butthis last case is very difficult to enumerate exactly. 

On an ordinary chessboard every square can be guarded by 8 rooks (thefewest possible) in 40, 320 ways, if no rook may attack another rook, butit is not known how many of these are fundamentally different. (Seesolution to No. 295, "The Eight Rooks. ") I have not enumerated the waysin which every rook shall be protected by another rook. 

On an ordinary chessboard every square can be guarded by 8 bishops (thefewest possible), if no bishop may attack another bishop. Ten bishopsare necessary if every bishop is to be protected. (See Nos. 297 and 298, "Bishops unguarded" and "Bishops guarded. ")

On an ordinary chessboard every square can be guarded by 12 knights ifall but 4 are unprotected. But if every knight must be protected, 14 arenecessary. (See No. 319, "The Knight-Guards. ")

Dealing with the queen on n² boards generally, where n is lessthan 8, the following results will be of interest:--

1 queen guards 2² board in 1 fundamental way. 

1 queen guards 3² board in 1 fundamental way. 

2 queens guard 4² board in 3 fundamental ways (protected). 

3 queens guard 4² board in 2 fundamental ways (not protected). 

3 queens guard 5² board in 37 fundamental ways (protected). 

3 queens guard 5² board in 2 fundamental ways (not protected). 

3 queens guard 6² board in 1 fundamental way (protected). 

4 queens guard 6² board in 17 fundamental ways (not protected). 

4 queens guard 7² board in 5 fundamental ways (protected). 

4 queens guard 7² board in 1 fundamental way (not protected). 

NON-ATTACKING CHESSBOARD ARRANGEMENTS. 

We know that n queens may always be placed on a square board of n²squares (if n be greater than 3) without any queen attacking anotherqueen. But no general formula for enumerating the number of differentways in which it may be done has yet been discovered; probably it isundiscoverable. The known results are as follows:--

Where n = 4 there is 1 fundamental solution and 2 in all. 

Where n = 5 there are 2 fundamental solutions and 10 in all. 

Where n = 6 there is 1 fundamental solution and 4 in all. 

Where n = 7 there are 6 fundamental solutions and 40 in all. 

Where n = 8 there are 12 fundamental solutions and 92 in all. 

Where n = 9 there are 46 fundamental solutions. 

Where n = 10 there are 92 fundamental solutions. 

Where n = 11 there are 341 fundamental solutions. 

Obviously n rooks may be placed without attack on an n² board in n!ways, but how many of these are fundamentally different I have onlyworked out in the four cases where n equals 2, 3, 4, and 5. The answershere are respectively 1, 2, 7, and 23. (See No. 296, "The Four Lions. ")

We can place 2n-2 bishops on an n² board in 2^{n} ways. (See No. 299, "Bishops in Convocation. ") For boards containing 2, 3, 4, 5, 6, 7, 8squares, on a side there are respectively 1, 2, 3, 6, 10, 20, 36fundamentally different arrangements. Where n is odd there are2^{½(n-1)} such arrangements, each giving 4 by reversals andreflections, and 2^{n-3} - 2^{½(n-3)} giving 8. Where n is even thereare 2^{½(n-2)}, each giving 4 by reversals and reflections, and 2^{n-3}- 2^{½(n-4)}, each giving 8. 

We can place ½(n²+1) knights on an n² board without attack, when nis odd, in 1 fundamental way; and ½n² knights on an n² board, whenn is even, in 1 fundamental way. In the first case we place all theknights on the same colour as the central square; in the second case weplace them all on black, or all on white, squares. 

THE TWO PIECES PROBLEM. 

On a board of n² squares, two queens, two rooks, two bishops, or twoknights can always be placed, irrespective of attack or not, in ½(n^{4}- n²) ways. The following formulæ will show in how many of these waysthe two pieces may be placed with attack and without:--

 With Attack. Without Attack. 

 2 Queens 5n³ - 6n² + n 3n^{4} - 10n³ + 9n² - 2n ------------------- ------------------------------ 3 6

 2 Rooks n³ - n² n^{4} - 2n³ + n² ---------------------- 2

 2 Bishops 4n³ - 6n² + 2n 3n^{4} - 4n³ + 3n² - 2n -------------------- ----------------------------- 6 6

 2 Knights 4n² - 12n + 8 n^{4} - 9n² + 24n -------------------- 2

(See No. 318, " Lion Hunting. ")

DYNAMICAL CHESS PUZZLES. 

 "Push on--keep moving. " THOS. MORTON: _Cure for the Heartache_. 

320. --THE ROOK'S TOUR. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | R | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+

]

The puzzle is to move the single rook over the whole board, so that itshall visit every square of the board once, and only once, and end itstour on the square from which it starts. You have to do this in as fewmoves as possible, and unless you are very careful you will take justone move too many. Of course, a square is regarded equally as "visited"whether you merely pass over it or make it a stopping-place, and we willnot quibble over the point whether the original square is actuallyvisited twice. We will assume that it is not. 

321. --THE ROOK'S JOURNEY. 

This puzzle I call "The Rook's Journey, " because the word "tour"(derived from a turner's wheel) implies that we return to the point fromwhich we set out, and we do not do this in the present case. We shouldnot be satisfied with a personally conducted holiday tour that ended byleaving us, say, in the middle of the Sahara. The rook here makestwenty-one moves, in the course of which journey it visits every squareof the board once and only once, stopping at the square marked 10 at theend of its tenth move, and ending at the square marked 21. Twoconsecutive moves cannot be made in the same direction--that is to say, you must make a turn after every move. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | | | | | | | | R | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | | | | | | | | +---+---+---+---+---+---+---+---+ | | 21| | 10| | | | | +---+---+---+---+---+---+---+---+

]

322. --THE LANGUISHING MAIDEN. 

[Illustration:

 --+-----+-----+-----+-----+-----+-----+-----+ | | | | | | | | | Kt | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | M | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

A wicked baron in the good old days imprisoned an innocent maiden in oneof the deepest dungeons beneath the castle moat. It will be seen fromour illustration that there were sixty-three cells in the dungeon, allconnected by open doors, and the maiden was chained in the cell in whichshe is shown. Now, a valiant knight, who loved the damsel, succeeded inrescuing her from the enemy. Having gained an entrance to the dungeon atthe point where he is seen, he succeeded in reaching the maiden afterentering every cell once and only once. Take your pencil and try totrace out such a route. When you have succeeded, then try to discover aroute in twenty-two straight paths through the cells. It can be done inthis number without entering any cell a second time. 

323. --A DUNGEON PUZZLE. 

[Illustration:

 +-----+-----+-----+-----+-----+-----+-----+-----+ | | | | | | | | | | . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . | | . | | . | . | . | . | | . | +--. --+-- --+--. --+--. --+--. --+--. --+-- --+--. --+ | . | | . | . | . | . | | . | | . . . . . . . . . . . . . . . . . . . . . . . . . . . . | | | . | | | | | . | | +-- --+--. --+-- --+-- --+-- --+-- --+--. --+-- --+ | | . | | | | | . | | | . . . . . . . . . . . . . . . . . . . . . . . . . . . . | | . | | . | . | . | . | | . | +--. --+-- --+--. --+--. --+--. --+--. --+-- --+--. --+ | . | | . | . | . | . | | . | | . . . . . . . . . . . . . . . . . . . . . . . . . . . . | | | | | | | . | . | | +-- --+-- --+-- --+-- --+-- --+--. --+--. --+-- --+ | | | | | | . | . | | | . . . . . . . . . . . . . . . . . . . . . . . . . . . . | | . | | . | . | . | . | | . | +--. --+-- --+--. --+--. --+--. --+--. --+-- --+--. --+ | . | | . | . | . | . | | . | | . . . . . . . . . . . . . . . . . . . . . . . . . . . . | | | . | | | | | . | | +-- --+--. --+-- --+-- --+-- --+-- --+--. --+-- --+ | | . | | | | | . | | | . . . . . . . . . . . . . . . . . . . . . . . . . . . . | | . | | . | . | . | . | | . | +--. --+-- --+--. --+--. --+--. --+--. --+-- --+--. --+ | . | | . | . | . | . | | . | | . . . . . . . . . . . . . . P . . . . . . . . . . . . . | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

A French prisoner, for his sins (or other people's), was confined in anunderground dungeon containing sixty-four cells, all communicating withopen doorways, as shown in our illustration. In order to reduce thetedium of his restricted life, he set himself various puzzles, and thisis one of them. Starting from the cell in which he is shown, how couldhe visit every cell once, and only once, and make as many turnings aspossible? His first attempt is shown by the dotted track. It will befound that there are as many as fifty-five straight lines in his path, but after many attempts he improved upon this. Can you get more thanfifty-five? You may end your path in any cell you like. Try the puzzlewith a pencil on chessboard diagrams, or you may regard them as rooks'moves on a board. 

324. --THE LION AND THE MAN. 

In a public place in Rome there once stood a prison divided intosixty-four cells, all open to the sky and all communicating with oneanother, as shown in the illustration. The sports that here took placewere watched from a high tower. The favourite game was to place aChristian in one corner cell and a lion in the diagonally oppositecorner and then leave them with all the inner doors open. The consequenteffect was sometimes most laughable. On one occasion the man was given asword. He was no coward, and was as anxious to find the lion as the lionundoubtedly was to find him. 

[Illustration:

 +-----+-----+-----+-----+-----+-----+-----+-----+ | | | | | | | | | | L | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ | | | | | | | | | | C | | | | | | | | | | +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

The man visited every cell once and only once in the fewest possiblestraight lines until he reached the lion's cell. The lion, curiouslyenough, also visited every cell once and only once in the fewestpossible straight lines until he finally reached the man's cell. Theystarted together and went at the same speed; yet, although theyoccasionally got glimpses of one another, they never once met. Thepuzzle is to show the route that each happened to take. 

325. --AN EPISCOPAL VISITATION. 

The white squares on the chessboard represent the parishes of a diocese. Place the bishop on any square you like, and so contrive that (using theordinary bishop's move of chess) he shall visit every one of hisparishes in the fewest possible moves. Of course, all the parishespassed through on any move are regarded as "visited. " You can visit anysquares more than once, but you are not allowed to move twice betweenthe same two adjoining squares. What are the fewest possible moves? Thebishop need not end his visitation at the parish from which he first setout. 

326. --A NEW COUNTER PUZZLE. 

Here is a new puzzle with moving counters, or coins, that at firstglance looks as if it must be absurdly simple. But it will be foundquite a little perplexity. I give it in this place for a reason that Iwill explain when we come to the next puzzle. Copy the simple diagram, enlarged, on a sheet of paper; then place two white counters on thepoints 1 and 2, and two red counters on 9 and 10, The puzzle is to makethe red and white change places. You may move the counters one at a timein any order you like, along the lines from point to point, with theonly restriction that a red and a white counter may never stand at onceon the same straight line. Thus the first move can only be from 1 or 2to 3, or from 9 or 10 to 7. 

[Illustration:

 4 8 / \ / \ 2 6 10 \ / \ / 3 7 / \ / \ 1 5 9

]

327. --A NEW BISHOP'S PUZZLE. 

[Illustration:

 +---+---+---+---+ | b | b | b | b | +---+---+---+---+ | | | | | +---+---+---+---+ | | | | | +---+---+---+---+ | B | B | B | B | +---+---+---+---+

]

This is quite a fascinating little puzzle. Place eight bishops (fourblack and four white) on the reduced chessboard, as shown in theillustration. The problem is to make the black bishops change placeswith the white ones, no bishop ever attacking another of the oppositecolour. They must move alternately--first a white, then a black, then awhite, and so on. When you have succeeded in doing it at all, try tofind the fewest possible moves. 

If you leave out the bishops standing on black squares, and only play onthe white squares, you will discover my last puzzle turned on its side. 

328. --THE QUEEN'S TOUR. 

The puzzle of making a complete tour of the chessboard with the queen inthe fewest possible moves (in which squares may be visited more thanonce) was first given by the late Sam Loyd in his _Chess Strategy_. Butthe solution shown below is the one he gave in _American Chess-Nuts_ in1868. I have recorded at least six different solutions in the minimumnumber of moves--fourteen--but this one is the best of all, for reasonsI will explain. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | | | | | | | | | | . . . . . . . . . . . . . . . . . . . . . . . . . . . . . | | . | | | | | | | . | +-. -+---+---+---+---+---+---+-. -+ | . | | | | | | | . | | . | . . . . . . . . . . . . . . . . . . . . . . . . . . | | . | . | | | | | | . | +-. -+---. ---+---+---+---+---+. . -+ | . | |. | | | | . . | | . | . . . . . . . . . . . . . . . . . | . | . | | . | . | . | | |. | . | . | +-. -+---. ---. ---+---. ---+. --+-. -+ | . | |. |. | . | . | . | | . | . | . | . | . | . | . | . | | . | . . | . | . |. | . |. . | . | +-. -+-. -. ---. ---. ---+. --. -. -+-. -+ | . | . |. |. . |. . . | . | . | | . | . | . | . | . . | . | . | . | | . | . | . |. . | . . |. | . | . | +-. -+-. -+---. ---. . --. ---+-. -+-. -+ | . | . | . |. . . . |. | . | . | | . | . | . | . . | . | . | . | . | | . | . |. | . . |. . | . | . | . | +-. -+-. -. ---+. --. ---. ---. -. -+-. -+ | . | . . | . . |. |. |. . | . | | . | . | . | . | . | . | . | . | | . |. . | . |. | . | . | . . | . | +-. -. -. -+. --. ---+---. ---. -. -. -. -+ | . . | . . . | | |. |. . |. . | | . | . . | . . . . . . . . . . . . . | . | . | | | . | | | | | | | +---+---+---+---+---+---+---+---+

]

If you will look at the lettered square you will understand that thereare only ten really differently placed squares on a chessboard--thoseenclosed by a dark line--all the others are mere reversals orreflections. For example, every A is a corner square, and every J acentral square. Consequently, as the solution shown has a turning-pointat the enclosed D square, we can obtain a solution starting from andending at any square marked D--by just turning the board about. Now, this scheme will give you a tour starting from any A, B, C, D, E, F, orH, while no other route that I know can be adapted to more than fivedifferent starting-points. There is no Queen's Tour in fourteen moves(remember a tour must be re-entrant) that may start from a G, I, or J. But we can have a non-re-entrant path over the whole board in fourteenmoves, starting from any given square. Hence the following puzzle:--

[Illustration:

 +---+---+---+---*---+---+---+---+ | A | B | C | G " G | C | B | A | *===*---+---+---*---+---+---+---+ | B " D | E | H " H | E | D | B | +---*===*---+---*---+---+---+---+ | C | E " F | I " I | F | E | C | +---+---*===*---*---+---+---+---+ | G | H | I " J " J | I | H | G | +---+---+---*===*---+---+---+---+ | G | H | I | J | J | I | H | G | +---+---+---+---+---+---+---+---+ | C | E | F | I | I | F | E | C | +---+---+---+---+---+---+---+---+ | B | D | E | H | H | E | D | B | +---+---+---+---+---+---+---+---+ | A | B | C | G | G | C | B | A | +---+---+---+---+---+---+---+---+

]

Start from the J in the enclosed part of the lettered diagram and visitevery square of the board in fourteen moves, ending wherever you like. 

329. --THE STAR PUZZLE. 

[Illustration:

 +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | ¤ | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | ¤ | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+ | * | * | * | * | * | * | * | * | +---+---+---+---+---+---+---+---+

]

Put the point of your pencil on one of the white stars and (without everlifting your pencil from the paper) strike out all the stars in fourteencontinuous straight strokes, ending at the second white star. Yourstraight strokes may be in any direction you like, only every turningmust be made on a star. There is no objection to striking out any starmore than once. 

In this case, where both your starting and ending squares are fixedinconveniently, you cannot obtain a solution by breaking a Queen's Tour, or in any other way by queen moves alone. But you are allowed to useoblique straight lines--such as from the upper white star direct to acorner star. 

330. --THE YACHT RACE. 

Now then, ye land-lubbers, hoist your baby-jib-topsails, break out yourspinnakers, ease off your balloon sheets, and get your head-sails set!

Our race consists in starting from the point at which the yacht is lyingin the illustration and touching every one of the sixty-four buoys infourteen straight courses, returning in the final tack to the buoy fromwhich we start. The seventh course must finish at the buoy from which aflag is flying. 

This puzzle will call for a lot of skilful seamanship on account of thesharp angles at which it will occasionally be necessary to tack. Thepoint of a lead pencil and a good nautical eye are all the outfit thatwe require. 

[Illustration]

This is difficult, because of the condition as to the flag-buoy, andbecause it is a re-entrant tour. But again we are allowed those obliquelines. 

331. --THE SCIENTIFIC SKATER. 

[Illustration]

It will be seen that this skater has marked on the ice sixty-four pointsor stars, and he proposes to start _from his present position_ near thecorner and enter every one of the points in fourteen straight lines. Howwill he do it? Of course there is no objection to his passing over anypoint more than once, but his last straight stroke must bring him backto the position from which he started. 

It is merely a matter of taking your pencil and starting from the spoton which the skater's foot is at present resting, and striking out allthe stars in fourteen continuous straight lines, returning to the pointfrom which you set out. 

332. --THE FORTY-NINE STARS. 

[Illustration]

The puzzle in this case is simply to take your pencil and, starting fromone black star, strike out all the stars in twelve straight strokes, ending at the other black star. It will be seen that the attempt shownin the illustration requires fifteen strokes. Can you do it in twelve?Every turning must be made on a star, and the lines must be parallel tothe sides and diagonals of the square, as shown. In this case we aredealing with a chessboard of reduced dimensions, but only queen moves(without going outside the boundary as in the last case) are required. 

333. --THE QUEEN'S JOURNEY. 

[Illustration]

Place the queen on her own square, as shown in the illustration, andthen try to discover the greatest distance that she can travel over theboard in five queen's moves without passing over any square a secondtime. Mark the queen's path on the board, and note carefully also thatshe must never cross her own track. It seems simple enough, but thereader may find that he has tripped. 

334. --ST. GEORGE AND THE DRAGON. 

[Illustration]

Here is a little puzzle on a reduced chessboard of forty-nine squares. St. George wishes to kill the dragon. Killing dragons was a well-knownpastime of his, and, being a knight, it was only natural that he shoulddesire to perform the feat in a series of knight's moves. Can you showhow, starting from that central square, he may visit once, and onlyonce, every square of the board in a chain of chess knight's moves, andend by capturing the dragon on his last move? Of course a variety ofdifferent ways are open to him, so try to discover a route that formssome pretty design when you have marked each successive leap by astraight line from square to square. 

335. --FARMER LAWRENCE'S CORNFIELDS. 

One of the most beautiful districts within easy distance of London for asummer ramble is that part of Buckinghamshire known as the Valley of theChess--at least, it was a few years ago, before it was discovered by thespeculative builder. At the beginning of the present century therelived, not far from Latimers, a worthy but eccentric farmer namedLawrence. One of his queer notions was that every person who lived nearthe banks of the river Chess ought to be in some way acquainted with thenoble game of the same name, and in order to impress this fact on hismen and his neighbours he adopted at times strange terminology. Forexample, when one of his ewes presented him with a lamb, he would saythat it had "queened a pawn"; when he put up a new barn against thehighway, he called it "castling on the king's side"; and when he sent aman with a gun to keep his neighbour's birds off his fields, he spoke ofit as "attacking his opponent's rooks. " Everybody in the neighbourhoodused to be amused at Farmer Lawrence's little jokes, and one boy (thewag of the village) who got his ears pulled by the old gentleman forstealing his "chestnuts" went so far as to call him "a silly oldchess-protector!"

One year he had a large square field divided into forty-nine squareplots, as shown in the illustration. The white squares were sown withwheat and the black squares with barley. When the harvest time cameround he gave orders that his men were first to cut the corn in thepatch marked 1, and that each successive cutting should be exactly aknight's move from the last one, the thirteenth cutting being in thepatch marked 13, the twenty-fifth in the patch marked 25, thethirty-seventh in the one marked 37, and the last, or forty-ninthcutting, in the patch marked 49. This was too much for poor Hodge, andeach day Farmer Lawrence had to go down to the field and show whichpiece had to be operated upon. But the problem will perhaps present nodifficulty to my readers. 

[Illustration]

336. --THE GREYHOUND PUZZLE. 

In this puzzle the twenty kennels do not communicate with one another bydoors, but are divided off by a low wall. The solitary occupant is thegreyhound which lives in the kennel in the top left-hand corner. When heis allowed his liberty he has to obtain it by visiting every kennel onceand only once in a series of knight's moves, ending at the bottomright-hand corner, which is open to the world. The lines in the abovediagram show one solution. The puzzle is to discover in how manydifferent ways the greyhound may thus make his exit from his cornerkennel. 

[Illustration]

337. --THE FOUR KANGAROOS. 

[Illustration]

In introducing a little Commonwealth problem, I must first explain thatthe diagram represents the sixty-four fields, all properly fenced offfrom one another, of an Australian settlement, though I need hardly saythat our kith and kin "down under" always _do_ set out their land inthis methodical and exact manner. It will be seen that in every one ofthe four corners is a kangaroo. Why kangaroos have a marked preferencefor corner plots has never been satisfactorily explained, and it wouldbe out of place to discuss the point here. I should also add thatkangaroos, as is well known, always leap in what we call "knight'smoves. " In fact, chess players would probably have adopted the betterterm "kangaroo's move" had not chess been invented before kangaroos. 

The puzzle is simply this. One morning each kangaroo went for hismorning hop, and in sixteen consecutive knight's leaps visited justfifteen different fields and jumped back to his corner. No field wasvisited by more than one of the kangaroos. The diagram shows how theyarranged matters. What you are asked to do is to show how they mighthave performed the feat without any kangaroo ever crossing thehorizontal line in the middle of the square that divides the board intotwo equal parts. 

338. --THE BOARD IN COMPARTMENTS. 

[Illustration]

We cannot divide the ordinary chessboard into four equal squarecompartments, and describe a complete tour, or even path, in eachcompartment. But we may divide it into four compartments, as in theillustration, two containing each twenty squares, and the other two eachtwelve squares, and so obtain an interesting puzzle. You are asked todescribe a complete re-entrant tour on this board, starting where youlike, but visiting every square in each successive compartment beforepassing into another one, and making the final leap back to the squarefrom which the knight set out. It is not difficult, but will be foundvery entertaining and not uninstructive. 

Whether a re-entrant "tour" or a complete knight's "path" is possible ornot on a rectangular board of given dimensions depends not only on itsdimensions, but also on its shape. A tour is obviously not possible on aboard containing an odd number of cells, such as 5 by 5 or 7 by 7, forthis reason: Every successive leap of the knight must be from a whitesquare to a black and a black to a white alternately. But if there be anodd number of cells or squares there must be one more square of onecolour than of the other, therefore the path must begin from a square ofthe colour that is in excess, and end on a similar colour, and as aknight's move from one colour to a similar colour is impossible thepath cannot be re-entrant. But a perfect tour may be made on arectangular board of any dimensions provided the number of squares beeven, and that the number of squares on one side be not less than 6 andon the other not less than 5. In other words, the smallest rectangularboard on which a re-entrant tour is possible is one that is 6 by 5. 

A complete knight's path (not re-entrant) over all the squares of aboard is never possible if there be only two squares on one side; nor isit possible on a square board of smaller dimensions than 5 by 5. So thaton a board 4 by 4 we can neither describe a knight's tour nor a completeknight's path; we must leave one square unvisited. Yet on a board 4 by 3(containing four squares fewer) a complete path may be described insixteen different ways. It may interest the reader to discover allthese. Every path that starts from and ends at different squares is herecounted as a different solution, and even reverse routes are calleddifferent. 

339. --THE FOUR KNIGHTS' TOURS. 

[Illustration]

I will repeat that if a chessboard be cut into four equal parts, asindicated by the dark lines in the illustration, it is not possible toperform a knight's tour, either re-entrant or not, on one of the parts. The best re-entrant attempt is shown, in which each knight has totrespass twice on other parts. The puzzle is to cut the boarddifferently into four parts, each of the same size and shape, so that are-entrant knight's tour may be made on each part. Cuts along the dottedlines will not do, as the four central squares of the board would beeither detached or hanging on by a mere thread. 

340. --THE CUBIC KNIGHT'S TOUR. 

Some few years ago I happened to read somewhere that Abnit Vandermonde, a clever mathematician, who was born in 1736 and died in 1793, haddevoted a good deal of study to the question of knight's tours. Beyondwhat may be gathered from a few fragmentary references, I am not awareof the exact nature or results of his investigations, but one thingattracted my attention, and that was the statement that he had proposedthe question of a tour of the knight over the six surfaces of a cube, each surface being a chessboard. Whether he obtained a solution or not Ido not know, but I have never seen one published. So I at once set towork to master this interesting problem. Perhaps the reader may like toattempt it. 

341. --THE FOUR FROGS. 

[Illustration]

In the illustration we have eight toadstools, with white frogs on 1 and3 and black frogs on 6 and 8. The puzzle is to move one frog at a time, in any order, along one of the straight lines from toadstool totoadstool, until they have exchanged places, the white frogs being lefton 6 and 8 and the black ones on 1 and 3. If you use four counters on asimple diagram, you will find this quite easy, but it is a little morepuzzling to do it in only seven plays, any number of successive moves byone frog counting as one play. Of course, more than one frog cannot beon a toadstool at the same time. 

342. --THE MANDARIN'S PUZZLE. 

The following puzzle has an added interest from the circumstance that acorrect solution of it secured for a certain young Chinaman the hand ofhis charming bride. The wealthiest mandarin within a radius of a hundredmiles of Peking was Hi-Chum-Chop, and his beautiful daughter, Peeky-Bo, had innumerable admirers. One of her most ardent lovers was Winky-Hi, and when he asked the old mandarin for his consent to their marriage, Hi-Chum-Chop presented him with the following puzzle and promised hisconsent if the youth brought him the correct answer within a week. Winky-Hi, following a habit which obtains among certain solvers to thisday, gave it to all his friends, and when he had compared theirsolutions he handed in the best one as his own. Luckily it was quiteright. The mandarin thereupon fulfilled his promise. The fatted pup waskilled for the wedding feast, and when Hi-Chum-Chop passed Winky-Hi theliver wing all present knew that it was a token of eternal goodwill, inaccordance with Chinese custom from time immemorial. 

The mandarin had a table divided into twenty-five squares, as shown inthe diagram. On each of twenty-four of these squares was placed anumbered counter, just as I have indicated. The puzzle is to get thecounters in numerical order by moving them one at a time in what we call"knight's moves. " Counter 1 should be where 16 is, 2 where 11 is, 4where 13 now is, and so on. It will be seen that all the counters onshaded squares are in their proper positions. Of course, two countersmay never be on a square at the same time. Can you perform the feat inthe fewest possible moves?

[Illustration]

In order to make the manner of moving perfectly clear I will point outthat the first knight's move can only be made by 1 or by 2 or by 10. Supposing 1 moves, then the next move must be by 23, 4, 8, or 21. Asthere is never more than one square vacant, the order in which thecounters move may be written out as follows: 1--21--14--18--22, etc. Arough diagram should be made on a larger scale for practice, andnumbered counters or pieces of cardboard used. 

343. --EXERCISE FOR PRISONERS. 

The following is the plan of the north wing of a certain gaol, showingthe sixteen cells all communicating by open doorways. Fifteen prisonerswere numbered and arranged in the cells as shown. They were allowed tochange their cells as much as they liked, but if two prisoners were everin the same cell together there was a severe punishment promised them. 

[Illustration]

Now, in order to reduce their growing obesity, and to combine physicalexercise with mental recreation, the prisoners decided, on thesuggestion of one of their number who was interested in knight's tours, to try to form themselves into a perfect knight's path without breakingthe prison regulations, and leaving the bottom right-hand corner cellvacant, as originally. The joke of the matter is that the arrangement atwhich they arrived was as follows:--

 8 3 12 1 11 14 9 6 4 7 2 13 15 10 5

The warders failed to detect the important fact that the men could notpossibly get into this position without two of them having been at sometime in the same cell together. Make the attempt with counters on aruled diagram, and you will find that this is so. Otherwise the solutionis correct enough, each member being, as required, a knight's move fromthe preceding number, and the original corner cell vacant. 

The puzzle is to start with the men placed as in the illustration andshow how it might have been done in the fewest moves, while giving acomplete rest to as many prisoners as possible. 

As there is never more than one vacant cell for a man to enter, it isonly necessary to write down the numbers of the men in the order inwhich they move. It is clear that very few men can be left throughout intheir cells undisturbed, but I will leave the solver to discover justhow many, as this is a very essential part of the puzzle. 

344. --THE KENNEL PUZZLE. 

[Illustration]

A man has twenty-five dog kennels all communicating with each other bydoorways, as shown in the illustration. He wishes to arrange his twentydogs so that they shall form a knight's string from dog No. 1 to dog No. 20, the bottom row of five kennels to be left empty, as at present. Thisis to be done by moving one dog at a time into a vacant kennel. The dogsare well trained to obedience, and may be trusted to remain in thekennels in which they are placed, except that if two are placed in thesame kennel together they will fight it out to the death. How is thepuzzle to be solved in the fewest possible moves without two dogs everbeing together?

345. --THE TWO PAWNS. 

[Illustration]

Here is a neat little puzzle in counting. In how many different ways maythe two pawns advance to the eighth square? You may move them in anyorder you like to form a different sequence. For example, you may movethe Q R P (one or two squares) first, or the K R P first, or one pawn asfar as you like before touching the other. Any sequence is permissible, only in this puzzle as soon as a pawn reaches the eighth square it isdead, and remains there unconverted. Can you count the number ofdifferent sequences? At first it will strike you as being verydifficult, but I will show that it is really quite simple when properlyattacked. 

VARIOUS CHESS PUZZLES. 

 "Chesse-play is a good and wittie exercise of the minde for some kinde of men. " Burton's _Anatomy of Melancholy_. 

346. --SETTING THE BOARD. 

I have a single chessboard and a single set of chessmen. In how manydifferent ways may the men be correctly set up for the beginning of agame? I find that most people slip at a particular point in making thecalculation. 

347. --COUNTING THE RECTANGLES. 

Can you say correctly just how many squares and other rectangles thechessboard contains? In other words, in how great a number of differentways is it possible to indicate a square or other rectangle enclosed bylines that separate the squares of the board?

348. --THE ROOKERY. 

[Illustration]

The White rooks cannot move outside the little square in which they areenclosed except on the final move, in giving checkmate. The puzzle ishow to checkmate Black in the fewest possible moves with No. 8 rook, theother rooks being left in numerical order round the sides of theirsquare with the break between 1 and 7. 

349. --STALEMATE. 

Some years ago the puzzle was proposed to construct an imaginary game ofchess, in which White shall be stalemated in the fewest possible moveswith all the thirty-two pieces on the board. Can you build up such aposition in fewer than twenty moves?

350. --THE FORSAKEN KING. 

[Illustration]

Set up the position shown in the diagram. Then the condition of thepuzzle is--White to play and checkmate in six moves. Notwithstanding thecomplexities, I will show how the manner of play may be condensed intoquite a few lines, merely stating here that the first two moves of Whitecannot be varied. 

351. --THE CRUSADER. 

The following is a prize puzzle propounded by me some years ago. Producea game of chess which, after sixteen moves, shall leave White with allhis sixteen men on their original squares and Black in possession of hisking alone (not necessarily on his own square). White is then to _force_mate in three moves. 

352. --IMMOVABLE PAWNS. 

Starting from the ordinary arrangement of the pieces as for a game, whatis the smallest possible number of moves necessary in order to arrive atthe following position? The moves for both sides must, of course, beplayed strictly in accordance with the rules of the game, though theresult will necessarily be a very weird kind of chess. 

[Illustration]

353. --THIRTY-SIX MATES. 

[Illustration]

Place the remaining eight White pieces in such a position that Whiteshall have the choice of thirty-six different mates on the move. Everymove that checkmates and leaves a different position is a differentmate. The pieces already placed must not be moved. 

354. --AN AMAZING DILEMMA. 

In a game of chess between Mr. Black and Mr. White, Black was indifficulties, and as usual was obliged to catch a train. So he proposedthat White should complete the game in his absence on condition that nomoves whatever should be made for Black, but only with the White pieces. Mr. White accepted, but to his dismay found it utterly impossible to winthe game under such conditions. Try as he would, he could not checkmatehis opponent. On which square did Mr. Black leave his king? The otherpieces are in their proper positions in the diagram. White may leaveBlack in check as often as he likes, for it makes no difference, as hecan never arrive at a checkmate position. 

[Illustration]

355. --CHECKMATE!

[Illustration]

Strolling into one of the rooms of a London club, I noticed a positionleft by two players who had gone. This position is shown in the diagram. It is evident that White has checkmated Black. But how did he do it?That is the puzzle. 

356. --QUEER CHESS. 

Can you place two White rooks and a White knight on the board so thatthe Black king (who must be on one of the four squares in the middle ofthe board) shall be in check with no possible move open to him? "Inother words, " the reader will say, "the king is to be shown checkmated. "Well, you can use the term if you wish, though I intentionally do notemploy it myself. The mere fact that there is no White king on the boardwould be a sufficient reason for my not doing so. 

357. --ANCIENT CHINESE PUZZLE. 

[Illustration]

My next puzzle is supposed to be Chinese, many hundreds of years old, and never fails to interest. White to play and mate, moving each of thethree pieces once, and once only. 

358. --THE SIX PAWNS. 

In how many different ways may I place six pawns on the chessboard sothat there shall be an even number of unoccupied squares in every rowand every column? We are not here considering the diagonals at all, andevery different six squares occupied makes a different solution, so wehave not to exclude reversals or reflections. 

359. --COUNTER SOLITAIRE. 

Here is a little game of solitaire that is quite easy, but not so easyas to be uninteresting. You can either rule out the squares on a sheetof cardboard or paper, or you can use a portion of your chessboard. Ihave shown numbered counters in the illustration so as to make thesolution easy and intelligible to all, but chess pawns or draughts willserve just as well in practice. 

[Illustration]

The puzzle is to remove all the counters except one, and this one thatis left must be No. 1. You remove a counter by jumping over anothercounter to the next space beyond, if that square is vacant, but youcannot make a leap in a diagonal direction. The following moves willmake the play quite clear: 1-9, 2-10, 1-2, and so on. Here 1 jumps over9, and you remove 9 from the board; then 2 jumps over 10, and you remove10; then 1 jumps over 2, and you remove 2. Every move is thus a capture, until the last capture of all is made by No. 1. 

360. --CHESSBOARD SOLITAIRE. 

[Illustration]

Here is an extension of the last game of solitaire. All you need is achessboard and the thirty-two pieces, or the same number of draughts orcounters. In the illustration numbered counters are used. The puzzle isto remove all the counters except two, and these two must haveoriginally been on the same side of the board; that is, the two leftmust either belong to the group 1 to 16 or to the other group, 17 to 32. You remove a counter by jumping over it with another counter to the nextsquare beyond, if that square is vacant, but you cannot make a leap in adiagonal direction. The following moves will make the play quite clear:3-11, 4-12, 3-4, 13-3. Here 3 jumps over 11, and you remove 11; 4 jumpsover 12, and you remove 12; and so on. It will be found a fascinatinglittle game of patience, and the solution requires the exercise of someingenuity. 

361. --THE MONSTROSITY. 

One Christmas Eve I was travelling by rail to a little place in one ofthe southern counties. The compartment was very full, and the passengerswere wedged in very tightly. My neighbour in one of the corner seats wasclosely studying a position set up on one of those little foldingchessboards that can be carried conveniently in the pocket, and I couldscarcely avoid looking at it myself. Here is the position:--

[Illustration]

My fellow-passenger suddenly turned his head and caught the look ofbewilderment on my face. 

"Do you play chess?" he asked. 

"Yes, a little. What is that? A problem?"

"Problem? No; a game. "

"Impossible!" I exclaimed rather rudely. "The position is a perfectmonstrosity!"

He took from his pocket a postcard and handed it to me. It bore anaddress at one side and on the other the words "43. K to Kt 8. "

"It is a correspondence game. " he exclaimed. "That is my friend's lastmove, and I am considering my reply. "

"But you really must excuse me; the position seems utterly impossible. How on earth, for example--"

"Ah!" he broke in smilingly. "I see; you are a beginner; you play towin. "

"Of course you wouldn't play to lose or draw!"

He laughed aloud. 

"You have much to learn. My friend and myself do not play for results ofthat antiquated kind. We seek in chess the wonderful, the whimsical, theweird. Did you ever see a position like that?"

I inwardly congratulated myself that I never had. 

"That position, sir, materializes the sinuous evolvements and syncretic, synthetic, and synchronous concatenations of two cerebralindividualities. It is the product of an amphoteric and intercalatoryinterchange of--"

"Have you seen the evening paper, sir?" interrupted the man opposite, holding out a newspaper. I noticed on the margin beside his thumb somepencilled writing. Thanking him, I took the paper and read--"Insane, butquite harmless. He is in my charge. "

After that I let the poor fellow run on in his wild way until both gotout at the next station. 

But that queer position became fixed indelibly in my mind, with Black'slast move 43. K to Kt 8; and a short time afterwards I found it actuallypossible to arrive at such a position in forty-three moves. Can thereader construct such a sequence? How did White get his rooks and king'sbishop into their present positions, considering Black can never havemoved his king's bishop? No odds were given, and every move wasperfectly legitimate. 

MEASURING, WEIGHING, AND PACKING PUZZLES. 

 "Measure still for measure. " _Measure for Measure_, v. 1. 

Apparently the first printed puzzle involving the measuring of a givenquantity of liquid by pouring from one vessel to others of knowncapacity was that propounded by Niccola Fontana, better known as"Tartaglia" (the stammerer), 1500-1559. It consists in dividing 24 oz. Of valuable balsam into three equal parts, the only measures availablebeing vessels holding 5, 11, and 13 ounces respectively. There are manydifferent solutions to this puzzle in six manipulations, or pouringsfrom one vessel to another. Bachet de Méziriac reprinted this and otherof Tartaglia's puzzles in his _Problèmes plaisans et délectables_(1612). It is the general opinion that puzzles of this class can only besolved by trial, but I think formulæ can be constructed for the solutiongenerally of certain related cases. It is a practically unexplored fieldfor investigation. 

The classic weighing problem is, of course, that proposed by Bachet. Itentails the determination of the least number of weights that wouldserve to weigh any integral number of pounds from 1 lb. To 40 lbs. Inclusive, when we are allowed to put a weight in either of the twopans. The answer is 1, 3, 9, and 27 lbs. Tartaglia had previouslypropounded the same puzzle with the condition that the weights may onlybe placed in one pan. The answer in that case is 1, 2, 4, 8, 16, 32 lbs. Major MacMahon has solved the problem quite generally. A full accountwill be found in Ball's _Mathematical Recreations_ (5th edition). 

Packing puzzles, in which we are required to pack a maximum number ofarticles of given dimensions into a box of known dimensions, are, Ibelieve, of quite recent introduction. At least I cannot recall anyexample in the books of the old writers. One would rather expect to findin the toy shops the idea presented as a mechanical puzzle, but I do notthink I have ever seen such a thing. The nearest approach to it wouldappear to be the puzzles of the jig-saw character, where there is onlyone depth of the pieces to be adjusted. 

362. --THE WASSAIL BOWL. 

One Christmas Eve three Weary Willies came into possession of what wasto them a veritable wassail bowl, in the form of a small barrel, containing exactly six quarts of fine ale. One of the men possessed afive-pint jug and another a three-pint jug, and the problem for them wasto divide the liquor equally amongst them without waste. Of course, theyare not to use any other vessels or measures. If you can show how it wasto be done at all, then try to find the way that requires the fewestpossible manipulations, every separate pouring from one vessel toanother, or down a man's throat, counting as a manipulation. 

363. --THE DOCTOR'S QUERY. 

"A curious little point occurred to me in my dispensary this morning, "said a doctor. "I had a bottle containing ten ounces of spirits of wine, and another bottle containing ten ounces of water. I poured a quarter ofan ounce of spirits into the water and shook them up together. Themixture was then clearly forty to one. Then I poured back aquarter-ounce of the mixture, so that the two bottles should again eachcontain the same quantity of fluid. What proportion of spirits to waterdid the spirits of wine bottle then contain?"

364. --THE BARREL PUZZLE. 

The men in the illustration are disputing over the liquid contents of abarrel. What the particular liquid is it is impossible to say, for weare unable to look into the barrel; so we will call it water. One mansays that the barrel is more than half full, while the other insiststhat it is not half full. What is their easiest way of settling thepoint? It is not necessary to use stick, string, or implement of anykind for measuring. I give this merely as one of the simplest possibleexamples of the value of ordinary sagacity in the solving of puzzles. What are apparently very difficult problems may frequently be solved ina similarly easy manner if we only use a little common sense. 

[Illustration]

365. --NEW MEASURING PUZZLE. 

Here is a new poser in measuring liquids that will be found interesting. A man has two ten-quart vessels full of wine, and a five-quart and afour-quart measure. He wants to put exactly three quarts into each ofthe two measures. How is he to do it? And how many manipulations(pourings from one vessel to another) do you require? Of course, wasteof wine, tilting, and other tricks are not allowed. 

366. --THE HONEST DAIRYMAN. 

An honest dairyman in preparing his milk for public consumption employeda can marked B, containing milk, and a can marked A, containing water. From can A he poured enough to double the contents of can B. Then hepoured from can B into can A enough to double its contents. Then hefinally poured from can A into can B until their contents were exactlyequal. After these operations he would send the can A to London, and thepuzzle is to discover what are the relative proportions of milk andwater that he provides for the Londoners' breakfast-tables. Do they getequal proportions of milk and water--or two parts of milk and one ofwater--or what? It is an interesting question, though, curiously enough, we are not told how much milk or water he puts into the cans at thestart of his operations. 

367. --WINE AND WATER. 

Mr. Goodfellow has adopted a capital idea of late. When he gives alittle dinner party and the time arrives to smoke, after the departureof the ladies, he sometimes finds that the conversation is apt to becometoo political, too personal, too slow, or too scandalous. Then he alwaysmanages to introduce to the company some new poser that he has secretedup his sleeve for the occasion. This invariably results in no end ofinteresting discussion and debate, and puts everybody in a good humour. 

Here is a little puzzle that he propounded the other night, and it isextraordinary how the company differed in their answers. He filled awine-glass half full of wine, and another glass twice the size one-thirdfull of wine. Then he filled up each glass with water and emptied thecontents of both into a tumbler. "Now, " he said, "what part of themixture is wine and what part water?" Can you give the correct answer?

368. --THE KEG OF WINE. 

Here is a curious little problem. A man had a ten-gallon keg full ofwine and a jug. One day he drew off a jugful of wine and filled up thekeg with water. Later on, when the wine and water had got thoroughlymixed, he drew off another jugful and again filled up the keg withwater. It was then found that the keg contained equal proportions ofwine and water. Can you find from these facts the capacity of the jug?

369. --MIXING THE TEA. 

"Mrs. Spooner called this morning, " said the honest grocer to hisassistant. "She wants twenty pounds of tea at 2s. 4½d. Per lb. Ofcourse we have a good 2s. 6d. Tea, a slightly inferior at 2s. 3d. , and acheap Indian at 1s. 9d. , but she is very particular always about herprices. "

"What do you propose to do?" asked the innocent assistant. 

"Do?" exclaimed the grocer. "Why, just mix up the three teas indifferent proportions so that the twenty pounds will work out fairly atthe lady's price. Only don't put in more of the best tea than you canhelp, as we make less profit on that, and of course you will use onlyour complete pound packets. Don't do any weighing. "

How was the poor fellow to mix the three teas? Could you have shown himhow to do it?

370. --A PACKING PUZZLE. 

As we all know by experience, considerable ingenuity is often requiredin packing articles into a box if space is not to be unduly wasted. Aman once told me that he had a large number of iron balls, all exactlytwo inches in diameter, and he wished to pack as many of these aspossible into a rectangular box 24+9/10 inches long, 22+4/5 incheswide, and 14 inches deep. Now, what is the greatest number of theballs that he could pack into that box?

371. --GOLD PACKING IN RUSSIA. 

The editor of the _Times_ newspaper was invited by a high Russianofficial to inspect the gold stored in reserve at St. Petersburg, inorder that he might satisfy himself that it was not another "Humbertsafe. " He replied that it would be of no use whatever, for although thegold might appear to be there, he would be quite unable from a mereinspection to declare that what he saw was really gold. A correspondentof the _Daily Mail_ thereupon took up the challenge, but, although hewas greatly impressed by what he saw, he was compelled to confess hisincompetence (without emptying and counting the contents of every boxand sack, and assaying every piece of gold) to give any assurance on thesubject. In presenting the following little puzzle, I wish it to be alsounderstood that I do not guarantee the real existence of the gold, andthe point is not at all material to our purpose. Moreover, if the readersays that gold is not usually "put up" in slabs of the dimensions that Igive, I can only claim problematic licence. 

Russian officials were engaged in packing 800 gold slabs, each measuring12½ inches long, 11 inches wide, and 1 inch deep. What are theinterior dimensions of a box of equal length and width, and necessarydepth, that will exactly contain them without any space being left over?Not more than twelve slabs may be laid on edge, according to the rulesof the government. It is an interesting little problem in packing, andnot at all difficult. 

372. --THE BARRELS OF HONEY. 

[Illustration]

Once upon a time there was an aged merchant of Bagdad who was muchrespected by all who knew him. He had three sons, and it was a rule ofhis life to treat them all exactly alike. Whenever one received apresent, the other two were each given one of equal value. One day thisworthy man fell sick and died, bequeathing all his possessions to histhree sons in equal shares. 

The only difficulty that arose was over the stock of honey. There wereexactly twenty-one barrels. The old man had left instructions that notonly should every son receive an equal quantity of honey, but shouldreceive exactly the same number of barrels, and that no honey should betransferred from barrel to barrel on account of the waste involved. Now, as seven of these barrels were full of honey, seven were half-full, andseven were empty, this was found to be quite a puzzle, especially aseach brother objected to taking more than four barrels of, the samedescription--full, half-full, or empty. Can you show how they succeededin making a correct division of the property?

CROSSING RIVER PROBLEMS

 "My boat is on the shore. " BYRON. 

This is another mediæval class of puzzles. Probably the earliest examplewas by Abbot Alcuin, who was born in Yorkshire in 735 and died at Toursin 804. And everybody knows the story of the man with the wolf, goat, and basket of cabbages whose boat would only take one of the three at atime with the man himself. His difficulties arose from his being unableto leave the wolf alone with the goat, or the goat alone with thecabbages. These puzzles were considered by Tartaglia and Bachet, andhave been later investigated by Lucas, De Fonteney, Delannoy, Tarry, andothers. In the puzzles I give there will be found one or two newconditions which add to the complexity somewhat. I also include a pulleyproblem that practically involves the same principles. 

[Illustration]

373. --CROSSING THE STREAM. 

During a country ramble Mr. And Mrs. Softleigh found themselves in apretty little dilemma. They had to cross a stream in a small boat whichwas capable of carrying only 150 lbs. Weight. But Mr. Softleigh and hiswife each weighed exactly 150 lbs. , and each of their sons weighed 75lbs. And then there was the dog, who could not be induced on any termsto swim. On the principle of "ladies first, " they at once sent Mrs. Softleigh over; but this was a stupid oversight, because she had to comeback again with the boat, so nothing was gained by that operation. Howdid they all succeed in getting across? The reader will find it mucheasier than the Softleigh family did, for their greatest enemy could nothave truthfully called them a brilliant quartette--while the dog was aperfect fool. 

374--CROSSING THE RIVER AXE. 

Many years ago, in the days of the smuggler known as "Rob Roy of theWest, " a piratical band buried on the coast of South Devon a quantity oftreasure which was, of course, abandoned by them in the usualinexplicable way. Some time afterwards its whereabouts was discovered bythree countrymen, who visited the spot one night and divided the spoilbetween them, Giles taking treasure to the value of £800, Jasper £500worth, and Timothy £300 worth. In returning they had to cross the riverAxe at a point where they had left a small boat in readiness. Here, however, was a difficulty they had not anticipated. The boat would onlycarry two men, or one man and a sack, and they had so little confidencein one another that no person could be left alone on the land or in theboat with more than his share of the spoil, though two persons (being acheck on each other) might be left with more than their shares. Thepuzzle is to show how they got over the river in the fewest possiblecrossings, taking their treasure with them. No tricks, such as ropes, "flying bridges, " currents, swimming, or similar dodges, may beemployed. 

375. --FIVE JEALOUS HUSBANDS. 

During certain local floods five married couples found themselvessurrounded by water, and had to escape from their unpleasant position ina boat that would only hold three persons at a time. Every husband wasso jealous that he would not allow his wife to be in the boat or oneither bank with another man (or with other men) unless he was himselfpresent. Show the quickest way of getting these five men and their wivesacross into safety. 

Call the men A, B, C, D, E, and their respective wives a, b, c, d, e. Togo over and return counts as two crossings. No tricks such as ropes, swimming, currents, etc. , are permitted. 

376. --THE FOUR ELOPEMENTS. 

Colonel B---- was a widower of a very taciturn disposition. Histreatment of his four daughters was unusually severe, almost cruel, andthey not unnaturally felt disposed to resent it. Being charming girlswith every virtue and many accomplishments, it is not surprising thateach had a fond admirer. But the father forbade the young men to call athis house, intercepted all letters, and placed his daughters understricter supervision than ever. But love, which scorns locks and keysand garden walls, was equal to the occasion, and the four youthsconspired together and planned a general elopement. 

At the foot of the tennis lawn at the bottom of the garden ran thesilver Thames, and one night, after the four girls had been safelyconducted from a dormitory window to _terra firma_, they all creptsoftly down to the bank of the river, where a small boat belonging tothe Colonel was moored. With this they proposed to cross to the oppositeside and make their way to a lane where conveyances were waiting tocarry them in their flight. Alas! here at the water's brink theirdifficulties already began. 

The young men were so extremely jealous that not one of them would allowhis prospective bride to remain at any time in the company of anotherman, or men, unless he himself were present also. Now, the boat wouldonly hold two persons, though it could, of course, be rowed by one, andit seemed impossible that the four couples would ever get across. Butmidway in the stream was a small island, and this seemed to present away out of the difficulty, because a person or persons could be leftthere while the boat was rowed back or to the opposite shore. If theyhad been prepared for their difficulty they could have easily worked outa solution to the little poser at any other time. But they were now sohurried and excited in their flight that the confusion they soon gotinto was exceedingly amusing--or would have been to any one exceptthemselves. 

As a consequence they took twice as long and crossed the river twice asoften as was really necessary. Meanwhile, the Colonel, who was a verylight sleeper, thought he heard a splash of oars. He quickly raised thealarm among his household, and the young ladies were found to bemissing. Somebody was sent to the police-station, and a number ofofficers soon aided in the pursuit of the fugitives, who, in consequenceof that delay in crossing the river, were quickly overtaken. The fourgirls returned sadly to their homes, and afterwards broke off theirengagements in disgust. 

For a considerable time it was a mystery how the party of eight managedto cross the river in that little boat without any girl being ever leftwith a man, unless her betrothed was also present. The favourite methodis to take eight counters or pieces of cardboard and mark them A, B, C, D, a, b, c, d, to represent the four men and their prospective brides, and carry them from one side of a table to the other in a matchbox (torepresent the boat), a penny being placed in the middle of the table asthe island. 

Readers are now asked to find the quickest method of getting the partyacross the river. How many passages are necessary from land to land? By"land" is understood either shore or island. Though the boat would notnecessarily call at the island every time of crossing, the possibilityof its doing so must be provided for. For example, it would not do for aman to be alone in the boat (though it were understood that he intendedmerely to cross from one bank to the opposite one) if there happened tobe a girl alone on the island other than the one to whom he was engaged. 

377. --STEALING THE CASTLE TREASURE. 

The ingenious manner in which a box of treasure, consisting principallyof jewels and precious stones, was stolen from Gloomhurst Castle hasbeen handed down as a tradition in the De Gourney family. The thievesconsisted of a man, a youth, and a small boy, whose only mode of escapewith the box of treasure was by means of a high window. Outside thewindow was fixed a pulley, over which ran a rope with a basket at eachend. When one basket was on the ground the other was at the window. Therope was so disposed that the persons in the basket could neither helpthemselves by means of it nor receive help from others. In short, theonly way the baskets could be used was by placing a heavier weight inone than in the other. 

Now, the man weighed 195 lbs. , the youth 105 lbs. , the boy 90 lbs. , andthe box of treasure 75 lbs. The weight in the descending basket couldnot exceed that in the other by more than 15 lbs. Without causing adescent so rapid as to be most dangerous to a human being, though itwould not injure the stolen property. Only two persons, or one personand the treasure, could be placed in the same basket at one time. Howdid they all manage to escape and take the box of treasure with them?

The puzzle is to find the shortest way of performing the feat, which initself is not difficult. Remember, a person cannot help himself byhanging on to the rope, the only way being to go down "with a bump, "with the weight in the other basket as a counterpoise. 

PROBLEMS CONCERNING GAMES. 

 "The little pleasure of the game. " MATTHEW PRIOR. 

Every game lends itself to the propounding of a variety of puzzles. Theycan be made, as we have seen, out of the chessboard and the peculiarmoves of the chess pieces. I will now give just a few examples ofpuzzles with playing cards and dominoes, and also go out of doors andconsider one or two little posers in the cricket field, at the footballmatch, and the horse race and motor-car race. 

378. --DOMINOES IN PROGRESSION. 

[Illustration]

It will be seen that I have played six dominoes, in the illustration, inaccordance with the ordinary rules of the game, 4 against 4, 1 against1, and so on, and yet the sum of the spots on the successive dominoes, 4, 5, 6, 7, 8, 9, are in arithmetical progression; that is, the numberstaken in order have a common difference of 1. In how many different waysmay we play six dominoes, from an ordinary box of twenty-eight, so thatthe numbers on them may lie in arithmetical progression? We must alwaysplay from left to right, and numbers in decreasing arithmeticalprogression (such as 9, 8, 7, 6, 5, 4) are not admissible. 

379. --THE FIVE DOMINOES. 

[Illustration]

Here is a new little puzzle that is not difficult, but will probably befound entertaining by my readers. It will be seen that the five dominoesare so arranged in proper sequence (that is, with 1 against 1, 2 against2, and so on), that the total number of pips on the two end dominoes isfive, and the sum of the pips on the three dominoes in the middle isalso five. There are just three other arrangements giving five for theadditions. They are: --

 (1--0) (0--0) (0--2) (2--1) (1--3) (4--0) (0--0) (0--2) (2--1) (1--0) (2--0) (0--0) (0--1) (1--3) (3--0)

Now, how many similar arrangements are there of five dominoes that shallgive six instead of five in the two additions?

380. --THE DOMINO FRAME PUZZLE. 

[Illustration]

It will be seen in the illustration that the full set of twenty-eightdominoes is arranged in the form of a square frame, with 6 against 6, 2against 2, blank against blank, and so on, as in the game. It will befound that the pips in the top row and left-hand column both add up 44. The pips in the other two sides sum to 59 and 32 respectively. Thepuzzle is to rearrange the dominoes in the same form so that all of thefour sides shall sum to 44. Remember that the dominoes must be correctlyplaced one against another as in the game. 

381. --THE CARD FRAME PUZZLE. 

In the illustration we have a frame constructed from the ten playingcards, ace to ten of diamonds. The children who made it wanted the pipson all four sides to add up alike, but they failed in their attempt andgave it up as impossible. It will be seen that the pips in the top row, the bottom row, and the left-hand side all add up 14, but the right-handside sums to 23. Now, what they were trying to do is quite possible. Canyou rearrange the ten cards in the same formation so that all four sidesshall add up alike? Of course they need not add up 14, but any numberyou choose to select. 

[Illustration]

382. --THE CROSS OF CARDS. 

[Illustration]

In this case we use only nine cards--the ace to nine of diamonds. Thepuzzle is to arrange them in the form of a cross, exactly in the wayshown in the illustration, so that the pips in the vertical bar and inthe horizontal bar add up alike. In the example given it will be foundthat both directions add up 23. What I want to know is, how manydifferent ways are there of rearranging the cards in order to bringabout this result? It will be seen that, without affecting the solution, we may exchange the 5 with the 6, the 5 with the 7, the 8 with the 3, and so on. Also we may make the horizontal and the vertical bars changeplaces. But such obvious manipulations as these are not to be regardedas different solutions. They are all mere variations of one fundamentalsolution. Now, how many of these fundamentally different solutions arethere? The pips need not, of course, always add up 23. 

383. --THE "T" CARD PUZZLE. 

[Illustration]

An entertaining little puzzle with cards is to take the nine cards of asuit, from ace to nine inclusive, and arrange them in the form of theletter "T, " as shown in the illustration, so that the pips in thehorizontal line shall count the same as those in the column. In theexample given they add up twenty-three both ways. Now, it is quite easyto get a single correct arrangement. The puzzle is to discover in justhow many different ways it may be done. Though the number is high, thesolution is not really difficult if we attack the puzzle in the rightmanner. The reverse way obtained by reflecting the illustration in amirror we will not count as different, but all other changes in therelative positions of the cards will here count. How many different waysare there?

384. --CARD TRIANGLES. 

Here you pick out the nine cards, ace to nine of diamonds, and arrangethem in the form of a triangle, exactly as shown in the illustration, sothat the pips add up the same on the three sides. In the example givenit will be seen that they sum to 20 on each side, but the particularnumber is of no importance so long as it is the same on all three sides. The puzzle is to find out in just how many different ways this can bedone. 

If you simply turn the cards round so that one of the other two sides isnearest to you this will not count as different, for the order will bethe same. Also, if you make the 4, 9, 5 change places with the 7, 3, 8, and at the same time exchange the 1 and the 6, it will not be different. But if you only change the 1 and the 6 it will be different, because theorder round the triangle is not the same. This explanation will preventany doubt arising as to the conditions. 

[Illustration]

385. --"STRAND" PATIENCE. 

The idea for this came to me when considering the game of Patience thatI gave in the _Strand Magazine_ for December, 1910, which has beenreprinted in Ernest Bergholt's _Second Book of Patience Games_, underthe new name of "King Albert. "

Make two piles of cards as follows: 9 D, 8 S, 7 D, 6 S, 5 D, 4 S, 3 D, 2S, 1 D, and 9 H, 8 C, 7 H, 6 C, 5 H, 4 C, 3 H, 2 C, 1 H, with the 9 ofdiamonds at the bottom of one pile and the 9 of hearts at the bottom ofthe other. The point is to exchange the spades with the clubs, so thatthe diamonds and clubs are still in numerical order in one pile and thehearts and spades in the other. There are four vacant spaces in additionto the two spaces occupied by the piles, and any card may be laid on aspace, but a card can only be laid on another of the next highervalue--an ace on a two, a two on a three, and so on. Patience isrequired to discover the shortest way of doing this. When there are fourvacant spaces you can pile four cards in seven moves, with only threespaces you can pile them in nine moves, and with two spaces you cannotpile more than two cards. When you have a grasp of these and similarfacts you will be able to remove a number of cards bodily and write down7, 9, or whatever the number of moves may be. The gradual shortening ofplay is fascinating, and first attempts are surprisingly lengthy. 

386. --A TRICK WITH DICE. 

[Illustration]

Here is a neat little trick with three dice. I ask you to throw the dicewithout my seeing them. Then I tell you to multiply the points of thefirst die by 2 and add 5; then multiply the result by 5 and add thepoints of the second die; then multiply the result by 10 and add thepoints of the third die. You then give me the total, and I can at oncetell you the points thrown with the three dice. How do I do it? As anexample, if you threw 1, 3, and 6, as in the illustration, the resultyou would give me would be 386, from which I could at once say what youhad thrown. 

387. --THE VILLAGE CRICKET MATCH. 

In a cricket match, Dingley Dell v. All Muggleton, the latter had thefirst innings. Mr. Dumkins and Mr. Podder were at the wickets, when thewary Dumkins made a splendid late cut, and Mr. Podder called on him torun. Four runs were apparently completed, but the vigilant umpires ateach end called, "three short, " making six short runs in all. Whatnumber did Mr. Dumkins score? When Dingley Dell took their turn at thewickets their champions were Mr. Luffey and Mr. Struggles. The lattermade a magnificent off-drive, and invited his colleague to "come along, "with the result that the observant spectators applauded them for whatwas supposed to have been three sharp runs. But the umpires declaredthat there had been two short runs at each end--four in all. To whatextent, if any, did this manoeuvre increase Mr. Struggles's total?

388. --SLOW CRICKET. 

In the recent county match between Wessex and Nincomshire the formerteam were at the wickets all day, the last man being put out a fewminutes before the time for drawing stumps. The play was so slow thatmost of the spectators were fast asleep, and, on being awakened by oneof the officials clearing the ground, we learnt that two men had beenput out leg-before-wicket for a combined score of 19 runs; four men werecaught for a combined score or 17 runs; one man was run out for a duck'segg; and the others were all bowled for 3 runs each. There were noextras. We were not told which of the men was the captain, but he madeexactly 15 more than the average of his team. What was the captain'sscore?

389. --THE FOOTBALL PLAYERS. 

"It is a glorious game!" an enthusiast was heard to exclaim. "At theclose of last season, of the footballers of my acquaintance four hadbroken their left arm, five had broken their right arm, two had theright arm sound, and three had sound left arms. " Can you discover fromthat statement what is the smallest number of players that the speakercould be acquainted with?

It does not at all follow that there were as many as fourteen men, because, for example, two of the men who had broken the left arm mightalso be the two who had sound right arms. 

390. --THE HORSE-RACE PUZZLE. 

There are no morals in puzzles. When we are solving the old puzzle ofthe captain who, having to throw half his crew overboard in a storm, arranged to draw lots, but so placed the men that only the Turks weresacrificed, and all the Christians left on board, we do not stop todiscuss the questionable morality of the proceeding. And when we aredealing with a measuring problem, in which certain thirsty pilgrims areto make an equitable division of a barrel of beer, we do not objectthat, as total abstainers, it is against our conscience to have anythingto do with intoxicating liquor. Therefore I make no apology forintroducing a puzzle that deals with betting. 

Three horses--Acorn, Bluebottle, and Capsule--start in a race. The oddsare 4 to 1, Acorn; 3 to 1, Bluebottle; 2 to 1, Capsule. Now, how muchmust I invest on each horse in order to win £13, no matter which horsecomes in first? Supposing, as an example, that I betted £5 on eachhorse. Then, if Acorn won, I should receive £20 (four times £5), andhave to pay £5 each for the other two horses; thereby winning £10. Butit will be found that if Bluebottle was first I should only win £5, andif Capsule won I should gain nothing and lose nothing. This will makethe question perfectly clear to the novice, who, like myself, is notinterested in the calling of the fraternity who profess to be engaged inthe noble task of "improving the breed of horses. "

391. --THE MOTOR-CAR RACE. 

Sometimes a quite simple statement of fact, if worded in an unfamiliarmanner, will cause considerable perplexity. Here is an example, and itwill doubtless puzzle some of my more youthful readers just a little. Ihappened to be at a motor-car race at Brooklands, when one spectatorsaid to another, while a number of cars were whirling round and roundthe circular track:--

"There's Gogglesmith--that man in the white car!"

"Yes, I see, " was the reply; "but how many cars are running in thisrace?"

Then came this curious rejoinder:--

"One-third of the cars in front of Gogglesmith added to three-quartersof those behind him will give you the answer. "

Now, can you tell how many cars were running in the race?

PUZZLE GAMES. 

 "He that is beaten may be said To lie in honour's truckle bed. " HUDIBRAS. 

It may be said generally that a game is a contest of skill for two ormore persons, into which we enter either for amusement or to win aprize. A puzzle is something to be done or solved by the individual. Forexample, if it were possible for us so to master the complexities of thegame of chess that we could be assured of always winning with the firstor second move, as the case might be, or of always drawing, then itwould cease to be a game and would become a puzzle. Of course among theyoung and uninformed, when the correct winning play is not understood, apuzzle may well make a very good game. Thus there is no doubt childrenwill continue to play "Noughts and Crosses, " though I have shown (No. 109, "_Canterbury Puzzles_") that between two players who boththoroughly understand the play, every game should be drawn. Neitherplayer could ever win except through the blundering of his opponent. ButI am writing from the point of view of the student of these things. 

The examples that I give in this class are apparently games, but, sinceI show in every case how one player may win if he only play correctly, they are in reality puzzles. Their interest, therefore, lies inattempting to discover the leading method of play. 

392. --THE PEBBLE GAME. 

Here is an interesting little puzzle game that I used to play with anacquaintance on the beach at Slocomb-on-Sea. Two players place an oddnumber of pebbles, we will say fifteen, between them. Then each takes inturn one, two, or three pebbles (as he chooses), and the winner is theone who gets the odd number. Thus, if you get seven and your opponenteight, you win. If you get six and he gets nine, he wins. Ought thefirst or second player to win, and how? When you have settled thequestion with fifteen pebbles try again with, say, thirteen. 

393. --THE TWO ROOKS. 

This is a puzzle game for two players. Each player has a single rook. The first player places his rook on any square of the board that he maychoose to select, and then the second player does the same. They nowplay in turn, the point of each play being to capture the opponent'srook. But in this game you cannot play through a line of attack withoutbeing captured. That is to say, if in the diagram it is Black's turn toplay, he cannot move his rook to his king's knight's square, or to hisking's rook's square, because he would enter the "line of fire" whenpassing his king's bishop's square. For the same reason he cannot moveto his queen's rook's seventh or eighth squares. Now, the game can neverend in a draw. Sooner or later one of the rooks must fall, unless, ofcourse, both players commit the absurdity of not trying to win. Thetrick of winning is ridiculously simple when you know it. Can you solvethe puzzle?

[Illustration]

394. --PUSS IN THE CORNER. 

[Illustration]

This variation of the last puzzle is also played by two persons. Oneputs a counter on No. 6, and the other puts one on No. 55, and they playalternately by removing the counter to any other number in a line. Ifyour opponent moves at any time on to one of the lines you occupy, oreven crosses one of your lines, you immediately capture him and win. Wewill take an illustrative game. 

A moves from 55 to 52; B moves from 6 to 13; A advances to 23; B goes to15; A retreats to 26; B retreats to 13; A advances to 21; B retreats to2; A advances to 7; B goes to 3; A moves to 6; B must now go to 4; Aestablishes himself at 11, and B must be captured next move because heis compelled to cross a line on which A stands. Play this over and youwill understand the game directly. Now, the puzzle part of the game isthis: Which player should win, and how many moves are necessary?

395. --A WAR PUZZLE GAME. 

[Illustration]

Here is another puzzle game. One player, representing the Britishgeneral, places a counter at B, and the other player, representing theenemy, places his counter at E. The Britisher makes the first advancealong one of the roads to the next town, then the enemy moves to one ofhis nearest towns, and so on in turns, until the British general getsinto the same town as the enemy and captures him. Although each mustalways move along a road to the next town only, and the second playermay do his utmost to avoid capture, the British general (as we shouldsuppose, from the analogy of real life) must infallibly win. But how?That is the question. 

396. --A MATCH MYSTERY. 

Here is a little game that is childishly simple in its conditions. Butit is well worth investigation. 

Mr. Stubbs pulled a small table between himself and his friend, Mr. Wilson, and took a box of matches, from which he counted out thirty. 

"Here are thirty matches, " he said. "I divide them into three unequalheaps. Let me see. We have 14, 11, and 5, as it happens. Now, the twoplayers draw alternately any number from any one heap, and he who drawsthe last match loses the game. That's all! I will play with you, Wilson. I have formed the heaps, so you have the first draw. "

"As I can draw any number, " Mr. Wilson said, "suppose I exhibit my usualmoderation and take all the 14 heap. "

"That is the worst you could do, for it loses right away. I take 6 fromthe 11, leaving two equal heaps of 5, and to leave two equal heaps is acertain win (with the single exception of 1, 1), because whatever you doin one heap I can repeat in the other. If you leave 4 in one heap, Ileave 4 in the other. If you then leave 2 in one heap, I leave 2 in theother. If you leave only 1 in one heap, then I take all the other heap. If you take all one heap, I take all but one in the other. No, you mustnever leave two heaps, unless they are equal heaps and more than 1, 1. Let's begin again. "

"Very well, then, " said Mr. Wilson. "I will take 6 from the 14, andleave you 8, 11, 5. "

Mr. Stubbs then left 8, 11, 3; Mr. Wilson, 8, 5, 3; Mr. Stubbs, 6, 5, 3;Mr. Wilson, 4, 5, 3; Mr. Stubbs, 4, 5, 1; Mr. Wilson, 4, 3, 1; Mr. Stubbs, 2, 3, 1; Mr. Wilson, 2, 1, 1; which Mr. Stubbs reduced to 1, 1, 1. 

"It is now quite clear that I must win, " said Mr. Stubbs, because youmust take 1, and then I take 1, leaving you the last match. You neverhad a chance. There are just thirteen different ways in which thematches may be grouped at the start for a certain win. In fact, thegroups selected, 14, 11, 5, are a certain win, because for whatever youropponent may play there is another winning group you can secure, and soon and on down to the last match. "

397. --THE MONTENEGRIN DICE GAME. 

It is said that the inhabitants of Montenegro have a little dice gamethat is both ingenious and well worth investigation. The two playersfirst select two different pairs of odd numbers (always higher than 3)and then alternately toss three dice. Whichever first throws the dice sothat they add up to one of his selected numbers wins. If they are bothsuccessful in two successive throws it is a draw and they try again. Forexample, one player may select 7 and 15 and the other 5 and 13. Then ifthe first player throws so that the three dice add up 7 or 15 he wins, unless the second man gets either 5 or 13 on his throw. 

The puzzle is to discover which two pairs of numbers should be selectedin order to give both players an exactly even chance. 

398. --THE CIGAR PUZZLE. 

I once propounded the following puzzle in a London club, and for aconsiderable period it absorbed the attention of the members. They couldmake nothing of it, and considered it quite impossible of solution. Andyet, as I shall show, the answer is remarkably simple. 

Two men are seated at a square-topped table. One places an ordinarycigar (flat at one end, pointed at the other) on the table, then theother does the same, and so on alternately, a condition being that nocigar shall touch another. Which player should succeed in placing thelast cigar, assuming that they each will play in the best possiblemanner? The size of the table top and the size of the cigar are notgiven, but in order to exclude the ridiculous answer that the tablemight be so diminutive as only to take one cigar, we will say that thetable must not be less than 2 feet square and the cigar not more than 4½inches long. With those restrictions you may take any dimensions youlike. Of course we assume that all the cigars are exactly alike inevery respect. Should the first player, or the second player, win?

MAGIC SQUARE PROBLEMS. 

 "By magic numbers. " CONGREVE, _The Mourning Bride. _

This is a very ancient branch of mathematical puzzledom, and it has animmense, though scattered, literature of its own. In their simple formof consecutive whole numbers arranged in a square so that every column, every row, and each of the two long diagonals shall add up alike, thesemagic squares offer three main lines of investigation: Construction, Enumeration, and Classification. Of recent years many ingenious methodshave been devised for the construction of magics, and the law of theirformation is so well understood that all the ancient mystery hasevaporated and there is no longer any difficulty in making squares ofany dimensions. Almost the last word has been said on this subject. Thequestion of the enumeration of all the possible squares of a given orderstands just where it did over two hundred years ago. Everybody knowsthat there is only one solution for the third order, three cells bythree; and Frénicle published in 1693 diagrams of all the arrangementsof the fourth order--880 in number--and his results have been verifiedover and over again. I may here refer to the general solution for thisorder, for numbers not necessarily consecutive, by E. Bergholt in_Nature_, May 26, 1910, as it is of the greatest importance to studentsof this subject. The enumeration of the examples of any higher order isa completely unsolved problem. 

As to classification, it is largely a matter of individualtaste--perhaps an æsthetic question, for there is beauty in the law andorder of numbers. A man once said that he divided the human race intotwo great classes: those who take snuff and those who do not. I am notsure that some of our classifications of magic squares are not almost asvalueless. However, lovers of these things seem somewhat agreed thatNasik magic squares (so named by Mr. Frost, a student of them, after thetown in India where he lived, and also called Diabolique andPandiagonal) and Associated magic squares are of special interest, so Iwill just explain what these are for the benefit of the novice. 

[Illustration: SIMPLE]

[Illustration: SEMI-NASIK]

[Illustration: ASSOCIATED]

[Illustration: NASIK]

I published in _The Queen_ for January 15, 1910, an article that wouldenable the reader to write out, if he so desired, all the 880 magics ofthe fourth order, and the following is the complete classification thatI gave. The first example is that of a Simple square that fulfils thesimple conditions and no more. The second example is a Semi-Nasik, whichhas the additional property that the opposite short diagonals of twocells each together sum to 34. Thus, 14 + 4 + 11 + 5 = 34 and 12 + 6 +13 + 3 = 34. The third example is not only Semi-Nasik but alsoAssociated, because in it every number, if added to the number that isequidistant, in a straight line, from the centre gives 17. Thus, 1 + 16, 2 + 15, 3 + 14, etc. The fourth example, considered the most "perfect"of all, is a Nasik. Here all the broken diagonals sum to 34. Thus, forexample, 15 + 14 + 2 + 3, and 10 + 4 + 7 + 13, and 15 + 5 + 2 + 12. As aconsequence, its properties are such that if you repeat the square inall directions you may mark off a square, 4 × 4, wherever you please, and it will be magic. 

The following table not only gives a complete enumeration under the fourforms described, but also a classification under the twelve graphictypes indicated in the diagrams. The dots at the end of each linerepresent the relative positions of those complementary pairs, 1 + 16, 2+ 15, etc. , which sum to 17. For example, it will be seen that the firstand second magic squares given are of Type VI. , that the third square isof Type III. , and that the fourth is of Type I. Edouard Lucas indicatedthese types, but he dropped exactly half of them and did not attempt theclassification. 

 NASIK (Type I. ) . . . . . 48 SEMI-NASIK (Type II. , Transpositions of Nasik) . 48 " (Type III. , Associated) 48 " (Type IV. ) . . . 96 " (Type V. ) . . . 96 192 ___ " (Type VI. ) . . . 96 384 ___ SIMPLE. (Type VI. ) . . . 208 " (Type VII. ) . . . 56 " (Type VIII. ). . . 56 " (Type IX. ) . . . 56 " (Type X. ) . . . 56 224 ___ " (Type XI. ) . . . 8 " (Type XII. ) . . . 8 16 448 ___ ___ ___ 880 ___

It is hardly necessary to say that every one of these squares willproduce seven others by mere reversals and reflections, which we do notcount as different. So that there are 7, 040 squares of this order, 880of which are fundamentally different. 

An infinite variety of puzzles may be made introducing new conditionsinto the magic square. In _The Canterbury Puzzles_ I have given examplesof such squares with coins, with postage stamps, with cutting-outconditions, and other tricks. I will now give a few variants involvingfurther novel conditions. 

399. --THE TROUBLESOME EIGHT. 

Nearly everybody knows that a "magic square" is an arrangement ofnumbers in the form of a square so that every row, every column, andeach of the two long diagonals adds up alike. For example, you wouldfind little difficulty in merely placing a different number in each ofthe nine cells in the illustration so that the rows, columns, anddiagonals shall all add up 15. And at your first attempt you willprobably find that you have an 8 in one of the corners. The puzzle is toconstruct the magic square, under the same conditions, with the 8 in theposition shown. 

[Illustration]

400. --THE MAGIC STRIPS. 

[Illustration]

I happened to have lying on my table a number of strips of cardboard, with numbers printed on them from 1 upwards in numerical order. The ideasuddenly came to me, as ideas have a way of unexpectedly coming, to makea little puzzle of this. I wonder whether many readers will arrive atthe same solution that I did. 

Take seven strips of cardboard and lay them together as above. Thenwrite on each of them the numbers 1, 2, 3, 4, 5, 6, 7, as shown, so thatthe numbers shall form seven rows and seven columns. 

Now, the puzzle is to cut these strips into the fewest possible piecesso that they may be placed together and form a magic square, the sevenrows, seven columns, and two diagonals adding up the same number. Nofigures may be turned upside down or placed on their sides--that is, allthe strips must lie in their original direction. 

Of course you could cut each strip into seven separate pieces, eachpiece containing a number, and the puzzle would then be very easy, but Ineed hardly say that forty-nine pieces is a long way from being thefewest possible. 

401. --EIGHT JOLLY GAOL BIRDS. 

[Illustration]

The illustration shows the plan of a prison of nine cells allcommunicating with one another by doorways. The eight prisoners havetheir numbers on their backs, and any one of them is allowed to exercisehimself in whichever cell may happen to be vacant, subject to the rulethat at no time shall two prisoners be in the same cell. The merrymonarch in whose dominions the prison was situated offered them specialcomforts one Christmas Eve if, without breaking that rule, they could soplace themselves that their numbers should form a magic square. 

Now, prisoner No. 7 happened to know a good deal about magic squares, sohe worked out a scheme and naturally selected the method that was mostexpeditious--that is, one involving the fewest possible moves from cellto cell. But one man was a surly, obstinate fellow (quite unfit for thesociety of his jovial companions), and he refused to move out of hiscell or take any part in the proceedings. But No. 7 was quite equal tothe emergency, and found that he could still do what was required in thefewest possible moves without troubling the brute to leave his cell. Thepuzzle is to show how he did it and, incidentally, to discover whichprisoner was so stupidly obstinate. Can you find the fellow?

402. --NINE JOLLY GAOL BIRDS. 

[Illustration]

Shortly after the episode recorded in the last puzzle occurred, a ninthprisoner was placed in the vacant cell, and the merry monarch thenoffered them all complete liberty on the following strange conditions. They were required so to rearrange themselves in the cells that theirnumbers formed a magic square without their movements causing any two ofthem ever to be in the same cell together, except that at the start oneman was allowed to be placed on the shoulders of another man, and thusadd their numbers together, and move as one man. For example, No. 8might be placed on the shoulders of No. 2, and then they would moveabout together as 10. The reader should seek first to solve the puzzlein the fewest possible moves, and then see that the man who is burdenedhas the least possible amount of work to do. 

403. --THE SPANISH DUNGEON. 

Not fifty miles from Cadiz stood in the middle ages a castle, all tracesof which have for centuries disappeared. Among other interestingfeatures, this castle contained a particularly unpleasant dungeondivided into sixteen cells, all communicating with one another, as shownin the illustration. 

Now, the governor was a merry wight, and very fond of puzzles withal. One day he went to the dungeon and said to the prisoners, "By myhalidame!" (or its equivalent in Spanish) "you shall all be set free ifyou can solve this puzzle. You must so arrange yourselves in the sixteencells that the numbers on your backs shall form a magic square in whichevery column, every row, and each of the two diagonals shall add up thesame. Only remember this: that in no case may two of you ever betogether in the same cell. "

One of the prisoners, after working at the problem for two or threedays, with a piece of chalk, undertook to obtain the liberty of himselfand his fellow-prisoners if they would follow his directions and movethrough the doorway from cell to cell in the order in which he shouldcall out their numbers. 

[Illustration]

He succeeded in his attempt, and, what is more remarkable, it would seemfrom the account of his method recorded in the ancient manuscript lyingbefore me, that he did so in the fewest possible moves. The reader isasked to show what these moves were. 

404. --THE SIBERIAN DUNGEONS. 

[Illustration]

The above is a trustworthy plan of a certain Russian prison in Siberia. All the cells are numbered, and the prisoners are numbered the same asthe cells they occupy. The prison diet is so fattening that thesepolitical prisoners are in perpetual fear lest, should their pardonarrive, they might not be able to squeeze themselves through the narrowdoorways and get out. And of course it would be an unreasonable thing toask any government to pull down the walls of a prison just to liberatethe prisoners, however innocent they might be. Therefore these men takeall the healthy exercise they can in order to retard their increasingobesity, and one of their recreations will serve to furnish us with thefollowing puzzle. 

Show, in the fewest possible moves, how the sixteen men may formthemselves into a magic square, so that the numbers on their backs shalladd up the same in each of the four columns, four rows, and twodiagonals without two prisoners having been at any time in the same celltogether. I had better say, for the information of those who have notyet been made acquainted with these places, that it is a peculiarity ofprisons that you are not allowed to go outside their walls. Any prisonermay go any distance that is possible in a single move. 

405. --CARD MAGIC SQUARES. 

[Illustration]

Take an ordinary pack of cards and throw out the twelve court cards. Now, with nine of the remainder (different suits are of no consequence)form the above magic square. It will be seen that the pips add upfifteen in every row in every column, and in each of the two longdiagonals. The puzzle is with the remaining cards (without disturbingthis arrangement) to form three more such magic squares, so that each ofthe four shall add up to a different sum. There will, of course, be fourcards in the reduced pack that will not be used. These four may be anythat you choose. It is not a difficult puzzle, but requires just alittle thought. 

406. --THE EIGHTEEN DOMINOES. 

The illustration shows eighteen dominoes arranged in the form of asquare so that the pips in every one of the six columns, six rows, andtwo long diagonals add up 13. This is the smallest summation possiblewith any selection of dominoes from an ordinary box of twenty-eight. Thegreatest possible summation is 23, and a solution for this number may beeasily obtained by substituting for every number its complement to 6. Thus for every blank substitute a 6, for every 1 a 5, for every 2 a 4, for 3 a 3, for 4 a 2, for 5 a 1, and for 6 a blank. But the puzzle is tomake a selection of eighteen dominoes and arrange them (in exactly theform shown) so that the summations shall be 18 in all the fourteendirections mentioned. 

[Illustration]

SUBTRACTING, MULTIPLYING, AND DIVIDING MAGICS. 

Although the adding magic square is of such great antiquity, curiouslyenough the multiplying magic does not appear to have been mentioneduntil the end of the eighteenth century, when it was referred toslightly by one writer and then forgotten until I revived it in_Tit-Bits_ in 1897. The dividing magic was apparently first discussed byme in _The Weekly Dispatch_ in June 1898. The subtracting magic is hereintroduced for the first time. It will now be convenient to deal withall four kinds of magic squares together. 

[Illustration: ADDING SUBTRACTING MULTIPLYING DIVIDING]

In these four diagrams we have examples in the third order of adding, subtracting, multiplying, and dividing squares. In the first theconstant, 15, is obtained by the addition of the rows, columns, and twodiagonals. In the second case you get the constant, 5, by subtractingthe first number in a line from the second, and the result from thethird. You can, of course, perform the operation in either direction;but, in order to avoid negative numbers, it is more convenient simply todeduct the middle number from the sum of the two extreme numbers. Thisis, in effect, the same thing. It will be seen that the constant of theadding square is n times that of the subtracting square derived fromit, where n is the number of cells in the side of square. And themanner of derivation here is simply to reverse the two diagonals. Bothsquares are "associated"--a term I have explained in the introductoryarticle to this department. 

The third square is a multiplying magic. The constant, 216, is obtainedby multiplying together the three numbers in any line. It is"associated" by multiplication, instead of by addition. It is herenecessary to remark that in an adding square it is not essential thatthe nine numbers should be consecutive. Write down any nine numbers inthis way--

 1 3 5 4 6 8 7 9 11

so that the horizontal differences are all alike and the verticaldifferences also alike (here 2 and 3), and these numbers will form anadding magic square. By making the differences 1 and 3 we, of course, get consecutive numbers--a particular case, and nothing more. Now, inthe case of the multiplying square we must take these numbers ingeometrical instead of arithmetical progression, thus--

 1 3 9 2 6 18 4 12 36

Here each successive number in the rows is multiplied by 3, and in thecolumns by 2. Had we multiplied by 2 and 8 we should get the regulargeometrical progression, 1, 2, 4, 8, 16, 32, 64, 128, and 256, but Iwish to avoid high numbers. The numbers are arranged in the square inthe same order as in the adding square. 

The fourth diagram is a dividing magic square. The constant 6 is hereobtained by dividing the second number in a line by the first (in eitherdirection) and the third number by the quotient. But, again, the processis simplified by dividing the product of the two extreme numbers by themiddle number. This square is also "associated" by multiplication. It isderived from the multiplying square by merely reversing the diagonals, and the constant of the multiplying square is the cube of that of thedividing square derived from it. 

The next set of diagrams shows the solutions for the fifth order ofsquare. They are all "associated" in the same way as before. Thesubtracting square is derived from the adding square by reversing thediagonals and exchanging opposite numbers in the centres of the borders, and the constant of one is again n times that of the other. Thedividing square is derived from the multiplying square in the same way, and the constant of the latter is the 5th power (that is the nth) ofthat of the former. 

[Illustration]

These squares are thus quite easy for odd orders. But the reader willprobably find some difficulty over the even orders, concerning which Iwill leave him to make his own researches, merely propounding two littleproblems. 

407. --TWO NEW MAGIC SQUARES. 

Construct a subtracting magic square with the first sixteen wholenumbers that shall be "associated" by _subtraction_. The constant is, ofcourse, obtained by subtracting the first number from the second inline, the result from the third, and the result again from the fourth. Also construct a dividing magic square of the same order that shall be"associated" by _division_. The constant is obtained by dividing thesecond number in a line by the first, the third by the quotient, and thefourth by the next quotient. 

408. --MAGIC SQUARES OF TWO DEGREES. 

While reading a French mathematical work I happened to come across, thefollowing statement: "A very remarkable magic square of 8, in twodegrees, has been constructed by M. Pfeffermann. In other words, he hasmanaged to dispose the sixty-four first numbers on the squares of achessboard in such a way that the sum of the numbers in every line, every column, and in each of the two diagonals, shall be the same; andmore, that if one substitutes for all the numbers their squares, thesquare still remains magic. " I at once set to work to solve thisproblem, and, although it proved a very hard nut, one was rewarded bythe discovery of some curious and beautiful laws that govern it. Thereader may like to try his hand at the puzzle. 

MAGIC SQUARES OF PRIMES. 

The problem of constructing magic squares with prime numbers only wasfirst discussed by myself in _The Weekly Dispatch_ for 22nd July and 5thAugust 1900; but during the last three or four years it has receivedgreat attention from American mathematicians. First, they have sought toform these squares with the lowest possible constants. Thus, the firstnine prime numbers, 1 to 23 inclusive, sum to 99, which (being divisibleby 3) is theoretically a suitable series; yet it has been demonstratedthat the lowest possible constant is 111, and the required series asfollows: 1, 7, 13, 31, 37, 43, 61, 67, and 73. Similarly, in the case ofthe fourth order, the lowest series of primes that are "theoreticallysuitable" will not serve. But in every other order, up to the 12thinclusive, magic squares have been constructed with the lowest series ofprimes theoretically possible. And the 12th is the lowest order in whicha straight series of prime numbers, unbroken, from 1 upwards has beenmade to work. In other words, the first 144 odd prime numbers haveactually been arranged in magic form. The following summary is takenfrom _The Monist_ (Chicago) for October 1913:--

 Order of Totals of Lowest Squares Square. Series. Constants. Made by-- (Henry E. 3rd 333 111 { Dudeney ( (1900). 

 (Ernest Bergholt 4th 408 102 { and C. D. ( Shuldham. 

 5th 1065 213 H. A. Sayles. 

 (C. D. Shuldham 6th 2448 408 { and J. ( N. Muncey. 

 7th 4893 699 do. 8th 8912 1114 do. 9th 15129 1681 do. 10th 24160 2416 J. N. Muncey. 11th 36095 3355 do. 12th 54168 4514 do. 

For further details the reader should consult the article itself, by W. S. Andrews and H. A. Sayles. 

These same investigators have also performed notable feats inconstructing associated and bordered prime magics, and Mr. Shuldham hassent me a remarkable paper in which he gives examples of Nasik squaresconstructed with primes for all orders from the 4th to the 10th, withthe exception of the 3rd (which is clearly impossible) and the 9th, which, up to the time of writing, has baffled all attempts. 

409. --THE BASKETS OF PLUMS. 

[Illustration]

This is the form in which I first introduced the question of magicsquares with prime numbers. I will here warn the reader that there is alittle trap. 

A fruit merchant had nine baskets. Every basket contained plums (allsound and ripe), and the number in every basket was different. Whenplaced as shown in the illustration they formed a magic square, so thatif he took any three baskets in a line in the eight possible directionsthere would always be the same number of plums. This part of the puzzleis easy enough to understand. But what follows seems at first sight alittle queer. 

The merchant told one of his men to distribute the contents of anybasket he chose among some children, giving plums to every child so thateach should receive an equal number. But the man found it quiteimpossible, no matter which basket he selected and no matter how manychildren he included in the treat. Show, by giving contents of the ninebaskets, how this could come about. 

410. --THE MANDARIN'S "T" PUZZLE. 

[Illustration]

Before Mr. Beauchamp Cholmondely Marjoribanks set out on his tour in theFar East, he prided himself on his knowledge of magic squares, a subjectthat he had made his special hobby; but he soon discovered that he hadnever really touched more than the fringe of the subject, and that thewily Chinee could beat him easily. I present a little problem that onelearned mandarin propounded to our traveller, as depicted on the lastpage. 

The Chinaman, after remarking that the construction of the ordinarymagic square of twenty-five cells is "too velly muchee easy, " asked ourcountryman so to place the numbers 1 to 25 in the square that everycolumn, every row, and each of the two diagonals should add up 65, withonly prime numbers on the shaded "T. " Of course the prime numbersavailable are 1, 2, 3, 5, 7, 11, 13, 17, 19, and 23, so you are atliberty to select any nine of these that will serve your purpose. Canyou construct this curious little magic square?

411. --A MAGIC SQUARE OF COMPOSITES. 

As we have just discussed the construction of magic squares with primenumbers, the following forms an interesting companion problem. Make amagic square with nine consecutive composite numbers--the smallestpossible. 

412. --THE MAGIC KNIGHT'S TOUR. 

Here is a problem that has never yet been solved, nor has itsimpossibility been demonstrated. Play the knight once to every square ofthe chessboard in a complete tour, numbering the squares in the ordervisited, so that when completed the square shall be "magic, " adding upto 260 in every column, every row, and each of the two long diagonals. Ishall give the best answer that I have been able to obtain, in whichthere is a slight error in the diagonals alone. Can a perfect solutionbe found? I am convinced that it cannot, but it is only a "piousopinion. "

MAZES AND HOW TO THREAD THEM. 

 "In wandering mazes lost. " _Paradise Lost. _

The Old English word "maze, " signifying a labyrinth, probably comes fromthe Scandinavian, but its origin is somewhat uncertain. The lateProfessor Skeat thought that the substantive was derived from the verb, and as in old times to be mazed or amazed was to be "lost in thought, "the transition to a maze in whose tortuous windings we are lost isnatural and easy. 

The word "labyrinth" is derived from a Greek word signifying thepassages of a mine. The ancient mines of Greece and elsewhere inspiredfear and awe on account of their darkness and the danger of getting lostin their intricate passages. Legend was afterwards built round thesemazes. The most familiar instance is the labyrinth made by Dædalus inCrete for King Minos. In the centre was placed the Minotaur, and no onewho entered could find his way out again, but became the prey of themonster. Seven youths and seven maidens were sent regularly by theAthenians, and were duly devoured, until Theseus slew the monster andescaped from the maze by aid of the clue of thread provided by Ariadne;which accounts for our using to-day the expression "threading a maze. "

The various forms of construction of mazes include complicated ranges ofcaverns, architectural labyrinths, or sepulchral buildings, tortuousdevices indicated by coloured marbles and tiled pavements, winding pathscut in the turf, and topiary mazes formed by clipped hedges. As a matterof fact, they may be said to have descended to us in precisely thisorder of variety. 

Mazes were used as ornaments on the state robes of Christian emperorsbefore the ninth century, and were soon adopted in the decoration ofcathedrals and other churches. The original idea was doubtless to employthem as symbols of the complicated folds of sin by which man issurrounded. They began to abound in the early part of the twelfthcentury, and I give an illustration of one of this period in the parishchurch at St. Quentin (Fig. 1). It formed a pavement of the nave, andits diameter is 34½ feet. The path here is the line itself. If you placeyour pencil at the point A and ignore the enclosing line, the line leadsyou to the centre by a long route over the entire area; but you neverhave any option as to direction during your course. As we shall find insimilar cases, these early ecclesiastical mazes were generally not of apuzzle nature, but simply long, winding paths that took you overpractically all the ground enclosed. 

[Illustration: FIG. 1. --Maze at St. Quentin. ]

[Illustration: FIG. 2. --Maze in Chartres Cathedral. ]

In the abbey church of St. Berlin, at St. Omer, is another of thesecurious floors, representing the Temple of Jerusalem, with stations forpilgrims. These mazes were actually visited and traversed by them as acompromise for not going to the Holy Land in fulfilment of a vow. Theywere also used as a means of penance, the penitent frequently beingdirected to go the whole course of the maze on hands and knees. 

[Illustration: FIG. 3. --Maze in Lucca Cathedral. ]

The maze in Chartres Cathedral, of which I give an illustration (Fig. 2), is 40 feet across, and was used by penitents following theprocession of Calvary. A labyrinth in Amiens Cathedral was octagonal, similar to that at St. Quentin, measuring 42 feet across. It bore thedate 1288, but was destroyed in 1708. In the chapter-house at Bayeux isa labyrinth formed of tiles, red, black, and encaustic, with a patternof brown and yellow. Dr. Ducarel, in his "_Tour through Part ofNormandy_" (printed in 1767), mentions the floor of the greatguard-chamber in the abbey of St. Stephen, at Caen, "the middle whereofrepresents a maze or labyrinth about 10 feet diameter, and so artfullycontrived that, were we to suppose a man following all the intricatemeanders of its volutes, he could not travel less than a mile before hegot from one end to the other. "

[Illustration: FIG. 4. --Maze at Saffron Walden, Essex. ]

Then these mazes were sometimes reduced in size and represented on asingle tile (Fig. 3). I give an example from Lucca Cathedral. It is onone of the porch piers, and is 19½ inches in diameter. A writer in1858 says that, "from the continual attrition it has received fromthousands of tracing fingers, a central group of Theseus and theMinotaur has now been very nearly effaced. " Other examples were, andperhaps still are, to be found in the Abbey of Toussarts, atChâlons-sur-Marne, in the very ancient church of St. Michele at Pavia, at Aix in Provence, in the cathedrals of Poitiers, Rheims, and Arras, inthe church of Santa Maria in Aquiro in Rome, in San Vitale at Ravenna, in the Roman mosaic pavement found at Salzburg, and elsewhere. Thesemazes were sometimes called "Chemins de Jerusalem, " as beingemblematical of the difficulties attending a journey to the earthlyJerusalem and of those encountered by the Christian before he can reachthe heavenly Jerusalem--where the centre was frequently called "Ciel. "

Common as these mazes were upon the Continent, it is probable that noexample is to be found in any English church; at least I am not aware ofthe existence of any. But almost every county has, or has had, itsspecimens of mazes cut in the turf. Though these are frequently known as"miz-mazes" or "mize-mazes, " it is not uncommon to find them locallycalled "Troy-towns, " "shepherds' races, " or "Julian's Bowers"--namesthat are misleading, as suggesting a false origin. From the facts alonethat many of these English turf mazes are clearly copied from those inthe Continental churches, and practically all are found close to someecclesiastical building or near the site of an ancient one, we mayregard it as certain that they were of church origin and not invented bythe shepherds or other rustics. And curiously enough, these turf mazesare apparently unknown on the Continent. They are distinctly mentionedby Shakespeare:--

 "The nine men's morris is filled up with mud, And the quaint mazes in the wanton green For lack of tread are undistinguishable. "

 _A Midsummer Night's Dream_, ii. 1. 

 "My old bones ache: here's a maze trod indeed, Through forth-rights and meanders!"

 _The Tempest_, iii. 3. 

[Illustration: FIG. 5. --Maze at Sneinton, Nottinghamshire. ]

There was such a maze at Comberton, in Cambridgeshire, and another, locally called the "miz-maze, " at Leigh, in Dorset. The latter was onthe highest part of a field on the top of a hill, a quarter of a milefrom the village, and was slightly hollow in the middle and enclosed bya bank about 3 feet high. It was circular, and was thirty paces indiameter. In 1868 the turf had grown over the little trenches, and itwas then impossible to trace the paths of the maze. The Comberton onewas at the same date believed to be perfect, but whether either or bothhave now disappeared I cannot say. Nor have I been able to verify theexistence or non-existence of the other examples of which I am able togive illustrations. I shall therefore write of them all in the pasttense, retaining the hope that some are still preserved. 

[Illustration: FIG. 6. --Maze at Alkborough, Lincolnshire. ]

In the next two mazes given--that at Saffron Walden, Essex (110 feet indiameter, Fig. 4), and the one near St. Anne's Well, at Sneinton, Nottinghamshire (Fig. 5), which was ploughed up on February 27th, 1797(51 feet in diameter, with a path 535 yards long)--the paths must ineach case be understood to be on the lines, black or white, as the casemay be. 

[Illustration: FIG. 7. --Maze at Boughton Green, Nottinghamshire. ]

I give in Fig. 6 a maze that was at Alkborough, Lincolnshire, overlooking the Humber. This was 44 feet in diameter, and theresemblance between it and the mazes at Chartres and Lucca (Figs. 2 and3) will be at once perceived. A maze at Boughton Green, inNottinghamshire, a place celebrated at one time for its fair (Fig. 7), was 37 feet in diameter. I also include the plan (Fig. 8) of one thatused to be on the outskirts of the village of Wing, near Uppingham, Rutlandshire. This maze was 40 feet in diameter. 

[Illustration: FIG. 8. --Maze at Wing, Rutlandshire. ]

[Illustration: FIG. 9. --Maze on St. Catherine's Hill, Winchester. ]

The maze that was on St. Catherine's Hill, Winchester, in the parish ofChilcombe, was a poor specimen (Fig. 9), since, as will be seen, therewas one short direct route to the centre, unless, as in Fig. 10 again, the path is the line itself from end to end. This maze was 86 feetsquare, cut in the turf, and was locally known as the "Mize-maze. " Itbecame very indistinct about 1858, and was then recut by the Warden ofWinchester, with the aid of a plan possessed by a lady living in theneighbourhood. 

[Illustration: FIG. 10. --Maze on Ripon Common. ]

A maze formerly existed on Ripon Common, in Yorkshire (Fig. 10). It wasploughed up in 1827, but its plan was fortunately preserved. Thisexample was 20 yards in diameter, and its path is said to have been 407yards long. 

[Illustration: FIG. 11. --Maze at Theobalds, Hertfordshire. ]

In the case of the maze at Theobalds, Hertfordshire, after you havefound the entrance within the four enclosing hedges, the path is forced(Fig. 11). As further illustrations of this class of maze, I give onetaken from an Italian work on architecture by Serlio, published in 1537(Fig. 12), and one by London and Wise, the designers of the HamptonCourt maze, from their book, _The Retired Gard'ner_, published in 1706(Fig. 13). Also, I add a Dutch maze (Fig. 14). 

[Illustration: FIG. 12. --Italian Maze of Sixteenth Century. ]

[Illustration: FIG. 13. --By the Designers of Hampton Court Maze. ]

[Illustration: FIG. 14. --A Dutch Maze. ]

So far our mazes have been of historical interest, but they havepresented no difficulty in threading. After the Reformation period wefind mazes converted into mediums for recreation, and they generallyconsisted of labyrinthine paths enclosed by thick and carefully trimmedhedges. These topiary hedges were known to the Romans, with whom the_topiarius_ was the ornamental gardener. This type of maze has of lateyears degenerated into the seaside "Puzzle Gardens. Teas, sixpence, including admission to the Maze. " The Hampton Court Maze, sometimescalled the "Wilderness, " at the royal palace, was designed, as I havesaid, by London and Wise for William III. , who had a liking for suchthings (Fig. 15). I have before me some three or four versions of it, all slightly different from one another; but the plan I select is takenfrom an old guide-book to the palace, and therefore ought to betrustworthy. The meaning of the dotted lines, etc. , will be explainedlater on. 

[Illustration: FIG. 15. --Maze at Hampton Court Palace. ]

[Illustration: FIG. 16. --Maze at Hatfield House, Herts. ]

[Illustration: FIG. 17. --Maze formerly at South Kensington. ]

[Illustration: FIG. 18. --A German Maze. ]

The maze at Hatfield House (Fig. 16), the seat of the Marquis ofSalisbury, like so many labyrinths, is not difficult on paper; but boththis and the Hampton Court Maze may prove very puzzling to actuallythread without knowing the plan. One reason is that one is so apt to godown the same blind alleys over and over again, if one proceeds withoutmethod. The maze planned by the desire of the Prince Consort for theRoyal Horticultural Society's Gardens at South Kensington was allowed togo to ruin, and was then destroyed--no great loss, for it was a feeblething. It will be seen that there were three entrances from the outside(Fig. 17), but the way to the centre is very easy to discover. I includea German maze that is curious, but not difficult to thread on paper(Fig. 18). The example of a labyrinth formerly existing at Pimperne, inDorset, is in a class by itself (Fig. 19). It was formed of small ridgesabout a foot high, and covered nearly an acre of ground; but it was, unfortunately, ploughed up in 1730. 

[Illustration: FIG. 19. --Maze at Pimperne, Dorset. ]

We will now pass to the interesting subject of how to thread any maze. While being necessarily brief, I will try to make the matter clear toreaders who have no knowledge of mathematics. And first of all we willassume that we are trying to enter a maze (that is, get to the "centre")of which we have no plan and about which we know nothing. The first ruleis this: If a maze has no parts of its hedges detached from the rest, then if we always keep in touch with the hedge with the right hand (oralways touch it with the left), going down to the stop in every blindalley and coming back on the other side, we shall pass through everypart of the maze and make our exit where we went in. Therefore we mustat one time or another enter the centre, and every alley will betraversed twice. 

[Illustration: FIG. 20. --M. Tremaux's Method of Solution. ]

[Illustration: FIG. 21. --How to thread the Hatfield Maze. ]

Now look at the Hampton Court plan. Follow, say to the right, the pathindicated by the dotted line, and what I have said is clearly correct ifwe obliterate the two detached parts, or "islands, " situated on eachside of the star. But as these islands are there, you cannot by thismethod traverse every part of the maze; and if it had been so plannedthat the "centre" was, like the star, between the two islands, you wouldnever pass through the "centre" at all. A glance at the Hatfield mazewill show that there are three of these detached hedges or islands atthe centre, so this method will never take you to the "centre" of thatone. But the rule will at least always bring you safely out again unlessyou blunder in the following way. Suppose, when you were going in thedirection of the arrow in the Hampton Court Maze, that you could notdistinctly see the turning at the bottom, that you imagined you were ina blind alley and, to save time, crossed at once to the opposite hedge, then you would go round and round that U-shaped island with your righthand still always on the hedge--for ever after!

[Illustration: FIG. 22. The Philadelphia Maze, and its Solution. ]

This blunder happened to me a few years ago in a little maze on the isleof Caldy, South Wales. I knew the maze was a small one, but after a verylong walk I was amazed to find that I did not either reach the "centre"or get out again. So I threw a piece of paper on the ground, and sooncame round to it; from which I knew that I had blundered over a supposedblind alley and was going round and round an island. Crossing to theopposite hedge and using more care, I was quickly at the centre and outagain. Now, if I had made a similar mistake at Hampton Court, anddiscovered the error when at the star, I should merely have passed fromone island to another! And if I had again discovered that I was on adetached part, I might with ill luck have recrossed to the first islandagain! We thus see that this "touching the hedge" method should alwaysbring us safely out of a maze that we have entered; it may happen totake us through the "centre, " and if we miss the centre we shall knowthere must be islands. But it has to be done with a little care, and inno case can we be sure that we have traversed every alley or that thereare no detached parts. 

[Illustration: FIG. 23. --Simplified Diagram of Fig. 22. ]

If the maze has many islands, the traversing of the whole of it may be amatter of considerable difficulty. Here is a method for solving anymaze, due to M. Trémaux, but it necessitates carefully marking in someway your entrances and exits where the galleries fork. I give a diagramof an imaginary maze of a very simple character that will serve ourpurpose just as well as something more complex (Fig. 20). The circles atthe regions where we have a choice of turnings we may call nodes. A"new" path or node is one that has not been entered before on the route;an "old" path or node is one that has already been entered, 1. No pathmay be traversed more than twice. 2. When you come to a new node, takeany path you like. 3. When by a new path you come to an old node or tothe stop of a blind alley, return by the path you came. 4. When by anold path you come to an old node, take a new path if there is one; ifnot, an old path. The route indicated by the dotted line in the diagramis taken in accordance with these simple rules, and it will be seenthat it leads us to the centre, although the maze consists of fourislands. 

[Illustration: FIG. 24. --Can you find the Shortest Way to Centre?]

Neither of the methods I have given will disclose to us the shortest wayto the centre, nor the number of the different routes. But we can easilysettle these points with a plan. Let us take the Hatfield maze (Fig. 21). It will be seen that I have suppressed all the blind alleys by theshading. I begin at the stop and work backwards until the path forks. These shaded parts, therefore, can never be entered without our havingto retrace our steps. Then it is very clearly seen that if we enter at Awe must come out at B; if we enter at C we must come out at D. Then wehave merely to determine whether A, B, E, or C, D, E, is the shorterroute. As a matter of fact, it will be found by rough measurement orcalculation that the shortest route to the centre is by way of C, D, E, F. 

[Illustration: FIG. 25. --Rosamund's Bower. ]

I will now give three mazes that are simply puzzles on paper, for, sofar as I know, they have never been constructed in any other way. Thefirst I will call the Philadelphia maze (Fig. 22). Fourteen years ago atravelling salesman, living in Philadelphia, U. S. A. , developed acuriously unrestrained passion for puzzles. He neglected his business, and soon his position was taken from him. His days and nights were nowpassed with the subject that fascinated him, and this little maze seemsto have driven him into insanity. He had been puzzling over it for sometime, and finally it sent him mad and caused him to fire a bulletthrough his brain. Goodness knows what his difficulties could have been!But there can be little doubt that he had a disordered mind, and that ifthis little puzzle had not caused him to lose his mental balance someother more or less trivial thing would in time have done so. There is nomoral in the story, unless it be that of the Irish maxim, which appliesto every occupation of life as much as to the solving of puzzles: "Takethings aisy; if you can't take them aisy, take them as aisy as you can. "And it is a bad and empirical way of solving any puzzle--by blowing yourbrains out. 

Now, how many different routes are there from A to B in this maze if wemust never in any route go along the same passage twice? The four openspaces where four passages end are not reckoned as "passages. " In thediagram (Fig. 22) it will be seen that I have again suppressed the blindalleys. It will be found that, in any case, we must go from A to C, andalso from F to B. But when we have arrived at C there are three ways, marked 1, 2, 3, of getting to D. Similarly, when we get to E there arethree ways, marked 4, 5, 6, of getting to F. We have also the dottedroute from C to E, the other dotted route from D to F, and the passagefrom D to E, indicated by stars. We can, therefore, express the positionof affairs by the little diagram annexed (Fig. 23). Here everycondition of route exactly corresponds to that in the circular maze, only it is much less confusing to the eye. Now, the number of routes, under the conditions, from A to B on this simplified diagram is 640, andthat is the required answer to the maze puzzle. 

Finally, I will leave two easy maze puzzles (Figs. 24, 25) for myreaders to solve for themselves. The puzzle in each case is to find theshortest possible route to the centre. Everybody knows the story of FairRosamund and the Woodstock maze. What the maze was like or whether itever existed except in imagination is not known, many writers believingthat it was simply a badly-constructed house with a large number ofconfusing rooms and passages. At any rate, my sketch lacks the authorityof the other mazes in this article. My "Rosamund's Bower" is simplydesigned to show that where you have the plan before you it oftenhappens that the easiest way to find a route into a maze is by workingbackwards and first finding a way out. 

THE PARADOX PARTY. 

 "Is not life itself a paradox?" C. L. DODGSON, _Pillow Problems_. 

"It is a wonderful age!" said Mr. Allgood, and everybody at the tableturned towards him and assumed an attitude of expectancy. 

This was an ordinary Christmas dinner of the Allgood family, with asprinkling of local friends. Nobody would have supposed that the aboveremark would lead, as it did, to a succession of curious puzzles andparadoxes, to which every member of the party contributed something ofinterest. The little symposium was quite unpremeditated, so we must notbe too critical respecting a few of the posers that were forthcoming. The varied character of the contributions is just what we would expecton such an occasion, for it was a gathering not of expert mathematiciansand logicians, but of quite ordinary folk. 

"It is a wonderful age!" repeated Mr. Allgood. "A man has just designeda square house in such a cunning manner that all the windows on the foursides have a south aspect. "

"That would appeal to me, " said Mrs. Allgood, "for I cannot endure aroom with a north aspect. "

"I cannot conceive how it is done, " Uncle John confessed. "I suppose heputs bay windows on the east and west sides; but how on earth can becontrive to look south from the north side? Does he use mirrors, orsomething of that kind?"

"No, " replied Mr. Allgood, "nothing of the sort. All the windows areflush with the walls, and yet you get a southerly prospect from everyone of them. You see, there is no real difficulty in designing the houseif you select the proper spot for its erection. Now, this house isdesigned for a gentleman who proposes to build it exactly at the NorthPole. If you think a moment you will realize that when you stand at theNorth Pole it is impossible, no matter which way you may turn, to lookelsewhere than due south! There are no such directions as north, east, or west when you are exactly at the North Pole. Everything is duesouth!"

"I am afraid, mother, " said her son George, after the laughter hadsubsided, "that, however much you might like the aspect, the situationwould be a little too bracing for you. "

"Ah, well!" she replied. "Your Uncle John fell also into the trap. I amno good at catches and puzzles. I suppose I haven't the right sort ofbrain. Perhaps some one will explain this to me. Only last week Iremarked to my hairdresser that it had been said that there are morepersons in the world than any one of them has hairs on his head. Hereplied, 'Then it follows, madam, that two persons, at least, must haveexactly the same number of hairs on their heads. ' If this is a fact, Iconfess I cannot see it. "

"How do the bald-headed affect the question?" asked Uncle John. 

"If there are such persons in existence, " replied Mrs. Allgood, "whohaven't a solitary hair on their heads discoverable under amagnifying-glass, we will leave them out of the question. Still, Idon't see how you are to prove that at least two persons have exactlythe same number to a hair. "

"I think I can make it clear, " said Mr. Filkins, who had dropped in forthe evening. "Assume the population of the world to be only one million. Any number will do as well as another. Then your statement was to theeffect that no person has more than nine hundred and ninety-ninethousand nine hundred and ninety-nine hairs on his head. Is that so?"

"Let me think, " said Mrs. Allgood. "Yes--yes--that is correct. "

"Very well, then. As there are only nine hundred and ninety-ninethousand nine hundred and ninety-nine _different_ ways of bearing hair, it is clear that the millionth person must repeat one of those ways. Doyou see?"

"Yes; I see that--at least I think I see it. "

"Therefore two persons at least must have the same number of hairs ontheir heads; and as the number of people on the earth so greatly exceedsthe number of hairs on any one person's head, there must, of course, bean immense number of these repetitions. "

"But, Mr. Filkins, " said little Willie Allgood, "why could not themillionth man have, say, ten thousand hairs and a half?"

"That is mere hair-splitting, Willie, and does not come into thequestion. "

"Here is a curious paradox, " said George. "If a thousand soldiers aredrawn up in battle array on a plane"--they understood him to mean"plain"--"only one man will stand upright. "

Nobody could see why. But George explained that, according to Euclid, aplane can touch a sphere only at one point, and that person only whostands at that point, with respect to the centre of the earth, willstand upright. 

"In the same way, " he remarked, "if a billiard-table were quitelevel--that is, a perfect plane--the balls ought to roll to the centre. "

Though he tried to explain this by placing a visiting-card on an orangeand expounding the law of gravitation, Mrs. Allgood declined to acceptthe statement. She could not see that the top of a true billiard-tablemust, theoretically, be spherical, just like a portion of theorange-peel that George cut out. Of course, the table is so small inproportion to the surface of the earth that the curvature is notappreciable, but it is nevertheless true in theory. A surface that wecall level is not the same as our idea of a true geometrical plane. 

"Uncle John, " broke in Willie Allgood, "there is a certain islandsituated between England and France, and yet that island is farther fromFrance than England is. What is the island?"

"That seems absurd, my boy; because if I place this tumbler, torepresent the island, between these two plates, it seems impossible thatthe tumbler can be farther from either of the plates than they are fromeach other. "

"But isn't Guernsey between England and France?" asked Willie. 

"Yes, certainly. "

"Well, then, I think you will find, uncle, that Guernsey is abouttwenty-six miles from France, and England is only twenty-one miles fromFrance, between Calais and Dover. "

"My mathematical master, " said George, "has been trying to induce me toaccept the axiom that 'if equals be multiplied by equals the productsare equal. '"

"It is self-evident, " pointed out Mr. Filkins. "For example, if 3 feetequal 1 yard, then twice 3 feet will equal 2 yards. Do you see?"

"But, Mr. Filkins, " asked George, "is this tumbler half full of waterequal to a similar glass half empty?"

"Certainly, George. "

"Then it follows from the axiom that a glass full must equal a glassempty. Is that correct?"

"No, clearly not. I never thought of it in that light. "

"Perhaps, " suggested Mr. Allgood, "the rule does not apply to liquids. "

"Just what I was thinking, Allgood. It would seem that we must make anexception in the case of liquids. "

"But it would be awkward, " said George, with a smile, "if we also had toexcept the case of solids. For instance, let us take the solid earth. One mile square equals one square mile. Therefore two miles square mustequal two square miles. Is this so?"

"Well, let me see! No, of course not, " Mr. Filkins replied, "because twomiles square is four square miles. "

"Then, " said George, "if the axiom is not true in these cases, when isit true?"

Mr. Filkins promised to look into the matter, and perhaps the readerwill also like to give it consideration at leisure. 

"Look here, George, " said his cousin Reginald Woolley: "by whatfractional part does four-fourths exceed three-fourths?"

"By one-fourth!" shouted everybody at once. 

"Try another one, " George suggested. 

"With pleasure, when you have answered that one correctly, " wasReginald's reply. 

"Do you mean to say that it isn't one-fourth?"

"Certainly I do. "

Several members of the company failed to see that the correct answer is"one-third, " although Reginald tried to explain that three of anything, if increased by one-third, becomes four. 

"Uncle John, how do you pronounce 't-o-o'?" asked Willie. 

"'Too, " my boy. "

"And how do you pronounce 't-w-o'?"

"That is also 'too. '"

"Then how do you pronounce the second day of the week?"

"Well, that I should pronounce 'Tuesday, ' not 'Toosday. '"

"Would you really? I should pronounce it 'Monday. '"

"If you go on like this, Willie, " said Uncle John, with mock severity, "you will soon be without a friend in the world. "

"Can any of you write down quickly in figures 'twelve thousand twelvehundred and twelve pounds'?" asked Mr. Allgood. 

His eldest daughter, Miss Mildred, was the only person who happened tohave a pencil at hand. 

"It can't be done, " she declared, after making an attempt on the whitetable-cloth; but Mr. Allgood showed her that it should be written, "£13, 212. "

"Now it is my turn, " said Mildred. "I have been waiting to ask you all aquestion. In the Massacre of the Innocents under Herod, a number of poorlittle children were buried in the sand with only their feet stickingout. How might you distinguish the boys from the girls?"

"I suppose, " said Mrs. Allgood, "it is a conundrum--something to do withtheir poor little 'souls. '"

But after everybody had given it up, Mildred reminded the company thatonly boys were put to death. 

"Once upon a time, " began George, "Achilles had a race with atortoise--"

"Stop, George!" interposed Mr. Allgood. "We won't have that one. I knewtwo men in my youth who were once the best of friends, but theyquarrelled over that infernal thing of Zeno's, and they never spoke toone another again for the rest of their lives. I draw the line at that, and the other stupid thing by Zeno about the flying arrow. I don'tbelieve anybody understands them, because I could never do so myself. "

"Oh, very well, then, father. Here is another. The Post-Office peoplewere about to erect a line of telegraph-posts over a high hill fromTurmitville to Wurzleton; but as it was found that a railway company wasmaking a deep level cutting in the same direction, they arranged to putup the posts beside the line. Now, the posts were to be a hundred yardsapart, the length of the road over the hill being five miles, and thelength of the level cutting only four and a half miles. How many postsdid they save by erecting them on the level?"

"That is a very simple matter of calculation, " said Mr. Filkins. "Findhow many times one hundred yards will go in five miles, and how manytimes in four and a half miles. Then deduct one from the other, and youhave the number of posts saved by the shorter route. "

"Quite right, " confirmed Mr. Allgood. "Nothing could be easier. "

"That is just what the Post-Office people said, " replied George, "but itis quite wrong. If you look at this sketch that I have just made, youwill see that there is no difference whatever. If the posts are ahundred yards apart, just the same number will be required on the levelas over the surface of the hill. "

[Illustration]

"Surely you must be wrong, George, " said Mrs. Allgood, "for if the postsare a hundred yards apart and it is half a mile farther over the hill, you have to put up posts on that extra half-mile. "

"Look at the diagram, mother. You will see that the distance from postto post is not the distance from base to base measured along the ground. I am just the same distance from you if I stand on this spot on thecarpet or stand immediately above it on the chair. "

But Mrs. Allgood was not convinced. 

Mr. Smoothly, the curate, at the end of the table, said at this pointthat he had a little question to ask. 

"Suppose the earth were a perfect sphere with a smooth surface, and agirdle of steel were placed round the Equator so that it touched atevery point. "

"'I'll put a girdle round about the earth in forty minutes, '" mutteredGeorge, quoting the words of Puck in _A Midsummer Night's Dream_. 

"Now, if six yards were added to the length of the girdle, what wouldthen be the distance between the girdle and the earth, supposing thatdistance to be equal all round?"

"In such a great length, " said Mr. Allgood, "I do not suppose thedistance would be worth mentioning. "

"What do you say, George?" asked Mr. Smoothly. 

"Well, without calculating I should imagine it would be a very minutefraction of an inch. "

Reginald and Mr. Filkins were of the same opinion. 

"I think it will surprise you all, " said the curate, "to learn thatthose extra six yards would make the distance from the earth all roundthe girdle very nearly a yard!"

"Very nearly a yard!" everybody exclaimed, with astonishment; but Mr. Smoothly was quite correct. The increase is independent of the originallength of the girdle, which may be round the earth or round an orange;in any case the additional six yards will give a distance of nearly ayard all round. This is apt to surprise the non-mathematical mind. 

"Did you hear the story of the extraordinary precocity of Mrs. Perkins'sbaby that died last week?" asked Mrs. Allgood. "It was only three monthsold, and lying at the point of death, when the grief-stricken motherasked the doctor if nothing could save it. 'Absolutely nothing!' saidthe doctor. Then the infant looked up pitifully into its mother's faceand said--absolutely nothing!"

"Impossible!" insisted Mildred. "And only three months old!"

"There have been extraordinary cases of infantile precocity, " said Mr. Filkins, "the truth of which has often been carefully attested. But areyou sure this really happened, Mrs. Allgood?"

"Positive, " replied the lady. "But do you really think it astonishingthat a child of three months should say absolutely nothing? What wouldyou expect it to say?"

"Speaking of death, " said Mr. Smoothly, solemnly, "I knew two men, father and son, who died in the same battle during the South AfricanWar. They were both named Andrew Johnson and buried side by side, butthere was some difficulty in distinguishing them on the headstones. Whatwould you have done?"

"Quite simple, " said Mr. Allgood. "They should have described one as'Andrew Johnson, Senior, ' and the other as 'Andrew Johnson, Junior. '"

"But I forgot to tell you that the father died first. "

"What difference can that make?"

"Well, you see, they wanted to be absolutely exact, and that was thedifficulty. "

"But I don't see any difficulty, " said Mr. Allgood, nor could anybodyelse. 

"Well, " explained Mr. Smoothly, "it is like this. If the father diedfirst, the son was then no longer 'Junior. ' Is that so?"

"To be strictly exact, yes. "

"That is just what they wanted--to be strictly exact. Now, if he was nolonger 'Junior, ' then he did not die 'Junior. " Consequently it must beincorrect so to describe him on the headstone. Do you see the point?"

"Here is a rather curious thing, " said Mr. Filkins, "that I have justremembered. A man wrote to me the other day that he had recentlydiscovered two old coins while digging in his garden. One was dated '51B. C. , ' and the other one marked 'George I. ' How do I know that he wasnot writing the truth?"

"Perhaps you know the man to be addicted to lying, " said Reginald. 

"But that would be no proof that he was not telling the truth in thisinstance. "

"Perhaps, " suggested Mildred, "you know that there were no coins made atthose dates. 

"On the contrary, they were made at both periods. "

"Were they silver or copper coins?" asked Willie. 

"My friend did not state, and I really cannot see, Willie, that it makesany difference. "

"I see it!" shouted Reginald. "The letters 'B. C. ' would never be used ona coin made before the birth of Christ. They never anticipated the eventin that way. The letters were only adopted later to denote datesprevious to those which we call 'A. D. ' That is very good; but I cannotsee why the other statement could not be correct. "

"Reginald is quite right, " said Mr. Filkins, "about the first coin. Thesecond one could not exist, because the first George would never bedescribed in his lifetime as 'George I. '"

"Why not?" asked Mrs. Allgood. "He _was_ George I. "

"Yes; but they would not know it until there was a George II. "

"Then there was no George II. Until George III. Came to the throne?"

"That does not follow. The second George becomes 'George II. ' on accountof there having been a 'George I. '"

"Then the first George was 'George I. ' on account of there having beenno king of that name before him. "

"Don't you see, mother, " said George Allgood, "we did not call QueenVictoria 'Victoria I. ;' but if there is ever a 'Victoria II. , ' then shewill be known that way. "

"But there _have_ been several Georges, and therefore he was 'George I. 'There _haven't_ been several Victorias, so the two cases are notsimilar. "

They gave up the attempt to convince Mrs. Allgood, but the reader will, of course, see the point clearly. 

"Here is a question, " said Mildred Allgood, "that I should like some ofyou to settle for me. I am accustomed to buy from our greengrocerbundles of asparagus, each 12 inches in circumference. I always put atape measure round them to make sure I am getting the full quantity. Theother day the man had no large bundles in stock, but handed me insteadtwo small ones, each 6 inches in circumference. 'That is the samething, ' I said, 'and, of course, the price will be the same;' but heinsisted that the two bundles together contained more than the largeone, and charged me a few pence extra. Now, what I want to know is, which of us was correct? Would the two small bundles contain the samequantity as the large one? Or would they contain more?"

"That is the ancient puzzle, " said Reginald, laughing, "of the sack ofcorn that Sempronius borrowed from Caius, which your greengrocer, perhaps, had been reading about somewhere. He caught you beautifully. "

"Then they were equal?"

"On the contrary, you were both wrong, and you were badly cheated. Youonly got half the quantity that would have been contained in a largebundle, and therefore ought to have been charged half the originalprice, instead of more. "

Yes, it was a bad swindle, undoubtedly. A circle with a circumferencehalf that of another must have its area a quarter that of the other. Therefore the two small bundles contained together only half as muchasparagus as a large one. 

"Mr. Filkins, can you answer this?" asked Willie. "There is a man in thenext village who eats two eggs for breakfast every morning. "

"Nothing very extraordinary in that, " George broke in. "If you told usthat the two eggs ate the man it would be interesting. "

"Don't interrupt the boy, George, " said his mother. 

"Well, " Willie continued, "this man neither buys, borrows, barters, begs, steals, nor finds the eggs. He doesn't keep hens, and the eggs arenot given to him. How does he get the eggs?"

"Does he take them in exchange for something else?" asked Mildred. 

"That would be bartering them, " Willie replied. 

"Perhaps some friend sends them to him, " suggested Mrs. Allgood. 

"I said that they were not given to him. "

"I know, " said George, with confidence. "A strange hen comes into hisplace and lays them. "

"But that would be finding them, wouldn't it?"

"Does he hire them?" asked Reginald. 

"If so, he could not return them after they were eaten, so that would bestealing them. "

"Perhaps it is a pun on the word 'lay, '" Mr. Filkins said. "Does he laythem on the table?"

"He would have to get them first, wouldn't he? The question was, Howdoes he get them?"

"Give it up!" said everybody. Then little Willie crept round to theprotection of his mother, for George was apt to be rough on suchoccasions. 

"The man keeps ducks!" he cried, "and his servant collects the eggsevery morning. "

"But you said he doesn't keep birds!" George protested. 

"I didn't, did I, Mr. Filkins? I said he doesn't keep hens. "

"But he finds them, " said Reginald. 

"No; I said his servant finds them. "

"Well, then, " Mildred interposed, "his servant gives them to him. "

"You cannot give a man his own property, can you?"

All agreed that Willie's answer was quite satisfactory. Then Uncle Johnproduced a little fallacy that "brought the proceedings to a close, " asthe newspapers say. 

413. --A CHESSBOARD FALLACY. 

[Illustration]

"Here is a diagram of a chessboard, " he said. "You see there aresixty-four squares--eight by eight. Now I draw a straight line from thetop left-hand corner, where the first and second squares meet, to thebottom right-hand corner. I cut along this line with the scissors, slideup the piece that I have marked B, and then clip off the little corner Cby a cut along the first upright line. This little piece will exactlyfit into its place at the top, and we now have an oblong with sevensquares on one side and nine squares on the other. There are, therefore, now only sixty-three squares, because seven multiplied by nine makessixty-three. Where on earth does that lost square go to? I have triedover and over again to catch the little beggar, but he always eludes me. For the life of me I cannot discover where he hides himself. "

"It seems to be like the other old chessboard fallacy, and perhaps theexplanation is the same, " said Reginald--"that the pieces do not exactlyfit. "

"But they _do_ fit, " said Uncle John. "Try it, and you will see. "

Later in the evening Reginald and George, were seen in a corner withtheir heads together, trying to catch that elusive little square, and itis only fair to record that before they retired for the night theysucceeded in securing their prey, though some others of the companyfailed to see it when captured. Can the reader solve the little mystery?

UNCLASSIFIED PROBLEMS. 

 "A snapper up of unconsidered trifles. " _Winter's Tale_, iv. 2. 

414. --WHO WAS FIRST?

Anderson, Biggs, and Carpenter were staying together at a place by theseaside. One day they went out in a boat and were a mile at sea when arifle was fired on shore in their direction. Why or by whom the shot wasfired fortunately does not concern us, as no information on these pointsis obtainable, but from the facts I picked up we can get material for acurious little puzzle for the novice. 

It seems that Anderson only heard the report of the gun, Biggs only sawthe smoke, and Carpenter merely saw the bullet strike the water nearthem. Now, the question arises: Which of them first knew of thedischarge of the rifle?

415. --A WONDERFUL VILLAGE. 

There is a certain village in Japan, situated in a very low valley, andyet the sun is nearer to the inhabitants every noon, by 3, 000 miles andupwards, than when he either rises or sets to these people. In what partof the country is the village situated?

416. --A CALENDAR PUZZLE. 

If the end of the world should come on the first day of a new century, can you say what are the chances that it will happen on a Sunday?

417. --THE TIRING IRONS. 

[Illustration]

The illustration represents one of the most ancient of all mechanicalpuzzles. Its origin is unknown. Cardan, the mathematician, wrote aboutit in 1550, and Wallis in 1693; while it is said still to be found inobscure English villages (sometimes deposited in strange places, such asa church belfry), made of iron, and appropriately called "tiring-irons, "and to be used by the Norwegians to-day as a lock for boxes and bags. Inthe toyshops it is sometimes called the "Chinese rings, " though thereseems to be no authority for the description, and it more frequentlygoes by the unsatisfactory name of "the puzzling rings. " The French callit "Baguenaudier. "

The puzzle will be seen to consist of a simple _loop_ of wire fixed in ahandle to be held in the left hand, and a certain number of _rings_secured by _wires_ which pass through holes in the _bar_ and are keptthere by their blunted ends. The wires work freely in the bar, butcannot come apart from it, nor can the wires be removed from the rings. The general puzzle is to detach the loop completely from all the rings, and then to put them all on again. 

Now, it will be seen at a glance that the first ring (to the right) canbe taken off at any time by sliding it over the end and dropping itthrough the loop; or it may be put on by reversing the operation. Withthis exception, the only ring that can ever be removed is the one thathappens to be a contiguous second on the loop at the right-hand end. Thus, with all the rings on, the second can be dropped at once; with thefirst ring down, you cannot drop the second, but may remove the third;with the first three rings down, you cannot drop the fourth, but mayremove the fifth; and so on. It will be found that the first and secondrings can be dropped together or put on together; but to preventconfusion we will throughout disallow this exceptional double move, andsay that only one ring may be put on or removed at a time. 

We can thus take off one ring in 1 move; two rings in 2 moves; threerings in 5 moves; four rings in 10 moves; five rings in 21 moves; and ifwe keep on doubling (and adding one where the number of rings is odd) wemay easily ascertain the number of moves for completely removing anynumber of rings. To get off all the seven rings requires 85 moves. Letus look at the five moves made in removing the first three rings, thecircles above the line standing for rings on the loop and those underfor rings off the loop. 

Drop the first ring; drop the third; put up the first; drop the second;and drop the first--5 moves, as shown clearly in the diagrams. The darkcircles show at each stage, from the starting position to the finish, which rings it is possible to drop. After move 2 it will be noticed thatno ring can be dropped until one has been put on, because the first andsecond rings from the right now on the loop are not together. After thefifth move, if we wish to remove all seven rings we must now drop thefifth. But before we can then remove the fourth it is necessary to puton the first three and remove the first two. We shall then have 7, 6, 4, 3 on the loop, and may therefore drop the fourth. When we have put on 2and 1 and removed 3, 2, 1, we may drop the seventh ring. The nextoperation then will be to get 6, 5, 4, 3, 2, 1 on the loop and remove 4, 3, 2, 1, when 6 will come off; then get 5, 4, 3, 2, 1 on the loop, andremove 3, 2, 1, when 5 will come off; then get 4, 3, 2, 1 on the loopand remove 2, 1, when 4 will come off; then get 3, 2, 1 on the loop andremove 1, when 3 will come off; then get 2, 1 on the loop, when 2 willcome off; and 1 will fall through on the 85th move, leaving the loopquite free. The reader should now be able to understand the puzzle, whether or not he has it in his hand in a practical form. 

[Illustration]

[Illustration:

 o o o o o * * {-------------

 o o o o * o 1{------------- o

 o o o o o 2{------------- o o

 o o o o * * 3{------------- o

 o o o o * 4{------------- o o

 o o * o 5{------------- o o o

]

The particular problem I propose is simply this. Suppose there arealtogether fourteen rings on the tiring-irons, and we proceed to takethem all off in the correct way so as not to waste any moves. What willbe the position of the rings after the 9, 999th move has been made?

418. --SUCH A GETTING UPSTAIRS. 

In a suburban villa there is a small staircase with eight steps, notcounting the landing. The little puzzle with which Tommy Smart perplexedhis family is this. You are required to start from the bottom and landtwice on the floor above (stopping there at the finish), having returnedonce to the ground floor. But you must be careful to use every tread thesame number of times. In how few steps can you make the ascent? It seemsa very simple matter, but it is more than likely that at your firstattempt you will make a great many more steps than are necessary. Ofcourse you must not go more than one riser at a time. 

Tommy knows the trick, and has shown it to his father, who professes tohave a contempt for such things; but when the children are in bed thepater will often take friends out into the hall and enjoy a good laughat their bewilderment. And yet it is all so very simple when you knowhow it is done. 

419. --THE FIVE PENNIES. 

Here is a really hard puzzle, and yet its conditions are so absurdlysimple. Every reader knows how to place four pennies so that they areequidistant from each other. All you have to do is to arrange three ofthem flat on the table so that they touch one another in the form of atriangle, and lay the fourth penny on top in the centre. Then, as everypenny touches every other penny, they are all at equal distances fromone another. Now try to do the same thing with five pennies--place themso that every penny shall touch every other penny--and you will find ita different matter altogether. 

420. --THE INDUSTRIOUS BOOKWORM. 

[Illustration]

Our friend Professor Rackbrane is seen in the illustration to bepropounding another of his little posers. He is explaining that since helast had occasion to take down those three volumes of a learned bookfrom their place on his shelves a bookworm has actually bored a holestraight through from the first page to the last. He says that theleaves are together three inches thick in each volume, and that everycover is exactly one-eighth of an inch thick, and he asks how long atunnel had the industrious worm to bore in preparing his new tuberailway. Can you tell him?

421. --A CHAIN PUZZLE. 

[Illustration]

This is a puzzle based on a pretty little idea first dealt with by thelate Mr. Sam Loyd. A man had nine pieces of chain, as shown in theillustration. He wanted to join these fifty links into one endlesschain. It will cost a penny to open any link and twopence to weld a linktogether again, but he could buy a new endless chain of the samecharacter and quality for 2s. 2d. What was the cheapest course for himto adopt? Unless the reader is cunning he may find himself a good wayout in his answer. 

422. --THE SABBATH PUZZLE. 

I have come across the following little poser in an old book. I wonderhow many readers will see the author's intended solution to the riddle. 

 Christians the week's _first_ day for Sabbath hold; The Jews the _seventh_, as they did of old; The Turks the _sixth_, as we have oft been told. How can these three, in the same place and day, Have each his own true Sabbath? tell, I pray. 

423. --THE RUBY BROOCH. 

The annals of Scotland Yard contain some remarkable cases of jewelrobberies, but one of the most perplexing was the theft of LadyLittlewood's rubies. There have, of course, been many greater robberiesin point of value, but few so artfully conceived. Lady Littlewood, ofRomley Manor, had a beautiful but rather eccentric heirloom in the formof a ruby brooch. While staying at her town house early in the eightiesshe took the jewel to a shop in Brompton for some slight repairs. 

"A fine collection of rubies, madam, " said the shopkeeper, to whom herladyship was a stranger. 

"Yes, " she replied; "but curiously enough I have never actually countedthem. My mother once pointed out to me that if you start from the centreand count up one line, along the outside and down the next line, thereare always eight rubies. So I should always know if a stone weremissing. "

[Illustration]

Six months later a brother of Lady Littlewood's, who had returned fromhis regiment in India, noticed that his sister was wearing the rubybrooch one night at a county ball, and on their return home asked tolook at it more closely. He immediately detected the fact that four ofthe stones were gone. 

"How can that possibly be?" said Lady Littlewood. "If you count up oneline from the centre, along the edge, and down the next line, in anydirection, there are always eight stones. This was always so and is sonow. How, therefore, would it be possible to remove a stone without mydetecting it?"

"Nothing could be simpler, " replied the brother. "I know the broochwell. It originally contained forty-five stones, and there are now onlyforty-one. Somebody has stolen four rubies, and then reset as small anumber of the others as possible in such a way that there shall alwaysbe eight in any of the directions you have mentioned. "

There was not the slightest doubt that the Brompton jeweller was thethief, and the matter was placed in the hands of the police. But the manwas wanted for other robberies, and had left the neighbourhood some timebefore. To this day he has never been found. 

The interesting little point that at first baffled the police, and whichforms the subject of our puzzle, is this: How were the forty-five rubiesoriginally arranged on the brooch? The illustration shows exactly howthe forty-one were arranged after it came back from the jeweller; butalthough they count eight correctly in any of the directions mentioned, there are four stones missing. 

424. --THE DOVETAILED BLOCK. 

[Illustration]

Here is a curious mechanical puzzle that was given to me some years ago, but I cannot say who first invented it. It consists of two solid blocksof wood securely dovetailed together. On the other two vertical sidesthat are not visible the appearance is precisely the same as on thoseshown. How were the pieces put together? When I published this littlepuzzle in a London newspaper I received (though they were unsolicited)quite a stack of models, in oak, in teak, in mahogany, rosewood, satinwood, elm, and deal; some half a foot in length, and others varyingin size right down to a delicate little model about half an inch square. It seemed to create considerable interest. 

425. --JACK AND THE BEANSTALK. 

[Illustration]

The illustration, by a British artist, is a sketch of Jack climbing thebeanstalk. Now, the artist has made a serious blunder in this drawing. Can you find out what it is?

426. --THE HYMN-BOARD POSER. 

The worthy vicar of Chumpley St. Winifred is in great distress. A littlechurch difficulty has arisen that all the combined intelligence of theparish seems unable to surmount. What this difficulty is I will statehereafter, but it may add to the interest of the problem if I first givea short account of the curious position that has been brought about. Itall has to do with the church hymn-boards, the plates of which havebecome so damaged that they have ceased to fulfil the purpose for whichthey were devised. A generous parishioner has promised to pay for a newset of plates at a certain rate of cost; but strange as it may seem, noagreement can be come to as to what that cost should be. The proposedmaker of the plates has named a price which the donor declares to beabsurd. The good vicar thinks they are both wrong, so he asks theschoolmaster to work out the little sum. But this individual declaresthat he can find no rule bearing on the subject in any of his arithmeticbooks. An application having been made to the local medicalpractitioner, as a man of more than average intellect at Chumpley, hehas assured the vicar that his practice is so heavy that he has not hadtime even to look at it, though his assistant whispers that the doctorhas been sitting up unusually late for several nights past. Widow Wilsonhas a smart son, who is reputed to have once won a prize forpuzzle-solving. He asserts that as he cannot find any solution to theproblem it must have something to do with the squaring of the circle, the duplication of the cube, or the trisection of an angle; at any rate, he has never before seen a puzzle on the principle, and he gives it up. 

[Illustration]

This was the state of affairs when the assistant curate (who, I shouldsay, had frankly confessed from the first that a profound study oftheology had knocked out of his head all the knowledge of mathematics heever possessed) kindly sent me the puzzle. 

A church has three hymn-boards, each to indicate the numbers of fivedifferent hymns to be sung at a service. All the boards are in use atthe same service. The hymn-book contains 700 hymns. A new set of numbersis required, and a kind parishioner offers to present a set painted onmetal plates, but stipulates that only the smallest number of platesnecessary shall be purchased. The cost of each plate is to be 6d. , andfor the painting of each plate the charges are to be: For one plate, 1s. ; for two plates alike, 11¾d. Each; for three plates alike, 11½d. Each, and so on, the charge being one farthing less per platefor each similarly painted plate. Now, what should be the lowest cost?

Readers will note that they are required to use every legitimate andpractical method of economy. The illustration will make clear the natureof the three hymn-boards and plates. The five hymns are here indicatedby means of twelve plates. These plates slide in separately at the back, and in the illustration there is room, of course, for three more plates. 

427. --PHEASANT-SHOOTING. 

A Cockney friend, who is very apt to draw the long bow, and is evidentlyless of a sportsman than he pretends to be, relates to me the followingnot very credible yarn:--

"I've just been pheasant-shooting with my friend the duke. We hadsplendid sport, and I made some wonderful shots. What do you think ofthis, for instance? Perhaps you can twist it into a puzzle. The duke andI were crossing a field when suddenly twenty-four pheasants rose on thewing right in front of us. I fired, and two-thirds of them dropped deadat my feet. Then the duke had a shot at what were left, and brought downthree-twenty-fourths of them, wounded in the wing. Now, out of thosetwenty-four birds, how many still remained?"

It seems a simple enough question, but can the reader give a correctanswer?

428. --THE GARDENER AND THE COOK. 

A correspondent, signing himself "Simple Simon, " suggested that I shouldgive a special catch puzzle in the issue of _The Weekly Dispatch_ forAll Fools' Day, 1900. So I gave the following, and it causedconsiderable amusement; for out of a very large body of competitors, many quite expert, not a single person solved it, though it ran fornearly a month. 

[Illustration]

"The illustration is a fancy sketch of my correspondent, 'Simple Simon, 'in the act of trying to solve the following innocent little arithmeticalpuzzle. A race between a man and a woman that I happened to witness oneAll Fools' Day has fixed itself indelibly on my memory. It happened at acountry-house, where the gardener and the cook decided to run a race toa point 100 feet straight away and return. I found that the gardener ran3 feet at every bound and the cook only 2 feet, but then she made threebounds to his two. Now, what was the result of the race?"

A fortnight after publication I added the following note: "It has beensuggested that perhaps there is a catch in the 'return, ' but there isnot. The race is to a point 100 feet away and home again--that is, adistance of 200 feet. One correspondent asks whether they take exactlythe same time in turning, to which I reply that they do. Another seemsto suspect that it is really a conundrum, and that the answer is that'the result of the race was a (matrimonial) tie. ' But I had no suchintention. The puzzle is an arithmetical one, as it purports to be. "

429. --PLACING HALFPENNIES. 

[Illustration]

Here is an interesting little puzzle suggested to me by Mr. W. T. Whyte. Mark off on a sheet of paper a rectangular space 5 inches by 3 inches, and then find the greatest number of halfpennies that can be placedwithin the enclosure under the following conditions. A halfpenny isexactly an inch in diameter. Place your first halfpenny where you like, then place your second coin at exactly the distance of an inch from thefirst, the third an inch distance from the second, and so on. Nohalfpenny may touch another halfpenny or cross the boundary. Ourillustration will make the matter perfectly clear. No. 2 coin is an inchfrom No. 1; No. 3 an inch from No. 2; No. 4 an inch from No. 3; butafter No. 10 is placed we can go no further in this attempt. Yet severalmore halfpennies might have been got in. How many can the reader place?

430. --FIND THE MAN'S WIFE. 

[Illustration]

One summer day in 1903 I was loitering on the Brighton front, watchingthe people strolling about on the beach, when the friend who was with mesuddenly drew my attention to an individual who was standing alone, andsaid, "Can you point out that man's wife? They are stopping at the samehotel as I am, and the lady is one of those in view. " After a fewminutes' observation, I was successful in indicating the lady correctly. My friend was curious to know by what method of reasoning I had arrivedat the result. This was my answer:--

"We may at once exclude that Sister of Mercy and the girl in the shortfrock; also the woman selling oranges. It cannot be the lady in widows'weeds. It is not the lady in the bath chair, because she is not stayingat your hotel, for I happened to see her come out of a private housethis morning assisted by her maid. The two ladies in red breakfasted atmy hotel this morning, and as they were not wearing outdoor dress Iconclude they are staying there. It therefore rests between the lady inblue and the one with the green parasol. But the left hand that holdsthe parasol is, you see, ungloved and bears no wedding-ring. Consequently I am driven to the conclusion that the lady in blue is theman's wife--and you say this is correct. "

Now, as my friend was an artist, and as I thought an amusing puzzlemight be devised on the lines of his question, I asked him to make me adrawing according to some directions that I gave him, and I havepleasure in presenting his production to my readers. It will be seenthat the picture shows six men and six ladies: Nos. 1, 3, 5, 7, 9, and11 are ladies, and Nos. 2, 4, 6, 8, 10, and 12 are men. These twelveindividuals represent six married couples, all strangers to one another, who, in walking aimlessly about, have got mixed up. But we are onlyconcerned with the man that is wearing a straw hat--Number 10. Thepuzzle is to find this man's wife. Examine the six ladies carefully, andsee if you can determine which one of them it is. 

I showed the picture at the time to a few friends, and they expressedvery different opinions on the matter. One said, "I don't believe hewould marry a girl like Number 7. " Another said, "I am sure a nice girllike Number 3 would not marry such a fellow!" Another said, "It must beNumber 1, because she has got as far away as possible from the brute!"It was suggested, again, that it must be Number 11, because "he seems tobe looking towards her;" but a cynic retorted, "For that very reason, ifhe is really looking at her, I should say that she is not his wife!"

I now leave the question in the hands of my readers. Which is reallyNumber 10's wife?

The illustration is of necessity considerably reduced from the largescale on which it originally appeared in _The Weekly Dispatch_ (24th May1903), but it is hoped that the details will be sufficiently clear toallow the reader to derive entertainment from its examination. In anycase the solution given will enable him to follow the points withinterest. 

SOLUTIONS. 

1. --A POST-OFFICE PERPLEXITY. 

The young lady supplied 5 twopenny stamps, 30 penny stamps, and 8twopence-halfpenny stamps, which delivery exactly fulfils the conditionsand represents a cost of five shillings. 

2. --YOUTHFUL PRECOCITY. 

The price of the banana must have been one penny farthing. Thus, 960bananas would cost £5, and 480 sixpences would buy 2, 304 bananas. 

3. --AT A CATTLE MARKET. 

Jakes must have taken 7 animals to market, Hodge must have taken 11, andDurrant must have taken 21. There were thus 39 animals altogether. 

4. --THE BEANFEAST PUZZLE. 

The cobblers spent 35s. , the tailors spent also 35s. , the hatters spent42s. , and the glovers spent 21s. Thus, they spent altogether £6, 13s. , while it will be found that the five cobblers spent as much as fourtailors, twelve tailors as much as nine hatters, and six hatters as muchas eight glovers. 

5. --A QUEER COINCIDENCE. 

Puzzles of this class are generally solved in the old books by thetedious process of "working backwards. " But a simple general solution isas follows: If there are n players, the amount held by every player atthe end will be m(2^n), the last winner must have held m(n + 1)at the start, the next m(2n + 1), the next m(4n + 1), the nextm(8n + 1), and so on to the first player, who must have heldm(2^{n - 1}n + 1). 

Thus, in this case, n = 7, and the amount held by every player at theend was 2^7 farthings. Therefore m = 1, and G started with 8 farthings, F with 15, E with 29, D with 57, C with 113, B with 225, and A with 449farthings. 

6. --A CHARITABLE BEQUEST. 

There are seven different ways in which the money may be distributed: 5women and 19 men, 10 women and 16 men, 15 women and 13 men, 20 women and10 men, 25 women and 7 men, 30 women and 4 men, and 35 women and 1 man. But the last case must not be counted, because the condition was thatthere should be "men, " and a single man is not men. Therefore the answeris six years. 

7. --THE WIDOW'S LEGACY. 

The widow's share of the legacy must be £205, 2s. 6d. And 10/13 of apenny. 

8. --INDISCRIMINATE CHARITY

The gentleman must have had 3s. 6d. In his pocket when he set out forhome. 

9. --THE TWO AEROPLANES. 

The man must have paid £500 and £750 for the two machines, makingtogether £1, 250; but as he sold them for only £1, 200, he lost £50 by thetransaction. 

10. --BUYING PRESENTS. 

Jorkins had originally £19, 18s. In his pocket, and spent £9, 19s. 

11. --THE CYCLISTS' FEAST. 

There were ten cyclists at the feast. They should have paid 8s. Each;but, owing to the departure of two persons, the remaining eight wouldpay 10s. Each. 

12. --A QUEER THING IN MONEY. 

The answer is as follows: £44, 444, 4s. 4d. = 28, and, reduced to pence, 10, 666, 612=28. 

It is a curious little coincidence that in the answer 10, 666, 612 thefour central figures indicate the only other answer, £66, 6s. 6d. 

13. --A NEW MONEY PUZZLE. 

The smallest sum of money, in pounds, shillings, pence, and farthings, containing all the nine digits once, and once only, is £2, 567, 18s. 9¾d. 

14. --SQUARE MONEY. 

The answer is 1½d. And 3d. Added together they make 4½d. , and1½d. Multiplied by 3 is also 4½d. 

15. --POCKET MONEY. 

The largest possible sum is 15s. 9d. , composed of a crown and ahalf-crown (or three half-crowns), four florins, and a threepenny piece. 

16. --THE MILLIONAIRE'S PERPLEXITY. 

The answer to this quite easy puzzle may, of course, be readily obtainedby trial, deducting the largest power of 7 that is contained in onemillion dollars, then the next largest power from the remainder, and soon. But the little problem is intended to illustrate a simple directmethod. The answer is given at once by converting 1, 000, 000 to theseptenary scale, and it is on this subject of scales of notation that Ipropose to write a few words for the benefit of those who have neversufficiently considered the matter. 

Our manner of figuring is a sort of perfected arithmetical shorthand, asystem devised to enable us to manipulate numbers as rapidly andcorrectly as possible by means of symbols. If we write the number 2, 341to represent two thousand three hundred and forty-one dollars, we wishto imply 1 dollar, added to four times 10 dollars, added to three times100 dollars, added to two times 1, 000 dollars. From the number in theunits place on the right, every figure to the left is understood torepresent a multiple of the particular power of 10 that its positionindicates, while a cipher (0) must be inserted where necessary in orderto prevent confusion, for if instead of 207 we wrote 27 it would beobviously misleading. We thus only require ten figures, because directlya number exceeds 9 we put a second figure to the left, directly itexceeds 99 we put a third figure to the left, and so on. It will be seenthat this is a purely arbitrary method. It is working in the denary (orten) scale of notation, a system undoubtedly derived from the fact thatour forefathers who devised it had ten fingers upon which they wereaccustomed to count, like our children of to-day. It is unnecessary forus ordinarily to state that we are using the denary scale, because thisis always understood in the common affairs of life. 

But if a man said that he had 6, 553 dollars in the septenary (or seven)scale of notation, you will find that this is precisely the same amountas 2, 341 in our ordinary denary scale. Instead of using powers of ten, he uses powers of 7, so that he never needs any figure higher than 6, and 6, 553 really stands for 3, added to five times 7, added to fivetimes 49, added to six times 343 (in the ordinary notation), or 2, 341. To reverse the operation, and convert 2, 341 from the denary to theseptenary scale, we divide it by 7, and get 334 and remainder 3; divide334 by 7, and get 47 and remainder 5; and so keep on dividing by 7 aslong as there is anything to divide. The remainders, read backwards, 6, 5, 5, 3, give us the answer, 6, 553. 

Now, as I have said, our puzzle may be solved at once by merelyconverting 1, 000, 000 dollars to the septenary scale. Keep on dividingthis number by 7 until there is nothing more left to divide, and theremainders will be found to be 11333311 which is 1, 000, 000 expressed inthe septenary scale. Therefore, 1 gift of 1 dollar, 1 gift of 7 dollars, 3 gifts of 49 dollars, 3 gifts of 343 dollars, 3 gifts of 2, 401 dollars, 3 gifts of 16, 807 dollars, 1 gift of 117, 649 dollars, and onesubstantial gift of 823, 543 dollars, satisfactorily solves our problem. And it is the only possible solution. It is thus seen that no "trials"are necessary; by converting to the septenary scale of notation we godirect to the answer. 

17. --THE PUZZLING MONEY BOXES. 

The correct answer to this puzzle is as follows: John put into hismoney-box two double florins (8s. ), William a half-sovereign and aflorin (12s. ), Charles a crown (5s. ), and Thomas a sovereign (20s. ). There are six coins in all, of a total value of 45s. If John had 2s. More, William 2s. Less, Charles twice as much, and Thomas half as muchas they really possessed, they would each have had exactly 10s. 

18. --THE MARKET WOMEN. 

The price received was in every case 105 farthings. Therefore thegreatest number of women is eight, as the goods could only be sold atthe following rates: 105 lbs. At 1 farthing, 35 at 3, 21 at 5, 15 at 7, 7 at 15, 5 at 21, 3 at 35, and 1 lb. At 105 farthings. 

19. --THE NEW YEAR'S EVE SUPPERS. 

The company present on the occasion must have consisted of seven pairs, ten single men, and one single lady. Thus, there were twenty-fivepersons in all, and at the prices stated they would pay exactly £5together. 

20. --BEEF AND SAUSAGES. 

The lady bought 48 lbs. Of beef at 2s. , and the same quantity ofsausages at 1s. 6d. , thus spending £8, 8s. Had she bought 42 lbs. Ofbeef and 56 lbs. Of sausages she would have spent £4, 4s. On each, andhave obtained 98 lbs. Instead of 96 lbs. --a gain in weight of 2 lbs. 

21. --A DEAL IN APPLES. 

I was first offered sixteen apples for my shilling, which would be atthe rate of ninepence a dozen. The two extra apples gave me eighteen fora shilling, which is at the rate of eightpence a dozen, or one penny adozen less than the first price asked. 

22. --A DEAL IN EGGS. 

The man must have bought ten eggs at fivepence, ten eggs at one penny, and eighty eggs at a halfpenny. He would then have one hundred eggs at acost of eight shillings and fourpence, and the same number of eggs oftwo of the qualities. 

23. --THE CHRISTMAS-BOXES. 

The distribution took place "some years ago, " when the fourpenny-piecewas in circulation. Nineteen persons must each have received nineteenpence. There are five different ways in which this sum may have beenpaid in silver coins. We need only use two of these ways. Thus iffourteen men each received four four-penny-pieces and onethreepenny-piece, and five men each received five threepenny-pieces andone fourpenny-piece, each man would receive nineteen pence, and therewould be exactly one hundred coins of a total value of £1, 10s. 1d. 

24. --A SHOPPING PERPLEXITY. 

The first purchase amounted to 1s. 5¾d. , the second to 1s. 11½d. , and together they make 3s. 5¼d. Not one of these three amounts can bepaid in fewer than six current coins of the realm. 

25. --CHINESE MONEY. 

As a ching-chang is worth twopence and four-fifteenths of a ching-chang, the remaining eleven-fifteenths of a ching-chang must be worth twopence. Therefore eleven ching-changs are worth exactly thirty pence, or half acrown. Now, the exchange must be made with seven round-holed coins andone square-holed coin. Thus it will be seen that 7 round-holed coins areworth seven-elevenths of 15 ching-changs, and 1 square-holed coin isworth one-eleventh of 16 ching-changs--that is, 77 rounds equal 105ching-changs and 11 squares equal 16 ching-changs. Therefore 77 roundsadded to 11 squares equal 121 ching-changs; or 7 rounds and 1 squareequal 11 ching-changs, or its equivalent, half a crown. This is moresimple in practice than it looks here. 

26. --THE JUNIOR CLERKS' PUZZLE. 

Although Snoggs's _reason_ for wishing to take his rise at £2, 10s. Half-yearly did not concern our puzzle, the _fact_ that he was dupinghis employer into paying him more than was intended did concern it. Manyreaders will be surprised to find that, although Moggs only received£350 in five years, the artful Snoggs actually obtained £362, 10s. Inthe same time. The rest is simplicity itself. It is evident that ifMoggs saved £87, 10s. And Snoggs £181, 5s. , the latter would be savingtwice as great a proportion of his salary as the former (namely, one-half as against one-quarter), and the two sums added together make£268, 15s. 

27. --GIVING CHANGE. 

The way to help the American tradesman out of his dilemma is this. Describing the coins by the number of cents that they represent, thetradesman puts on the counter 50 and 25; the buyer puts down 100, 3, and2; the stranger adds his 10, 10, 5, 2, and 1. Now, considering that thecost of the purchase amounted to 34 cents, it is clear that out of thispooled money the tradesman has to receive 109, the buyer 71, and thestranger his 28 cents. Therefore it is obvious at a glance that the100-piece must go to the tradesman, and it then follows that the50-piece must go to the buyer, and then the 25-piece can only go to thestranger. Another glance will now make it clear that the two 10-centpieces must go to the buyer, because the tradesman now only wants 9 andthe stranger 3. Then it becomes obvious that the buyer must take the 1cent, that the stranger must take the 3 cents, and the tradesman the 5, 2, and 2. To sum up, the tradesman takes 100, 5, 2, and 2; the buyer, 50, 10, 10, and 1; the stranger, 25 and 3. It will be seen that not oneof the three persons retains any one of his own coins. 

28. --DEFECTIVE OBSERVATION. 

Of course the date on a penny is on the same side as Britannia--the"tail" side. Six pennies may be laid around another penny, all flat onthe table, so that every one of them touches the central one. The numberof threepenny-pieces that may be laid on the surface of a half-crown, sothat no piece lies on another or overlaps the edge of the half-crown, isone. A second threepenny-piece will overlap the edge of the larger coin. Few people guess fewer than three, and many persons give an absurdlyhigh number. 

29. --THE BROKEN COINS. 

If the three broken coins when perfect were worth 253 pence, and are nowin their broken condition worth 240 pence, it should be obvious that13/253 of the original value has been lost. And as the same fraction ofeach coin has been broken away, each coin has lost 13/253 of itsoriginal bulk. 

30. --TWO QUESTIONS IN PROBABILITIES. 

In tossing with the five pennies all at the same time, it is obviousthat there are 32 different ways in which the coins may fall, becausethe first coin may fall in either of two ways, then the second coin mayalso fall in either of two ways, and so on. Therefore five 2'smultiplied together make 32. Now, how are these 32 ways made up? Herethey are:--

 (a) 5 heads 1 way (b) 5 tails 1 way (c) 4 heads and 1 tail 5 ways (d) 4 tails and 1 head 5 ways (e) 3 heads and 2 tails 10 ways (f) 3 tails and 2 heads 10 ways

Now, it will be seen that the only favourable cases are a, b, c, and d--12 cases. The remaining 20 cases are unfavourable, because theydo not give at least four heads or four tails. Therefore the chances areonly 12 to 20 in your favour, or (which is the same thing) 3 to 5. Putanother way, you have only 3 chances out of 8. 

The amount that should be paid for a draw from the bag that containsthree sovereigns and one shilling is 15s. 3d. Many persons will saythat, as one's chances of drawing a sovereign were 3 out of 4, oneshould pay three-fourths of a pound, or 15s. , overlooking the fact thatone must draw at least a shilling--there being no blanks. 

31. --DOMESTIC ECONOMY. 

Without the hint that I gave, my readers would probably have beenunanimous in deciding that Mr. Perkins's income must have been £1, 710. But this is quite wrong. Mrs. Perkins says, "We have spent a third ofhis yearly income in rent, " etc. , etc. --that is, in two years they havespent an amount in rent, etc. , equal to one-third of his yearly income. Note that she does _not_ say that they have spent _each year_ this sum, whatever it is, but that _during the two years_ that amount has beenspent. The only possible answer, according to the exact reading of herwords, is, therefore, that his income was £180 per annum. Thus theamount spent in two years, during which his income has amounted to £360, will be £60 in rent, etc. , £90 in domestic expenses, £20 in other ways, leaving the balance of £190 in the bank as stated. 

32. --THE EXCURSION TICKET PUZZLE. 

Nineteen shillings and ninepence may be paid in 458, 908, 622 differentways. 

I do not propose to give my method of solution. Any such explanationwould occupy an amount of space out of proportion to its interest orvalue. If I could give within reasonable limits a general solution forall money payments, I would strain a point to find room; but such asolution would be extremely complex and cumbersome, and I do notconsider it worth the labour of working out. 

Just to give an idea of what such a solution would involve, I willmerely say that I find that, dealing only with those sums of money thatare multiples of threepence, if we only use bronze coins any sum can bepaid in (n + 1)² ways where n always represents the number ofpence. If threepenny-pieces are admitted, there are

 2n³ + 15n² + 33n --------------------- + 1 ways. 18

If sixpences are also used there are

 n^{4} + 22n³ + 159n² + 414n + 216 --------------------------------- 216

ways, when the sum is a multiple of sixpence, and the constant, 216, changes to 324 when the money is not such a multiple. And so theformulas increase in complexity in an accelerating ratio as we go on tothe other coins. 

I will, however, add an interesting little table of the possible ways ofchanging our current coins which I believe has never been given in abook before. Change may be given for a

 Farthing in 0 way. Halfpenny in 1 way. Penny in 3 ways. Threepenny-piece in 16 ways. Sixpence in 66 ways. Shilling in 402 ways. Florin in 3, 818 ways. Half-crown in 8, 709 ways. Double florin in 60, 239 ways. Crown in 166, 651 ways. Half-sovereign in 6, 261, 622 ways. Sovereign in 500, 291, 833 ways. 

It is a little surprising to find that a sovereign may be changed inover five hundred million different ways. But I have no doubt as to thecorrectness of my figures. 

33. --A PUZZLE IN REVERSALS. 

(i) £13. (2) £23, 19s. 11d. The words "the number of pounds exceeds thatof the pence" exclude such sums of money as £2, 16s. 2d. And all sumsunder £1. 

34. --THE GROCER AND DRAPER. 

The grocer was delayed half a minute and the draper eight minutes and ahalf (seventeen times as long as the grocer), making together nineminutes. Now, the grocer took twenty-four minutes to weigh out thesugar, and, with the half-minute delay, spent 24 min. 30 sec. Over thetask; but the draper had only to make _forty-seven_ cuts to divide theroll of cloth, containing forty-eight yards, into yard pieces! This tookhim 15 min. 40 sec. , and when we add the eight minutes and a half delaywe get 24 min. 10 sec. , from which it is clear that the draper won therace by twenty seconds. The majority of solvers make forty-eight cuts todivide the roll into forty-eight pieces!

35. --JUDKINS'S CATTLE. 

As there were five droves with an equal number of animals in each drove, the number must be divisible by 5; and as every one of the eight dealersbought the same number of animals, the number must be divisible by 8. Therefore the number must be a multiple of 40. The highest possiblemultiple of 40 that will work will be found to be 120, and this numbercould be made up in one of two ways--1 ox, 23 pigs, and 96 sheep, or 3oxen, 8 pigs, and 109 sheep. But the first is excluded by the statementthat the animals consisted of "oxen, pigs, and sheep, " because a singleox is not oxen. Therefore the second grouping is the correct answer. 

36. --BUYING APPLES. 

As there were the same number of boys as girls, it is clear that thenumber of children must be even, and, apart from a careful and exactreading of the question, there would be three different answers. Theremight be two, six, or fourteen children. In the first of these casesthere are ten different ways in which the apples could be bought. But wewere told there was an equal number of "boys and girls, " and one boy andone girl are not boys and girls, so this case has to be excluded. In thecase of fourteen children, the only possible distribution is that eachchild receives one halfpenny apple. But we were told that each child wasto receive an equal distribution of "apples, " and one apple is notapples, so this case has also to be excluded. We are therefore drivenback on our third case, which exactly fits in with all the conditions. Three boys and three girls each receive 1 halfpenny apple and 2third-penny apples. The value of these 3 apples is one penny andone-sixth, which multiplied by six makes sevenpence. Consequently, thecorrect answer is that there were six children--three girls and threeboys. 

37. --BUYING CHESTNUTS. 

In solving this little puzzle we are concerned with the exactinterpretation of the words used by the buyer and seller. I will givethe question again, this time adding a few words to make the matter moreclear. The added words are printed in italics. 

"A man went into a shop to buy chestnuts. He said he wanted apennyworth, and was given five chestnuts. 'It is not enough; I ought tohave a sixth _of a chestnut more_, ' he remarked. 'But if I give you onechestnut more, ' the shopman replied, 'you will have _five-sixths_ toomany. ' Now, strange to say, they were both right. How many chestnutsshould the buyer receive for half a crown?"

The answer is that the price was 155 chestnuts for half a crown. Dividethis number by 30, and we find that the buyer was entitled to 5+1/6chestnuts in exchange for his penny. He was, therefore, right when hesaid, after receiving five only, that he still wanted a sixth. And thesalesman was also correct in saying that if he gave one chestnut more(that is, six chestnuts in all) he would be giving five-sixths of achestnut in excess. 

38. --THE BICYCLE THIEF. 

People give all sorts of absurd answers to this question, and yet it isperfectly simple if one just considers that the salesman cannot possiblyhave lost more than the cyclist actually stole. The latter rode awaywith a bicycle which cost the salesman eleven pounds, and the ten pounds"change;" he thus made off with twenty-one pounds, in exchange for aworthless bit of paper. This is the exact amount of the salesman's loss, and the other operations of changing the cheque and borrowing from afriend do not affect the question in the slightest. The loss ofprospective profit on the sale of the bicycle is, of course, not directloss of money out of pocket. 

39. --THE COSTERMONGER'S PUZZLE. 

Bill must have paid 8s. Per hundred for his oranges--that is, 125 for10s. At 8s. 4d. Per hundred, he would only have received 120 oranges for10s. This exactly agrees with Bill's statement. 

40. --MAMMA'S AGE. 

The age of Mamma must have been 29 years 2 months; that of Papa, 35years; and that of the child, Tommy, 5 years 10 months. Added together, these make seventy years. The father is six times the age of the son, and, after 23 years 4 months have elapsed, their united ages will amountto 140 years, and Tommy will be just half the age of his father. 

41. --THEIR AGES. 

The gentleman's age must have been 54 years and that of his wife 45years. 

42. --THE FAMILY AGES. 

The ages were as follows: Billie, 3½ years; Gertrude, 1¾ year;Henrietta, 5¼ years; Charlie, 10½; years; and Janet, 21 years. 

43. --MRS. TIMPKINS'S AGE. 

The age of the younger at marriage is always the same as the number ofyears that expire before the elder becomes twice her age, if he wasthree times as old at marriage. In our case it was eighteen yearsafterwards; therefore Mrs. Timpkins was eighteen years of age on thewedding-day, and her husband fifty-four. 

44. --A CENSUS PUZZLE. 

Miss Ada Jorkins must have been twenty-four and her little brotherJohnnie three years of age, with thirteen brothers and sisters between. There was a trap for the solver in the words "seven times older thanlittle Johnnie. " Of course, "seven times older" is equal to eight timesas old. It is surprising how many people hastily assume that it is thesame as "seven times as old. " Some of the best writers have committedthis blunder. Probably many of my readers thought that the ages 24½and 3½ were correct. 

45. --MOTHER AND DAUGHTER. 

In four and a half years, when the daughter will be sixteen years and ahalf and the mother forty-nine and a half years of age. 

46. --MARY AND MARMADUKE. 

Marmaduke's age must have been twenty-nine years and two-fifths, andMary's nineteen years and three-fifths. When Marmaduke was aged nineteenand three-fifths, Mary was only nine and four-fifths; so Marmaduke wasat that time twice her age. 

47. --ROVER'S AGE. 

Rover's present age is ten years and Mildred's thirty years. Five yearsago their respective ages were five and twenty-five. Remember that wesaid "four times older than the dog, " which is the same as "five timesas old. " (See answer to No. 44. )

48. --CONCERNING TOMMY'S AGE. 

Tommy Smart's age must have been nine years and three-fifths. Ann's agewas sixteen and four-fifths, the mother's thirty-eight and two-fifths, and the father's fifty and two-fifths. 

49. --NEXT-DOOR NEIGHBOURS. 

Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 13; Mr. Simkin 42; Mrs. Simkin 40; Sophy 10; and Sammy 8. 

50. --THE BAG OF NUTS. 

It will be found that when Herbert takes twelve, Robert and Christopherwill take nine and fourteen respectively, and that they will havetogether taken thirty-five nuts. As 35 is contained in 770 twenty-twotimes, we have merely to multiply 12, 9, and 14 by 22 to discover thatHerbert's share was 264, Robert's 198, and Christopher's 308. Then, asthe total of their ages is 17½ years or half the sum of 12, 9, and 14, their respective ages must be 6, 4½, and 7 years. 

51. --HOW OLD WAS MARY?

The age of Mary to that of Ann must be as 5 to 3. And as the sum oftheir ages was 44, Mary was 27½ and Ann 16½. One is exactly 11 yearsolder than the other. I will now insert in brackets in the originalstatement the various ages specified: "Mary is (27½) twice as old as Annwas (13¾) when Mary was half as old (24¾) as Ann will be (49½) when Annis three times as old (49½) as Mary was (16½) when Mary was (16½) threetimes as old as Ann (5½). " Now, check this backwards. When Mary wasthree times as old as Ann, Mary was 16½ and Ann 5½ (11 years younger). Then we get 49½ for the age Ann will be when she is three times as oldas Mary was then. When Mary was half this she was 24¾. And at that timeAnn must have been 13¾ (11 years younger). Therefore Mary is now twiceas old--27½, and Ann 11 years younger--16½. 

52. --QUEER RELATIONSHIPS. 

If a man marries a woman, who dies, and he then marries his deceasedwife's sister and himself dies, it may be correctly said that he had(previously) married the sister of his widow. 

The youth was not the nephew of Jane Brown, because he happened to beher son. Her surname was the same as that of her brother, because shehad married a man of the same name as herself. 

53. --HEARD ON THE TUBE RAILWAY. 

The gentleman was the second lady's uncle. 

54. --A FAMILY PARTY. 

The party consisted of two little girls and a boy, their father andmother, and their father's father and mother. 

55. --A MIXED PEDIGREE. 

[Illustration:

 Thos. Bloggs m . . . . . | +------------------------+------------+ | | | | | | | W. Snoggs m Kate Bloggs. | | | | | | | . . M Henry Bloggs. | Joseph Bloggs m | | | | +--------+-------------+ | | | | | | | | | Jane John Alf. Mary Bloggs m Snoggs Snoggs m Bloggs

]

The letter m stands for "married. " It will be seen that John Snoggscan say to Joseph Bloggs, "You are my _father's brother-in-law_, becausemy father married your sister Kate; you are my _brother'sfather-in-law_, because my brother Alfred married your daughter Mary;and you are my _father-in-law's brother_, because my wife Jane was yourbrother Henry's daughter. "

56. --WILSON'S POSER. 

If there are two men, each of whom marries the mother of the other, andthere is a son of each marriage, then each of such sons will be at thesame time uncle and nephew of the other. There are other ways in whichthe relationship may be brought about, but this is the simplest. 

57. --WHAT WAS THE TIME?

The time must have been 9. 36 p. M. A quarter of the time since noon is 2hr. 24 min. , and a half of the time till noon next day is 7 hr. 12 min. These added together make 9 hr. 36 min. 

58. --A TIME PUZZLE. 

Twenty-six minutes. 

59. --A PUZZLING WATCH. 

If the 65 minutes be counted on the face of the same watch, then theproblem would be impossible: for the hands must coincide every 65+5/11minutes as shown by its face, and it matters not whether it runs fast orslow; but if it is measured by true time, it gains 5/11 of a minute in65 minutes, or 60/143 of a minute per hour. 

60. --THE WAPSHAW'S WHARF MYSTERY. 

There are eleven different times in twelve hours when the hour andminute hands of a clock are exactly one above the other. If we divide 12hours by 11 we get 1 hr. 5 min. 27+3/11 sec. , and this is the time aftertwelve o'clock when they are first together, and also the time thatelapses between one occasion of the hands being together and the next. They are together for the second time at 2 hr. 10 min. 54+6/11 sec. (twice the above time); next at 3 hr. 16 min. 21+9/11 sec. ; next at 4hr. 21 min. 49+1/11 sec. This last is the only occasion on which the twohands are together with the second hand "just past the forty-ninthsecond. " This, then, is the time at which the watch must have stopped. Guy Boothby, in the opening sentence of his _Across the World for aWife_, says, "It was a cold, dreary winter's afternoon, and by the timethe hands of the clock on my mantelpiece joined forces and stood attwenty minutes past four, my chambers were well-nigh as dark asmidnight. " It is evident that the author here made a slip, for, as wehave seen above, he is 1 min. 49+1/11 sec. Out in his reckoning. 

61. --CHANGING PLACES. 

There are thirty-six pairs of times when the hands exactly change placesbetween three p. M. And midnight. The number of pairs of times from anyhour (n) to midnight is the sum of 12 - (n + 1) natural numbers. Inthe case of the puzzle n = 3; therefore 12 - (3 + 1) = 8 and 1 + 2 + 3+ 4 + 5 + 6 + 7 + 8 = 36, the required answer. 

The first pair of times is 3 hr. 21+57/143 min. And 4 hr. 16+112/143min. , and the last pair is 10 hr. 59+83/143 min. And 11 hr. 54+138/143min. I will not give all the remainder of the thirty-six pairs of times, but supply a formula by which any of the sixty-six pairs that occur frommidday to midnight may be at once found:--

 720b + 60a 720a + 60b min. A hr ---------- min. And b hr. --------------- 143 143

For the letter a may be substituted any hour from 0, 1, 2, 3 up to 10(where nought stands for 12 o'clock midday); and b may represent anyhour, later than a, up to 11. 

By the aid of this formula there is no difficulty in discovering theanswer to the second question: a = 8 and b = 11 will give the pair 8 hr. 58+106/143 min. And 11 hr. 44+128/143 min. , the latter being the timewhen the minute hand is nearest of all to the point IX--in fact, it isonly 15/143 of a minute distant. 

Readers may find it instructive to make a table of all the sixty-sixpairs of times when the hands of a clock change places. An easy way isas follows: Make a column for the first times and a second column forthe second times of the pairs. By making a = 0 and b = 1 in the aboveexpressions we find the first case, and enter hr. 5+5/143 min. At thehead of the first column, and 1 hr. 0+60/143 min. At the head of thesecond column. Now, by successively adding 5+5/143 min. In the first, and 1 hr. 0+60/143 min. In the second column, we get all the _eleven_pairs in which the first time is a certain number of minutes afternought, or mid-day. Then there is a "jump" in the times, but you canfind the next pair by making a = 1 and b = 2, and then by successivelyadding these two times as before you will get all the _ten_ pairs after1 o'clock. Then there is another "jump, " and you will be able to get byaddition all the _nine_ pairs after 2 o'clock. And so on to the end. Iwill leave readers to investigate for themselves the nature and cause ofthe "jumps. " In this way we get under the successive hours, 11 + 10 + 9+ 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agreeswith the formula in the first paragraph of this article. 

Some time ago the principal of a Civil Service Training College, whoconducts a "Civil Service Column" in one of the periodicals, had thequery addressed to him, "How soon after XII o'clock will a clock withboth hands of the same length be ambiguous?" His first answer was, "Sometime past one o'clock, " but he varied the answer from issue to issue. Atlength some of his readers convinced him that the answer is, "At 5+5/143min. Past XII;" and this he finally gave as correct, together with thereason for it that at that time _the time indicated is the samewhichever hand you may assume as hour hand!_

62. --THE CLUB CLOCK. 

The positions of the hands shown in the illustration could only indicatethat the clock stopped at 44 min. 51+1143/1427 sec. After eleveno'clock. The second hand would next be "exactly midway between the othertwo hands" at 45 min. 52+496/1427 sec. After eleven o'clock. If we hadbeen dealing with the points on the circle to which the three hands aredirected, the answer would be 45 min. 22+106/1427 sec. After eleven; butthe question applied to the hands, and the second hand would not bebetween the others at that time, but outside them. 

63. --THE STOP-WATCH. 

The time indicated on the watch was 5+5/11 min. Past 9, when the secondhand would be at 27+3/11 sec. The next time the hands would be similardistances apart would be 54+6/11 min. Past 2, when the second hand wouldbe at 32+8/11 sec. But you need only hold the watch (or our previousillustration of it) in front of a mirror, when you will see the secondtime reflected in it! Of course, when reflected, you will read XI as I, X as II, and so on. 

64. --THE THREE CLOCKS. 

As a mere arithmetical problem this question presents no difficulty. Inorder that the hands shall all point to twelve o'clock at the same time, it is necessary that B shall gain at least twelve hours and that C shalllose twelve hours. As B gains a minute in a day of twenty-four hours, and C loses a minute in precisely the same time, it is evident that onewill have gained 720 minutes (just twelve hours) in 720 days, and theother will have lost 720 minutes in 720 days. Clock A keeping perfecttime, all three clocks must indicate twelve o'clock simultaneously atnoon on the 720th day from April 1, 1898. What day of the month willthat be?

I published this little puzzle in 1898 to see how many people were awareof the fact that 1900 would not be a leap year. It was surprising howmany were then ignorant on the point. Every year that can be divided byfour without a remainder is bissextile or leap year, with the exceptionthat one leap year is cut off in the century. 1800 was not a leap year, nor was 1900. On the other hand, however, to make the calendar morenearly agree with the sun's course, every fourth hundred year is stillconsidered bissextile. Consequently, 2000, 2400, 2800, 3200, etc. , willall be leap years. May my readers live to see them. We therefore findthat 720 days from noon of April 1, 1898, brings us to noon of March 22, 1900. 

65. --THE RAILWAY STATION CLOCK. 

The time must have been 43+7/11 min. Past two o'clock. 

66. --THE VILLAGE SIMPLETON. 

The day of the week on which the conversation took place was Sunday. Forwhen the day after to-morrow (Tuesday) is "yesterday, " "to-day" will beWednesday; and when the day before yesterday (Friday) was "to-morrow, ""to-day" was Thursday. There are two days between Thursday and Sunday, and between Sunday and Wednesday. 

67. --AVERAGE SPEED. 

The average speed is twelve miles an hour, not twelve and a half, asmost people will hastily declare. Take any distance you like, say sixtymiles. This would have taken six hours going and four hours returning. The double journey of 120 miles would thus take ten hours, and theaverage speed is clearly twelve miles an hour. 

68. --THE TWO TRAINS. 

One train was running just twice as fast as the other. 

69. --THE THREE VILLAGES. 

Calling the three villages by their initial letters, it is clear thatthe three roads form a triangle, A, B, C, with a perpendicular, measuring twelve miles, dropped from C to the base A, B. This dividesour triangle into two right-angled triangles with a twelve-mile side incommon. It is then found that the distance from A to C is 15 miles, fromC to B 20 miles, and from A to B 25 (that is 9 and 16) miles. Thesefigures are easily proved, for the square of 12 added to the square of 9equals the square of 15, and the square of 12 added to the square of 16equals the square of 20. 

70. --DRAWING HER PENSION. 

The distance must be 6¾ miles. 

71. --SIR EDWYN DE TUDOR. 

The distance must have been sixty miles. If Sir Edwyn left at noon androde 15 miles an hour, he would arrive at four o'clock--an hour toosoon. If he rode 10 miles an hour, he would arrive at six o'clock--anhour too late. But if he went at 12 miles an hour, he would reach thecastle of the wicked baron exactly at five o'clock--the time appointed. 

72. --THE HYDROPLANE QUESTION. 

The machine must have gone at the rate of seven-twenty-fourths of a mileper minute and the wind travelled five-twenty-fourths of a mile perminute. Thus, going, the wind would help, and the machine would dotwelve-twenty-fourths, or half a mile a minute, and returning onlytwo-twenty-fourths, or one-twelfth of a mile per minute, the wind beingagainst it. The machine without any wind could therefore do the tenmiles in thirty-four and two-sevenths minutes, since it could do sevenmiles in twenty-four minutes. 

73. --DONKEY RIDING. 

The complete mile was run in nine minutes. From the facts stated wecannot determine the time taken over the first and second quarter-milesseparately, but together they, of course, took four and a half minutes. The last two quarters were run in two and a quarter minutes each. 

74. --THE BASKET OF POTATOES. 

Multiply together the number of potatoes, the number less one, and twicethe number less one, then divide by 3. Thus 50, 49, and 99 multipliedtogether make 242, 550, which, divided by 3, gives us 80, 850 yards as thecorrect answer. The boy would thus have to travel 45 miles andfifteen-sixteenths--a nice little recreation after a day's work. 

75. --THE PASSENGER'S FARE. 

Mr. Tompkins should have paid fifteen shillings as his correct share ofthe motor-car fare. He only shared half the distance travelled for £3, and therefore should pay half of thirty shillings, or fifteen shillings. 

76. --THE BARREL OF BEER. 

Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, whichtogether sum to 29, whose digital root is 2. As the contents of thebarrels sold must be a number divisible by 3, if one buyer purchasedtwice as much as the other, we must find a barrel with root 2, 5, or 8to set on one side. There is only one barrel, that containing 20gallons, that fulfils these conditions. So the man must have kept these20 gallons of beer for his own use and sold one man 33 gallons (the18-gallon and 15-gallon barrels) and sold the other man 66 gallons (the16, 19, and 31 gallon barrels). 

77. --DIGITS AND SQUARES. 

The top row must be one of the four following numbers: 192, 219, 273, 327. The first was the example given. 

78. --ODD AND EVEN DIGITS. 

As we have to exclude complex and improper fractions and recurringdecimals, the simplest solution is this: 79 + 5+1/3 and 84 + 2/6, bothequal 84+1/3. Without any use of fractions it is obviously impossible. 

79. --THE LOCKERS PUZZLE. 

The smallest possible total is 356 = 107 + 249, and the largest sumpossible is 981 = 235 + 746, or 657+324. The middle sum may be either720 = 134 + 586, or 702 = 134 + 568, or 407 = 138 + 269. The total inthis case must be made up of three of the figures 0, 2, 4, 7, but nosum other than the three given can possibly be obtained. We havetherefore no choice in the case of the first locker, an alternative inthe case of the third, and any one of three arrangements in the caseof the middle locker. Here is one solution:--

 107 134 235 249 586 746 --- --- --- 356 720 981

Of course, in each case figures in the first two lines may be exchangedvertically without altering the total, and as a result there are just3, 072 different ways in which the figures might be actually placed onthe locker doors. I must content myself with showing one littleprinciple involved in this puzzle. The sum of the digits in the total isalways governed by the digit omitted. 9/9 - 7/10 - 5/11 -3/12 - 1/13 -8/14 - 6/15 - 4/16 - 2/17 - 0/18. Whichever digit shown here in theupper line we omit, the sum of the digits in the total will be foundbeneath it. Thus in the case of locker A we omitted 8, and the figuresin the total sum up to 14. If, therefore, we wanted to get 356, we mayknow at once to a certainty that it can only be obtained (if at all) bydropping the 8. 

80. --THE THREE GROUPS. 

There are nine solutions to this puzzle, as follows, and no more:--

 12 × 483 = 5, 796 27 × 198 = 5, 346 42 × 138 = 5, 796 39 × 186 = 7, 254 18 × 297 = 5, 346 48 × 159 = 7, 632 28 × 157 = 4, 396 4 × 1, 738 = 6, 952 4 × 1, 963 = 7, 852

The seventh answer is the one that is most likely to be overlooked bysolvers of the puzzle. 

81. --THE NINE COUNTERS. 

In this case a certain amount of mere "trial" is unavoidable. But thereare two kinds of "trials"--those that are purely haphazard, and thosethat are methodical. The true puzzle lover is never satisfied with merehaphazard trials. The reader will find that by just reversing thefigures in 23 and 46 (making the multipliers 32 and 64) both productswill be 5, 056. This is an improvement, but it is not the correct answer. We can get as large a product as 5, 568 if we multiply 174 by 32 and 96by 58, but this solution is not to be found without the exercise of somejudgment and patience. 

82. --THE TEN COUNTERS. 

As I pointed out, it is quite easy so to arrange the counters that theyshall form a pair of simple multiplication sums, each of which will givethe same product--in fact, this can be done by anybody in five minuteswith a little patience. But it is quite another matter to find that pairwhich gives the largest product and that which gives the smallestproduct. 

Now, in order to get the smallest product, it is necessary to select asmultipliers the two smallest possible numbers. If, therefore, we place 1and 2 as multipliers, all we have to do is to arrange the remainingeight counters in such a way that they shall form two numbers, one ofwhich is just double the other; and in doing this we must, of course, try to make the smaller number as low as possible. Of course the lowestnumber we could get would be 3, 045; but this will not work, neither will3, 405, 3, 45O, etc. , and it may be ascertained that 3, 485 is the lowestpossible. One of the required answers is 3, 485 × 2 = 6, 970, and 6, 970 ×1 = 6, 970. 

The other part of the puzzle (finding the pair with the highest product)is, however, the real knotty point, for it is not at all easy todiscover whether we should let the multiplier consist of one or of twofigures, though it is clear that we must keep, so far as we can, thelargest figures to the left in both multiplier and multiplicand. It willbe seen that by the following arrangement so high a number as 58, 560 maybe obtained. Thus, 915 × 64 = 58, 560, and 732 × 80 = 58, 560. 

83. --DIGITAL MULTIPLICATION. 

The solution that gives the smallest possible sum of digits in thecommon product is 23 × 174 = 58 × 69 = 4, 002, and the solution thatgives the largest possible sum of digits, 9×654 =18×327=5, 886. In thefirst case the digits sum to 6 and in the second case to 27. There is noway of obtaining the solution but by actual trial. 

84. --THE PIERROT'S PUZZLE. 

There are just six different solutions to this puzzle, as follows:--

 8 multiplied by 473 equals 3784 9 " 351 " 3159 15 " 93 " 1395 21 " 87 " 1287 27 " 81 " 2187 35 " 41 " 1435

It will be seen that in every case the two multipliers contain exactlythe same figures as the product. 

85. --THE CAB NUMBERS. 

The highest product is, I think, obtained by multiplying 8, 745, 231 by96--namely, 839, 542, 176. 

Dealing here with the problem generally, I have shown in the last puzzlethat with three digits there are only two possible solutions, and withfour digits only six different solutions. 

These cases have all been given. With five digits there are justtwenty-two solutions, as follows:--

 3 × 4128 = 12384 3 × 4281 = 12843 3 × 7125 = 21375 3 × 7251 = 21753 2541 × 6 = 15246 651 × 24 = 15624 678 × 42 = 28476 246 × 51 = 12546 57 × 834 = 47538 75 × 231 = 17325 624 × 78 = 48672 435 × 87 = 37845 ------ 9 × 7461 = 67149 72 × 936 = 67392 ------ 2 × 8714 = 17428 2 × 8741 = 17482 65 × 281 = 18265 65 × 983 = 63985 ------ 4973 × 8 = 39784 6521 × 8 = 52168 14 × 926 = 12964 86 × 251 = 21586

Now, if we took every possible combination and tested it bymultiplication, we should need to make no fewer than 30, 240 trials, or, if we at once rejected the number 1 as a multiplier, 28, 560 trials--atask that I think most people would be inclined to shirk. But let usconsider whether there be no shorter way of getting at the resultsrequired. I have already explained that if you add together the digitsof any number and then, as often as necessary, add the digits of theresult, you must ultimately get a number composed of one figure. Thislast number I call the "digital root. " It is necessary in every solutionof our problem that the root of the sum of the digital roots of ourmultipliers shall be the same as the root of their product. There areonly four ways in which this can happen: when the digital roots of themultipliers are 3 and 6, or 9 and 9, or 2 and 2, or 5 and 8. I havedivided the twenty-two answers above into these four classes. It is thusevident that the digital root of any product in the first two classesmust be 9, and in the second two classes 4. 

Owing to the fact that no number of five figures can have a digital sumless than 15 or more than 35, we find that the figures of our productmust sum to either 18 or 27 to produce the root 9, and to either 22 or31 to produce the root 4. There are 3 ways of selecting five differentfigures that add up to 18, there are 11 ways of selecting five figuresthat add up to 27, there are 9 ways of selecting five figures that addup to 22, and 5 ways of selecting five figures that add up to 31. Thereare, therefore, 28 different groups, and no more, from any one of whicha product may be formed. 

We next write out in a column these 28 sets of five figures, and proceedto tabulate the possible factors, or multipliers, into which they may besplit. Roughly speaking, there would now appear to be about 2, 000possible cases to be tried, instead of the 30, 240 mentioned above; butthe process of elimination now begins, and if the reader has a quick eyeand a clear head he can rapidly dispose of the large bulk of thesecases, and there will be comparatively few test multiplicationsnecessary. It would take far too much space to explain my own method indetail, but I will take the first set of figures in my table and showhow easily it is done by the aid of little tricks and dodges that shouldoccur to everybody as he goes along. 

My first product group of five figures is 84, 321. Here, as we have seen, the root of each factor must be 3 or a multiple of 3. As there is no 6or 9, the only single multiplier is 3. Now, the remaining four figurescan be arranged in 24 different ways, but there is no need to make 24multiplications. We see at a glance that, in order to get a five-figureproduct, either the 8 or the 4 must be the first figure to the left. Butunless the 2 is preceded on the right by the 8, it will produce whenmultiplied either a 6 or a 7, which must not occur. We are, therefore, reduced at once to the two cases, 3 × 4, 128 and 3 x 4, 281, both of whichgive correct solutions. Suppose next that we are trying the two-figurefactor, 21. Here we see that if the number to be multiplied is under 500the product will either have only four figures or begin with 10. Therefore we have only to examine the cases 21 × 843 and 21 × 834. Butwe know that the first figure will be repeated, and that the secondfigure will be twice the first figure added to the second. Consequently, as twice 3 added to 4 produces a nought in our product, the first caseis at once rejected. It only remains to try the remaining case bymultiplication, when we find it does not give a correct answer. If weare next trying the factor 12, we see at the start that neither the 8nor the 3 can be in the units place, because they would produce a 6, andso on. A sharp eye and an alert judgment will enable us thus to runthrough our table in a much shorter time than would be expected. Theprocess took me a little more than three hours. 

I have not attempted to enumerate the solutions in the cases of six, seven, eight, and nine digits, but I have recorded nearly fifty exampleswith nine digits alone. 

86. --QUEER MULTIPLICATION. 

If we multiply 32547891 by 6, we get the product, 195287346. In bothcases all the nine digits are used once and once only. 

87. --THE NUMBER CHECKS PUZZLE. 

Divide the ten checks into the following three groups: 7 1 5--4 6--3 2 89 0, and the first multiplied by the second produces the third. 

88. --DIGITAL DIVISION. 

It is convenient to consider the digits as arranged to form fractions ofthe respective values, one-half, one-third, one-fourth, one-fifth, one-sixth, one-seventh, one-eighth, and one-ninth. I will first give theeight answers, as follows:--

 6729/13458 = 1/2

 5823/17469 = 1/3

 3942/15768 = 1/4

 2697/13485 = 1/5

 2943/17658 = 1/6

 2394/16758 = 1/7

 3187/25496 = 1/8

 6381/57429 = 1/9

The sum of the numerator digits and the denominator digits will, ofcourse, always be 45, and the "digital root" is 9. Now, if we separatethe nine digits into any two groups, the sum of the two digital rootswill always be 9. In fact, the two digital roots must be either 9--9, 8--1, 7--2, 6--3, or 5--4. In the first case the actual sum is 18, butthen the digital root of this number is itself 9. The solutions in thecases of one-third, one-fourth, one-sixth, one-seventh, and one-ninthmust be of the form 9--9; that is to say, the digital roots of bothnumerator and denominator will be 9. In the cases of one-half andone-fifth, however, the digital roots are 6--3, but of course the higherroot may occur either in the numerator or in the denominator; thus2697/13485, 2769/13845, 2973/14865, 3729/18645, where, in the first twoarrangements, the roots of the numerator and denominator arerespectively 6--3, and in the last two 3--6. The most curious case ofall is, perhaps, one-eighth, for here the digital roots may be of anyone of the five forms given above. 

The denominators of the fractions being regarded as the numeratorsmultiplied by 2, 3, 4, 5, 6, 7, 8, and 9 respectively, we must payattention to the "carryings over. " In order to get five figures in theproduct there will, of course, always be a carry-over after multiplyingthe last figure to the left, and in every case higher than 4 we mustcarry over at least three times. Consequently in cases from one-fifth toone-ninth we cannot produce different solutions by a mere change ofposition of pairs of figures, as, for example, we may with 5832/17496and 5823/17469, where the 2/6 and 3/9 change places. It is true that thesame figures may often be differently arranged, as shown in the twopairs of values for one-fifth that I have given in the last paragraph, but here it will be found there is a general readjustment of figures andnot a simple changing of the positions of pairs. There are other littlepoints that would occur to every solver--such as that the figure 5cannot ever appear to the extreme right of the numerator, as this wouldresult in our getting either a nought or a second 5 in the denominator. Similarly 1 cannot ever appear in the same position, nor 6 in thefraction one-sixth, nor an even figure in the fraction one-fifth, and soon. The preliminary consideration of such points as I have touched uponwill not only prevent our wasting a lot of time in trying to produceimpossible forms, but will lead us more or less directly to the desiredsolutions. 

89. --ADDING THE DIGITS. 

The smallest possible sum of money is £1, 8s. 9¾d. , the digits of whichadd to 25. 

90. --THE CENTURY PUZZLE. 

The problem of expressing the number 100 as a mixed number or fraction, using all the nine digits once, and once only, has, like all thesedigital puzzles, a fascinating side to it. The merest tyro can bypatient trial obtain correct results, and there is a singular pleasurein discovering and recording each new arrangement akin to the delight ofthe botanist in finding some long-sought plant. It is simply a matter ofarranging those nine figures correctly, and yet with the thousands ofpossible combinations that confront us the task is not so easy as mightat first appear, if we are to get a considerable number of results. Hereare eleven answers, including the one I gave as a specimen:--

 2148 1752 1428 1578 96 ----, 96 ----, 96 ----, 94 ----, 537 438 357 263

 7524 5823 5742 3546 91 ----, 91 ----, 91 ----, 82 ----, 836 647 638 197

 7524 5643 69258 81 ----, 81 ----, 3 -----. 396 297 714

Now, as all the fractions necessarily represent whole numbers, it willbe convenient to deal with them in the following form: 96 + 4, 94 + 6, 91 + 9, 82 + 18, 81 + 19, and 3 + 97. 

With any whole number the digital roots of the fraction that brings itup to 100 will always be of one particular form. Thus, in the case of 96+ 4, one can say at once that if any answers are obtainable, then theroots of both the numerator and the denominator of the fraction will be6. Examine the first three arrangements given above, and you will findthat this is so. In the case of 94 + 6 the roots of the numerator anddenominator will be respectively 3--2, in the case of 91 + 9 and of 82 +18 they will be 9--8, in the case of 81 + 19 they will be 9--9, and inthe case of 3 + 97 they will be 3--3. Every fraction that can beemployed has, therefore, its particular digital root form, and you areonly wasting your time in unconsciously attempting to break through thislaw. 

Every reader will have perceived that certain whole numbers areevidently impossible. Thus, if there is a 5 in the whole number, therewill also be a nought or a second 5 in the fraction, which are barred bythe conditions. Then multiples of 10, such as 90 and 80, cannot ofcourse occur, nor can the whole number conclude with a 9, like 89 and79, because the fraction, equal to 11 or 21, will have 1 in the lastplace, and will therefore repeat a figure. Whole numbers that repeat afigure, such as 88 and 77, are also clearly useless. These cases, as Ihave said, are all obvious to every reader. But when I declare that suchcombinations as 98 + 2, 92 + 8, 86 + 14, 83 + 17, 74 + 26, etc. , etc. , are to be at once dismissed as impossible, the reason is not so evident, and I unfortunately cannot spare space to explain it. 

But when all those combinations have been struck out that are known tobe impossible, it does not follow that all the remaining "possibleforms" will actually work. The elemental form may be right enough, butthere are other and deeper considerations that creep in to defeat ourattempts. For example, 98 + 2 is an impossible combination, because weare able to say at once that there is no possible form for the digitalroots of the fraction equal to 2. But in the case of 97 + 3 there is apossible form for the digital roots of the fraction, namely, 6--5, andit is only on further investigation that we are able to determine thatthis form cannot in practice be obtained, owing to curiousconsiderations. The working is greatly simplified by a process ofelimination, based on such considerations as that certainmultiplications produce a repetition of figures, and that the wholenumber cannot be from 12 to 23 inclusive, since in every such casesufficiently small denominators are not available for forming thefractional part. 

91. --MORE MIXED FRACTIONS. 

The point of the present puzzle lies in the fact that the numbers 15 and18 are not capable of solution. There is no way of determining thiswithout trial. Here are answers for the ten possible numbers:--

 9+5472/1368 = 13; 9+6435/1287 = 14; 12+3576/894 = 16; 6+13258/947 = 20; 15+9432/786 = 27; 24+9756/813 = 36; 27+5148/396 = 40; 65+1892/473 = 69; 59+3614/278 = 72; 75+3648/192 = 94. 

I have only found the one arrangement for each of the numbers 16, 20, and 27; but the other numbers are all capable of being solved in morethan one way. As for 15 and 18, though these may be easily solved as asimple fraction, yet a "mixed fraction" assumes the presence of a wholenumber; and though my own idea for dodging the conditions is thefollowing, where the fraction is both complex and mixed, it will befairer to keep exactly to the form indicated:--

 3952 ---- 746 = 15; 3 ---- 1

 5742 ---- 638 = 18. 9 ---- 1

I have proved the possibility of solution for all numbers up to 100, except 1, 2, 3, 4, 15, and 18. The first three are easily shown to beimpossible. I have also noticed that numbers whose digital root is8--such as 26, 35, 44, 53, etc. --seem to lend themselves to the greatestnumber of answers. For the number 26 alone I have recorded no fewer thantwenty-five different arrangements, and I have no doubt that there aremany more. 

92. --DIGITAL SQUARE NUMBERS. 

So far as I know, there are no published tables of square numbers thatgo sufficiently high to be available for the purposes of this puzzle. The lowest square number containing all the nine digits once, and onceonly, is 139, 854, 276, the square of 11, 826. The highest square numberunder the same conditions is, 923, 187, 456, the square of 30, 384. 

93. --THE MYSTIC ELEVEN. 

Most people know that if the sum of the digits in the odd places of anynumber is the same as the sum of the digits in the even places, then thenumber is divisible by 11 without remainder. Thus in 896743012 the odddigits, 20468, add up 20, and the even digits, 1379, also add up 20. Therefore the number may be divided by 11. But few seem to know that ifthe difference between the sum of the odd and the even digits is 11, ora multiple of 11, the rule equally applies. This law enables us to find, with a very little trial, that the smallest number containing nine ofthe ten digits (calling nought a digit) that is divisible by 11 is102, 347, 586, and the highest number possible, 987, 652, 413. 

94. --THE DIGITAL CENTURY. 

There is a very large number of different ways in which arithmeticalsigns may be placed between the nine digits, arranged in numericalorder, so as to give an expression equal to 100. In fact, unless thereader investigated the matter very closely, he might not suspect thatso many ways are possible. It was for this reason that I added thecondition that not only must the fewest possible signs be used, but alsothe fewest possible strokes. In this way we limit the problem to asingle solution, and arrive at the simplest and therefore (in this case)the best result. 

Just as in the case of magic squares there are methods by which we maywrite down with the greatest ease a large number of solutions, but notall the solutions, so there are several ways in which we may quicklyarrive at dozens of arrangements of the "Digital Century, " withoutfinding all the possible arrangements. There is, in fact, very littleprinciple in the thing, and there is no certain way of demonstratingthat we have got the best possible solution. All I can say is that thearrangement I shall give as the best is the best I have up to thepresent succeeded in discovering. I will give the reader a fewinteresting specimens, the first being the solution usually published, and the last the best solution that I know. 

 Signs. Strokes. 1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 × 9) = 100 ( 9 18)

 - (1 × 2) - 3 - 4 - 5 + (6 × 7) + (8 × 9) = 100 (12 20)

 1 + (2 × 3) + (4 × 5) - 6 + 7 + (8 × 9) = 100 (11 21)

 (1 + 2 - 3 - 4)(5 - 6 - 7 - 8 - 9) = 100 ( 9 12)

 1 + (2 × 3) + 4 + 5 + 67 + 8 + 9 =100 (8 16)

 (1 × 2) + 34 + 56 + 7 - 8 + 9 = 100 (7 13)

 12 + 3 - 4 + 5 + 67 + 8 + 9 = 100 (6 11)

 123 - 4 - 5 - 6 - 7 + 8 - 9 = 100 (6 7)

 123 + 4 - 5 + 67 - 8 - 9 = 100 (4 6)

 123 + 45 - 67 + 8 - 9 = 100 (4 6)

 123 - 45 - 67 + 89 = 100 (3 4)

It will be noticed that in the above I have counted the bracket as onesign and two strokes. The last solution is singularly simple, and I donot think it will ever be beaten. 

95. --THE FOUR SEVENS. 

The way to write four sevens with simple arithmetical signs so that theyrepresent 100 is as follows:--

 7 7 -- × -- = 100. . 7 . 7

Of course the fraction, 7 over decimal 7, equals 7 divided by 7/10, which is the same as 70 divided by 7, or 10. Then 10 multiplied by 10 is100, and there you are! It will be seen that this solution appliesequally to any number whatever that you may substitute for 7. 

96. --THE DICE NUMBERS. 

The sum of all the numbers that can be formed with any given set of fourdifferent figures is always 6, 666 multiplied by the sum of the fourfigures. Thus, 1, 2, 3, 4 add up 10, and ten times 6, 666 is 66, 660. Now, there are thirty-five different ways of selecting four figures from theseven on the dice--remembering the 6 and 9 trick. The figures of allthese thirty-five groups add up to 600. Therefore 6, 666 multiplied by600 gives us 3, 999, 600 as the correct answer. 

Let us discard the dice and deal with the problem generally, using thenine digits, but excluding nought. Now, if you were given simply the sumof the digits--that is, if the condition were that you could use anyfour figures so long as they summed to a given amount--then we have toremember that several combinations of four digits will, in many cases, make the same sum. 

 10 11 12 13 14 15 16 17 18 19 20 1 1 2 3 5 6 8 9 11 11 12

 21 22 23 24 25 26 27 28 29 30 11 11 9 8 6 5 3 2 1 1

Here the top row of numbers gives all the possible sums of fourdifferent figures, and the bottom row the number of different ways inwhich each sum may be made. For example 13 may be made in three ways:1237, 1246, and 1345. It will be found that the numbers in the bottomrow add up to 126, which is the number of combinations of nine figurestaken four at a time. From this table we may at once calculate theanswer to such a question as this: What is the sum of all the numberscomposed of our different digits (nought excluded) that add up to 14?Multiply 14 by the number beneath t in the table, 5, and multiply theresult by 6, 666, and you will have the answer. It follows that, to knowthe sum of all the numbers composed of four different digits, if youmultiply all the pairs in the two rows and then add the resultstogether, you will get 2, 520, which, multiplied by 6, 666, gives theanswer 16, 798, 320. 

The following general solution for any number of digits will doubtlessinterest readers. Let n represent number of digits, then 5 (10^n - 1) 8!divided by (9 - n)! equals the required sum. Note that 0! equals 1. Thismay be reduced to the following practical rule: Multiply together 4 × 7× 6 × 5 . . . To (n - 1) factors; now add (n + 1) ciphers to the right, and from this result subtract the same set of figures with a singlecipher to the right. Thus for n = 4 (as in the case last mentioned), 4 ×7 × 6 = 168. Therefore 16, 800, 000 less 1, 680 gives us 16, 798, 320 inanother way. 

97. --THE SPOT ON THE TABLE. 

The ordinary schoolboy would correctly treat this as a quadraticequation. Here is the actual arithmetic. Double the product of the twodistances from the walls. This gives us 144, which is the square of 12. The sum of the two distances is 17. If we add these two numbers, 12 and17, together, and also subtract one from the other, we get the twoanswers that 29 or 5 was the radius, or half-diameter, of the table. Consequently, the full diameter was 58 in. Or 10 in. But a table of thelatter dimensions would be absurd, and not at all in accordance with theillustration. Therefore the table must have been 58 in. In diameter. Inthis case the spot was on the edge nearest to the corner of the room--towhich the boy was pointing. If the other answer were admissible, thespot would be on the edge farthest from the corner of the room. 

98. --ACADEMIC COURTESIES. 

There must have been ten boys and twenty girls. The number of bows girlto girl was therefore 380, of boy to boy 90, of girl with boy 400, andof boys and girls to teacher 30, making together 900, as stated. It willbe remembered that it was not said that the teacher himself returned thebows of any child. 

99. --THE THIRTY-THREE PEARLS. 

The value of the large central pearl must have been £3, 000. The pearl atone end (from which they increased in value by £100) was £1, 400; thepearl at the other end, £600. 

100. --THE LABOURER'S PUZZLE. 

The man said, "I am going twice as deep, " not "as deep again. " That isto say, he was still going twice as deep as he had gone already, so thatwhen finished the hole would be three times its present depth. Then theanswer is that at present the hole is 3 ft. 6 in. Deep and the man 2 ft. 4 in. Above ground. When completed the hole will be 10 ft. 6 in. Deep, and therefore the man will then be 4 ft. 8 in. Below the surface, ortwice the distance that he is now above ground. 

101. --THE TRUSSES OF HAY. 

Add together the ten weights and divide by 4, and we get 289 lbs. As theweight of the five trusses together. If we call the five trusses in theorder of weight A, B, C, D, and E, the lightest being A and the heaviestE, then the lightest, no lbs. , must be the weight of A and B; and thenext lightest, 112 lbs. , must be the weight of A and C. Then the twoheaviest, D and E, must weigh 121 lbs. , and C and E must weigh 120 lbs. We thus know that A, B, D, and E weigh together 231 lbs. , which, deducted from 289 lbs. (the weight of the five trusses), gives us theweight of C as 58 lbs. Now, by mere subtraction, we find the weight ofeach of the five trusses--54 lbs. , 56 lbs. , 58 lbs. , 59 lbs. , and 62lbs. Respectively. 

102. --MR. GUBBINS IN A FOG. 

The candles must have burnt for three hours and three-quarters. Onecandle had one-sixteenth of its total length left and the otherfour-sixteenths. 

103. --PAINTING THE LAMP-POSTS. 

Pat must have painted six more posts than Tim, no matter how manylamp-posts there were. For example, suppose twelve on each side; thenPat painted fifteen and Tim nine. If a hundred on each side, Pat paintedone hundred and three, and Tim only ninety-seven

104. --CATCHING THE THIEF. 

The constable took thirty steps. In the same time the thief would takeforty-eight, which, added to his start of twenty-seven, carried himseventy-five steps. This distance would be exactly equal to thirty stepsof the constable. 

105. --THE PARISH COUNCIL ELECTION, The voter can vote for one candidate in 23 ways, for two in 253 ways, for three in 1, 771, for four in 8, 855, for five in 33, 649, for six in100, 947, for seven in 245, 157, for eight in 490, 314, and for ninecandidates in 817, 190 different ways. Add these together, and we get thetotal of 1, 698, 159 ways of voting. 

106. --THE MUDDLETOWN ELECTION. 

The numbers of votes polled respectively by the Liberal, theConservative, the Independent, and the Socialist were 1, 553, 1, 535, 1, 407, and 978 All that was necessary was to add the sum of the threemajorities (739) to the total poll of 5, 473 (making 6, 212) and divide by4, which gives us 1, 553 as the poll of the Liberal. Then the polls ofthe other three candidates can, of course, be found by deducting thesuccessive majorities from the last-mentioned number. 

107. --THE SUFFRAGISTS' MEETING. 

Eighteen were present at the meeting and eleven left. If twelve hadgone, two-thirds would have retired. If only nine had gone, the meetingwould have lost half its members. 

108. --THE LEAP-YEAR LADIES. 

The correct and only answer is that 11, 616 ladies made proposals ofmarriage. Here are all the details, which the reader can check forhimself with the original statements. Of 10, 164 spinsters, 8, 085 marriedbachelors, 627 married widowers, 1, 221 were declined by bachelors, and231 declined by widowers. Of the 1, 452 widows, 1, 155 married bachelors, and 297 married widowers. No widows were declined. The problem is notdifficult, by algebra, when once we have succeeded in correctly statingit. 

109. --THE GREAT SCRAMBLE. 

The smallest number of sugar plums that will fulfil the conditions is26, 880. The five boys obtained respectively: Andrew, 2, 863; Bob, 6, 335;Charlie, 2, 438; David, 10, 294; Edgar, 4, 950. There is a little trapconcealed in the words near the end, "one-fifth of the same, " that seemsat first sight to upset the whole account of the affair. But a littlethought will show that the words could only mean "one-fifth offive-eighths", the fraction last mentioned--that is, one-eighth of thethree-quarters that Bob and Andrew had last acquired. 

110. --THE ABBOT'S PUZZLE. 

The only answer is that there were 5 men, 25 women, and 70 children. There were thus 100 persons in all, 5 times as many women as men, and asthe men would together receive 15 bushels, the women 50 bushels, and thechildren 35 bushels, exactly 100 bushels would be distributed. 

111. --REAPING THE CORN. 

The whole field must have contained 46. 626 square rods. The side of thecentral square, left by the farmer, is 4. 8284 rods, so it contains23. 313 square rods. The area of the field was thus something more than aquarter of an acre and less than one-third; to be more precise, . 2914 ofan acre. 

112. --A PUZZLING LEGACY. 

As the share of Charles falls in through his death, we have merely todivide the whole hundred acres between Alfred and Benjamin in theproportion of one-third to one-fourth--that is in the proportion offour-twelfths to three-twelfths, which is the same as four to three. Therefore Alfred takes four-sevenths of the hundred acres and Benjaminthree-sevenths. 

113. --THE TORN NUMBER. 

The other number that answers all the requirements of the puzzle is9, 801. If we divide this in the middle into two numbers and add themtogether we get 99, which, multiplied by itself, produces 9, 801. It istrue that 2, 025 may be treated in the same way, only this number isexcluded by the condition which requires that no two figures should bealike. 

The general solution is curious. Call the number of figures in each halfof the torn label n. Then, if we add 1 to each of the exponents of theprime factors (other than 3) of 10^n - 1 (1 being regarded as a factorwith the constant exponent, 1), their product will be the number ofsolutions. Thus, for a label of six figures, n = 3. The factors of 10^n- 1 are 1¹ × 37¹ (not considering the 3³), and the product of 2 × 2 =4, the number of solutions. This always includes the special cases 98 -01, 00 - 01, 998 - 01, 000 - 001, etc. The solutions are obtained asfollows:--Factorize 10³ - 1 in all possible ways, always keeping thepowers of 3 together, thus, 37 × 27, 999 × 1. Then solve the equation37x = 27y + 1. Here x = 19 and y = 26. Therefore, 19 × 37 = 703, thesquare of which gives one label, 494, 209. A complementary solution(through 27x = 37x + 1) can at once be found by 10^n - 703 = 297, thesquare of which gives 088, 209 for second label. (These non-significantnoughts to the left must be included, though they lead to peculiar caseslike 00238 - 04641 = 4879², where 0238 - 4641 would not work. ) Thespecial case 999 × 1 we can write at once 998, 001, according to the lawshown above, by adding nines on one half and noughts on the other, andits complementary will be 1 preceded by five noughts, or 000001. Thus weget the squares of 999 and 1. These are the four solutions. 

114. --CURIOUS NUMBERS. 

The three smallest numbers, in addition to 48, are 1, 680, 57, 120, and1, 940, 448. It will be found that 1, 681 and 841, 57, 121 and 28, 561, 1, 940, 449 and 970, 225, are respectively the squares of 41 and 29, 239and 169, 1, 393 and 985. 

115. --A PRINTER'S ERROR. 

The answer is that 2^5 . 9^2 is the same as 2592, and this is the onlypossible solution to the puzzle. 

116. --THE CONVERTED MISER. 

As we are not told in what year Mr. Jasper Bullyon made the generousdistribution of his accumulated wealth, but are required to find thelowest possible amount of money, it is clear that we must look for ayear of the most favourable form. 

There are four cases to be considered--an ordinary year with fifty-twoSundays and with fifty-three Sundays, and a leap-year with fifty-two andfifty-three Sundays respectively. Here are the lowest possible amountsin each case:--

 313 weekdays, 52 Sundays £112, 055 312 weekdays, 53 Sundays 19, 345 314 weekdays, 52 Sundays No solution possible. 313 weekdays, 53 Sundays £69, 174

The lowest possible amount, and therefore the correct answer, is£19, 345, distributed in an ordinary year that began on a Sunday. Thelast year of this kind was 1911. He would have paid £53 on every day ofthe year, or £62 on every weekday, with £1 left over, as required, inthe latter event. 

117. --A FENCE PROBLEM. 

Though this puzzle presents no great difficulty to any one possessing aknowledge of algebra, it has perhaps rather interesting features. 

Seeing, as one does in the illustration, just one corner of the proposedsquare, one is scarcely prepared for the fact that the field, in orderto comply with the conditions, must contain exactly 501, 760 acres, thefence requiring the same number of rails. Yet this is the correctanswer, and the only answer, and if that gentleman in Iowa carries outhis intention, his field will be twenty-eight miles long on each side, and a little larger than the county of Westmorland. I am not aware thatany limit has ever been fixed to the size of a "field, " though they donot run so large as this in Great Britain. Still, out in Iowa, where mycorrespondent resides, they do these things on a very big scale. I have, however, reason to believe that when he finds the sort of task he hasset himself, he will decide to abandon it; for if that cow decides toroam to fresh woods and pastures new, the milkmaid may have to start outa week in advance in order to obtain the morning's milk. 

Here is a little rule that will always apply where the length of therail is half a pole. Multiply the number of rails in a hurdle by four, and the result is the exact number of miles in the side of a squarefield containing the same number of acres as there are rails in thecomplete fence. Thus, with a one-rail fence the field is four milessquare; a two-rail fence gives eight miles square; a three-rail fence, twelve miles square; and so on, until we find that a seven-rail fencemultiplied by four gives a field of twenty-eight miles square. In thecase of our present problem, if the field be made smaller, then thenumber of rails will exceed the number of acres; while if the field bemade larger, the number of rails will be less than the acres of thefield. 

118. --CIRCLING THE SQUARES. 

Though this problem might strike the novice as being rather difficult, it is, as a matter of fact, quite easy, and is made still easier byinserting four out of the ten numbers. 

First, it will be found that squares that are diametrically oppositehave a common difference. For example, the difference between the squareof 14 and the square of 2, in the diagram, is 192; and the differencebetween the square of 16 and the square of 8 is also 192. This must beso in every case. Then it should be remembered that the differencebetween squares of two consecutive numbers is always twice the smallernumber plus 1, and that the difference between the squares of any twonumbers can always be expressed as the difference of the numbersmultiplied by their sum. Thus the square of 5 (25) less the square of 4(16) equals (2 × 4) + 1, or 9; also, the square of 7 (49) less thesquare of 3 (9) equals (7 + 3) × (7 - 3), or 40. 

Now, the number 192, referred to above, may be divided into fivedifferent pairs of even factors: 2 × 96, 4 × 48, 6 × 32, 8 × 24, and 12× 16, and these divided by 2 give us, 1 × 48, 2 × 24, 3 × 16, 4 × 12, and 6 × 8. The difference and sum respectively of each of these pairs inturn produce 47, 49; 22, 26; 13, 19; 8, 16; and 2, 14. These are therequired numbers, four of which are already placed. The six numbers thathave to be added may be placed in just six different ways, one of whichis as follows, reading round the circle clockwise: 16, 2, 49, 22, 19, 8, 14, 47, 26, 13. 

I will just draw the reader's attention to one other little point. Inall circles of this kind, the difference between diametrically oppositenumbers increases by a certain ratio, the first numbers (with theexception of a circle of 6) being 4 and 6, and the others formed bydoubling the next preceding but one. Thus, in the above case, the firstdifference is 2, and then the numbers increase by 4, 6, 8, and 12. Ofcourse, an infinite number of solutions may be found if we admitfractions. The number of squares in a circle of this kind must, however, be of the form 4n + 6; that is, it must be a number composed of 6 plus amultiple of 4. 

119. --RACKBRANE'S LITTLE LOSS. 

The professor must have started the game with thirteen shillings, Mr. Potts with four shillings, and Mrs. Potts with seven shillings. 

120. --THE FARMER AND HIS SHEEP. 

The farmer had one sheep only! If he divided this sheep (which is bestdone by weight) into two parts, making one part two-thirds and the otherpart one-third, then the difference between these two numbers is thesame as the difference between their squares--that is, one-third. Anytwo fractions will do if the denominator equals the sum of the twonumerators. 

121. --HEADS OR TAILS. 

Crooks must have lost, and the longer he went on the more he would lose. In two tosses he would be left with three-quarters of his money, in fourtosses with nine-sixteenths of his money, in six tosses withtwenty-seven sixty-fourths of his money, and so on. The order of thewins and losses makes no difference, so long as their number is in theend equal. 

122. --THE SEE-SAW PUZZLE. 

The boy's weight must have been about 39. 79 lbs. A brick weighed 3 lbs. Therefore 16 bricks weighed 48 lbs. And 11 bricks 33 lbs. Multiply 48 by33 and take the square root. 

123. --A LEGAL DIFFICULTY. 

It was clearly the intention of the deceased to give the son twice asmuch as the mother, or the daughter half as much as the mother. Therefore the most equitable division would be that the mother shouldtake two-sevenths, the son four-sevenths, and the daughter one-seventh. 

124. --A QUESTION OF DEFINITION. 

There is, of course, no difference in _area_ between a mile square and asquare mile. But there may be considerable difference in _shape_. A milesquare can be no other shape than square; the expression describes asurface of a certain specific size and shape. A square mile may be ofany shape; the expression names a unit of area, but does not prescribeany particular shape. 

125. --THE MINERS' HOLIDAY. 

Bill Harris must have spent thirteen shillings and sixpence, which wouldbe three shillings more than the average for the seven men--half aguinea. 

126. --SIMPLE MULTIPLICATION. 

The number required is 3, 529, 411, 764, 705, 882, which may be multiplied by3 and divided by 2, by the simple expedient of removing the 3 from oneend of the row to the other. If you want a longer number, you canincrease this one to any extent by repeating the sixteen figures in thesame order. 

127. --SIMPLE DIVISION. 

Subtract every number in turn from every other number, and we get 358(twice), 716, 1, 611, 1, 253, and 895. Now, we see at a glance that, as358 equals 2 × 179, the only number that can divide in every casewithout a remainder will be 179. On trial we find that this is such adivisor. Therefore, 179 is the divisor we want, which always leaves aremainder 164 in the case of the original numbers given. 

128. --A PROBLEM IN SQUARES. 

The sides of the three boards measure 31 in. , 41 in. , and 49 in. Thecommon difference of area is exactly five square feet. Three numberswhose squares are in A. P. , with a common difference of 7, are 113/120, 337/120, 463/120; and with a common difference of 13 are 80929/19380, 106921/19380, and 127729/19380. In the case of whole square numbers thecommon difference will always be divisible by 24, so it is obvious thatour squares must be fractional. Readers should now try to solve the casewhere the common difference is 23. It is rather a hard nut. 

129. --THE BATTLE OF HASTINGS. 

Any number (not itself a square number) may be multiplied by a squarethat will give a product 1 less than another square. The given numbermust not itself be a square, because a square multiplied by a squareproduces a square, and no square plus 1 can be a square. My remarksthroughout must be understood to apply to whole numbers, becausefractional soldiers are not of much use in war. 

Now, of all the numbers from 2 to 99 inclusive, 61 happens to be themost awkward one to work, and the lowest possible answer to our puzzleis that Harold's army consisted of 3, 119, 882, 982, 860, 264, 400 men. Thatis, there would be 51, 145, 622, 669, 840, 400 men (the square of226, 153, 980) in each of the sixty-one squares. Add one man (Harold), andthey could then form one large square with 1, 766, 319, 049 men on everyside. The general problem, of which this is a particular case, is knownas the "Pellian Equation"--apparently because Pell neither firstpropounded the question nor first solved it! It was issued as achallenge by Fermat to the English mathematicians of his day. It isreadily solved by the use of continued fractions. 

Next to 61, the most difficult number under 100 is 97, where 97 ×6, 377, 352² + 1 = a square. 

The reason why I assumed that there must be something wrong with thefigures in the chronicle is that we can confidently say that Harold'sarmy did not contain over three trillion men! If this army (not tomention the Normans) had had the whole surface of the earth (seaincluded) on which to encamp, each man would have had slightly more thana quarter of a square inch of space in which to move about! Put anotherway: Allowing one square foot of standing-room per man, each smallsquare would have required all the space allowed by a globe three timesthe diameter of the earth. 

130. --THE SCULPTOR'S PROBLEM. 

A little thought will make it clear that the answer must be fractional, and that in one case the numerator will be greater and in the other caseless than the denominator. As a matter of fact, the height of the largercube must be 8/7 ft. , and of the smaller 3/7 ft. , if we are to have theanswer in the smallest possible figures. Here the lineal measurement is11/7 ft. --that is, 1+4/7 ft. What are the cubic contents of the twocubes? First 8/7 × 3/7 × 8/7 = 512/343, and secondly 3/7 × 3/7 × 3/7 =27/343. Add these together and the result is 539/343, which reduces to11/7 or 1+4/7 ft. We thus see that the answers in cubic feet and linealfeet are precisely the same. 

The germ of the idea is to be found in the works of Diophantus ofAlexandria, who wrote about the beginning of the fourth century. Thesefractional numbers appear in triads, and are obtained from threegenerators, a, b, c, where a is the largest and c the smallest. 

Then ab + c² = denominator, and a² - c², b² - c², and a² - b² will bethe three numerators. Thus, using the generators 3, 2, 1, we get 8/7, 3/7, 5/7 and we can pair the first and second, as in the abovesolution, or the first and third for a second solution. Thedenominator must always be a prime number of the form 6n + 1, orcomposed of such primes. Thus you can have 13, 19, etc. , asdenominators, but not 25, 55, 187, etc. 

When the principle is understood there is no difficulty in writing downthe dimensions of as many sets of cubes as the most exacting collectormay require. If the reader would like one, for example, with plenty ofnines, perhaps the following would satisfy him: 99999999/99990001 and19999/99990001. 

131. --THE SPANISH MISER. 

There must have been 386 doubloons in one box, 8, 450 in another, and16, 514 in the third, because 386 is the smallest number that can occur. If I had asked for the smallest aggregate number of coins, the answerwould have been 482, 3, 362, and 6, 242. It will be found in either casethat if the contents of any two of the three boxes be combined, theyform a square number of coins. It is a curious coincidence (nothingmore, for it will not always happen) that in the first solution thedigits of the three numbers add to 17 in every case, and in the secondsolution to 14. It should be noted that the middle one of the threenumbers will always be half a square. 

132. --THE NINE TREASURE BOXES. 

Here is the answer that fulfils the conditions:--

 A = 4 B = 3, 364 C = 6, 724 D = 2, 116 E = 5, 476 F = 8, 836 G = 9, 409 H = 12, 769 I = 16, 129

Each of these is a square number, the roots, taken in alphabeticalorder, being 2, 58, 82, 46, 74, 94, 97, 113, and 127, while the requireddifference between A and B, B and C, D and E. Etc. , is in every case3, 360. 

133. --THE FIVE BRIGANDS. 

The sum of 200 doubloons might have been held by the five brigands inany one of 6, 627 different ways. Alfonso may have held any number from 1to 11. If he held 1 doubloon, there are 1, 005 different ways ofdistributing the remainder; if he held 2, there are 985 ways; if 3, there are 977 ways; if 4, there are 903 ways; if 5 doubloons, 832 ways;if 6 doubloons, 704 ways; if 7 doubloons, 570 ways; if 8 doubloons, 388ways; if 9 doubloons, 200 ways; if 10 doubloons, 60 ways; and if Alfonsoheld 11 doubloons, the remainder could be distributed in 3 differentways. More than 11 doubloons he could not possibly have had. It willscarcely be expected that I shall give all these 6, 627 ways at length. What I propose to do is to enable the reader, if he should feel sodisposed, to write out all the answers where Alfonso has one and thesame amount. Let us take the cases where Alfonso has 6 doubloons, andsee how we may obtain all the 704 different ways indicated above. Hereare two tables that will serve as keys to all these answers:--

 Table I. Table II. A = 6. A = 6. B = n. B = n. C = (63 - 5n) + m. C = 1 + m. D = (128 + 4n) - 4m. D = (376 - 16n) - 4m. E = 3 + 3m. E = (15n - 183) + 3m. 

In the first table we may substitute for n any whole number from 1 to 12inclusive, and m may be nought or any whole number from 1 to (31 + n)inclusive. In the second table n may have the value of any whole numberfrom 13 to 23 inclusive, and m may be nought or any whole number from 1to (93 - 4n) inclusive. The first table thus gives (32 + n) answers forevery value of n; and the second table gives (94 - 4n) answers for everyvalue of n. The former, therefore, produces 462 and the latter 242answers, which together make 704, as already stated. 

Let us take Table I. , and say n = 5 and m = 2; also in Table II. Take n= 13 and m = 0. Then we at once get these two answers:--

 A = 6 A = 6 B = 5 B = 13 C = 40 C = 1 D = 140 D = 168 E = 9 E = 12 --- --- 200 doubloons 200 doubloons. 

These will be found to work correctly. All the rest of the 704 answers, where Alfonso always holds six doubloons, may be obtained in this wayfrom the two tables by substituting the different numbers for theletters m and n. 

Put in another way, for every holding of Alfonso the number of answersis the sum of two arithmetical progressions, the common difference inone case being 1 and in the other -4. Thus in the case where Alfonsoholds 6 doubloons one progression is 33 + 34 + 35 + 36 + . . . + 43 + 44, and the other 42 + 38 + 34 + 30 + . . . + 6 + 2. The sum of the firstseries is 462, and of the second 242--results which again agree with thefigures already given. The problem may be said to consist in finding thefirst and last terms of these progressions. I should remark that whereAlfonso holds 9, 10, or 11 there is only one progression, of the secondform. 

134. --THE BANKER'S PUZZLE. 

In order that a number of sixpences may not be divisible into a numberof equal piles, it is necessary that the number should be a prime. Ifthe banker can bring about a prime number, he will win; and I will showhow he can always do this, whatever the customer may put in the box, andthat therefore the banker will win to a certainty. The banker must firstdeposit forty sixpences, and then, no matter how many the customer mayadd, he will desire the latter to transfer from the counter the squareof the number next below what the customer put in. Thus, banker puts 40, customer, we will say, adds 6, then transfers from the counter 25 (thesquare of 5), which leaves 71 in all, a prime number. Try again. Bankerputs 40, customer adds 12, then transfers 121 (the square of 11), asdesired, which leaves 173, a prime number. The key to the puzzle is thecurious fact that any number up to 39, if added to its square and thesum increased by 41, makes a prime number. This was first discovered byEuler, the great mathematician. It has been suggested that the bankermight desire the customer to transfer sufficient to raise the contentsof the box to a given number; but this would not only make the thing anabsurdity, but breaks the rule that neither knows what the other putsin. 

135. --THE STONEMASON'S PROBLEM. 

The puzzle amounts to this. Find the smallest square number that may beexpressed as the sum of more than three consecutive cubes, the cube 1being barred. As more than three heaps were to be supplied, thiscondition shuts out the otherwise smallest answer, 23³ + 24³ + 25³ =204². But it admits the answer, 25³ + 26³ + 27³ + 28³ + 29³ = 315². Thecorrect answer, however, requires more heaps, but a smaller aggregatenumber of blocks. Here it is: 14³ + 15³ + . . . Up to 25³ inclusive, ortwelve heaps in all, which, added together, make 97, 344 blocks of stonethat may be laid out to form a square 312 × 312. I will just remark thatone key to the solution lies in what are called triangular numbers. (Seepp. 13, 25, and 166. )

136. --THE SULTAN'S ARMY. 

The smallest primes of the form 4n + 1 are 5, 13, 17, 29, and 37, andthe smallest of the form 4n - 1 are 3, 7, 11, 19, and 23. Now, primes ofthe first form can always be expressed as the sum of two squares, and inonly one way. Thus, 5 = 4 + 1; 13 = 9 + 4; 17 = 16 + 1; 29 = 25 + 4; 37= 36 + 1. But primes of the second form can never be expressed as thesum of two squares in any way whatever. 

In order that a number may be expressed as the sum of two squares inseveral different ways, it is necessary that it shall be a compositenumber containing a certain number of primes of our first form. Thus, 5or 13 alone can only be so expressed in one way; but 65, (5 × 13), canbe expressed in two ways, 1, 105, (5 × 13 × 17), in four ways, 32, 045, (5× 13 × 17 × 29), in eight ways. We thus get double as many ways forevery new factor of this form that we introduce. Note, however, that Isay _new_ factor, for the _repetition_ of factors is subject to anotherlaw. We cannot express 25, (5 × 5), in two ways, but only in one; yet125, (5 × 5 × 5), can be given in two ways, and so can 625, (5 × 5 × 5 ×5); while if we take in yet another 5 we can express the number as thesum of two squares in three different ways. 

If a prime of the second form gets into your composite number, then thatnumber cannot be the sum of two squares. Thus 15, (3 × 5), will notwork, nor will 135, (3 × 3 × 3 × 5); but if we take in an even number of3's it will work, because these 3's will themselves form a squarenumber, but you will only get one solution. Thus, 45, (3 × 3 × 5, or 9 ×5) = 36 + 9. Similarly, the factor 2 may always occur, or any power of2, such as 4, 8, 16, 32; but its introduction or omission will neveraffect the number of your solutions, except in such a case as 50, whereit doubles a square and therefore gives you the two answers, 49 + 1 and25 + 25. 

Now, directly a number is decomposed into its prime factors, it ispossible to tell at a glance whether or not it can be split into twosquares; and if it can be, the process of discovery in how many ways isso simple that it can be done in the head without any effort. The numberI gave was 130. I at once saw that this was 2 × 5 × 13, and consequentlythat, as 65 can be expressed in two ways (64 + 1 and 49 + 16), 130 canalso be expressed in two ways, the factor 2 not affecting the question. 

The smallest number that can be expressed as the sum of two squares intwelve different ways is 160, 225, and this is therefore the smallestarmy that would answer the Sultan's purpose. The number is composed ofthe factors 5 × 5 × 13 × 17 × 29, each of which is of the required form. If they were all different factors, there would be sixteen ways; but asone of the factors is repeated, there are just twelve ways. Here are thesides of the twelve pairs of squares: (400 and 15), (399 and 32), (393and 76), (392 and 81), (384 and 113), (375 and 140), (360 and 175), (356and 183), (337 and 216), (329 and 228), (311 and 252), (265 and 300). Square the two numbers in each pair, add them together, and their sumwill in every case be 160, 225. 

137. --A STUDY IN THRIFT. 

Mrs. Sandy McAllister will have to save a tremendous sum out of herhousekeeping allowance if she is to win that sixth present that hercanny husband promised her. And the allowance must be a very liberal oneif it is to admit of such savings. The problem required that we shouldfind five numbers higher than 36 the units of which may be displayed soas to form a square, a triangle, two triangles, and three triangles, using the complete number in every one of the four cases. 

Every triangular number is such that if we multiply it by 8 and add 1the result is an odd square number. For example, multiply 1, 3, 6, 10, 15 respectively by 8 and add 1, and we get 9, 25, 49, 81, 121, which arethe squares of the odd numbers 3, 5, 7, 9, 11. Therefore in every casewhere 8x² + 1 = a square number, x² is also a triangular. This pointis dealt with in our puzzle, "The Battle of Hastings. " I will now merelyshow again how, when the first solution is found, the others may bediscovered without any difficulty. First of all, here are the figures:--

 8 × 1² + 1 = 3² 8 × 6² + 1 = 17² 8 × 35² + 1 = 99² 8 × 204² + 1 = 577² 8 × 1189² + 1 = 3363² 8 × 6930² + 1 = 19601² 8 × 40391² + 1 = 114243²

The successive pairs of numbers are found in this way:--

 (1 × 3) + (3 × 1) = 6 (8 × 1) + (3 × 3) = 17 (1 × 17) + (3 × 6) = 35 (8 × 6) + (3 × 17) = 99 (1 × 99) + (3 × 35) = 204 (8 × 35) + (3 × 99) = 577

and so on. Look for the numbers in the table above, and the method willexplain itself. 

Thus we find that the numbers 36, 1225, 41616, 1413721, 48024900, and1631432881 will form squares with sides of 6, 35, 204, 1189, 6930, and40391; and they will also form single triangles with sides of 8, 49, 288, 1681, 9800, and 57121. These numbers may be obtained from the lastcolumn in the first table above in this way: simply divide the numbersby 2 and reject the remainder. Thus the integral halves of 17, 99, and577 are 8, 49, and 288. 

All the numbers we have found will form either two or three triangles atwill. The following little diagram will show you graphically at a glancethat every square number must necessarily be the sum of two triangulars, and that the side of one triangle will be the same as the side of thecorresponding square, while the other will be just 1 less. 

[Illustration

 +-----------+ +---------+ |. . . . . /. | |. . . . /. | |. . . . /. . | |. . . /. . | |. . . /. . . | |. . /. . . | |. . /. . . . | |. /. . . . | |. /. . . . . | /. . . . . | /. . . . . . | +---------+ +-----------+

]

Thus a square may always be divided easily into two triangles, and thesum of two consecutive triangulars will always make a square. In numbersit is equally clear, for if we examine the first triangulars--1, 3, 6, 10, 15, 21, 28--we find that by adding all the consecutive pairs in turnwe get the series of square numbers--9, 16, 25, 36, 49, etc. 

The method of forming three triangles from our numbers is equallydirect, and not at all a matter of trial. But I must content myself withgiving actual figures, and just stating that every triangular higherthan 6 will form three triangulars. I give the sides of the triangles, and readers will know from my remarks when stating the puzzle how tofind from these sides the number of counters or coins in each, and socheck the results if they so wish. 

 +----------------------+-----------+---------------+-----------------------+ | Number | Side of | Side of | Sides of Two | Sides of Three | | | Square. | Triangle. | Triangles. | Triangles. | +------------+---------+-----------+---------------+-----------------------+ | 36 | 6 | 8 | 6 + 5 | 5 + 5 + 3 | | 1225 | 35 | 49 | 36 + 34 | 33 + 32 + 16 | | 41616 | 204 | 288 | 204 + 203 | 192 + 192 + 95 | | 1413721 | 1189 | 1681 | 1189 + 1188 | 1121 + 1120 + 560 | | 48024900 | 6930 | 9800 | 6930 + 6929 | 6533 + 6533 + 3267 | | 1631432881 | 40391 | 57121 | 40391 + 40390 | 38081 + 38080 + 19040 | +------------+---------+-----------+---------------+-----------------------+

I should perhaps explain that the arrangements given in the last twocolumns are not the only ways of forming two and three triangles. Thereare others, but one set of figures will fully serve our purpose. We thussee that before Mrs. McAllister can claim her sixth £5 present she mustsave the respectable sum of £1, 631, 432, 881. 

138. --THE ARTILLERYMEN'S DILEMMA. 

We were required to find the smallest number of cannon balls that wecould lay on the ground to form a perfect square, and could pile into asquare pyramid. I will try to make the matter clear to the merestnovice. 

 1 2 3 4 5 6 7 1 3 6 10 15 21 28 1 4 10 20 35 56 84 1 5 14 30 55 91 140

Here in the first row we place in regular order the natural numbers. Each number in the second row represents the sum of the numbers in therow above, from the beginning to the number just over it. Thus 1, 2, 3, 4, added together, make 10. The third row is formed in exactly the sameway as the second. In the fourth row every number is formed by addingtogether the number just above it and the preceding number. Thus 4 and10 make 14, 20 and 35 make 55. Now, all the numbers in the second roware triangular numbers, which means that these numbers of cannon ballsmay be laid out on the ground so as to form equilateral triangles. Thenumbers in the third row will all form our triangular pyramids, whilethe numbers in the fourth row will all form square pyramids. 

Thus the very process of forming the above numbers shows us that everysquare pyramid is the sum of two triangular pyramids, one of which hasthe same number of balls in the side at the base, and the other one ballfewer. If we continue the above table to twenty-four places, we shallreach the number 4, 900 in the fourth row. As this number is the squareof 70, we can lay out the balls in a square, and can form a squarepyramid with them. This manner of writing out the series until we cometo a square number does not appeal to the mathematical mind, but itserves to show how the answer to the particular puzzle may be easilyarrived at by anybody. As a matter of fact, I confess my failure todiscover any number other than 4, 900 that fulfils the conditions, norhave I found any rigid proof that this is the only answer. The problemis a difficult one, and the second answer, if it exists (which I do notbelieve), certainly runs into big figures. 

For the benefit of more advanced mathematicians I will add that thegeneral expression for square pyramid numbers is (2n³ + 3n² + n)/6. For this expression to be also a square number (the special case of 1excepted) it is necessary that n = p² - 1 = 6t², where 2p² - 1 = q²(the "Pellian Equation"). In the case of our solution above, n = 24, p =5, t = 2, q = 7. 

139. --THE DUTCHMEN'S WIVES. 

The money paid in every case was a square number of shillings, becausethey bought 1 at 1s. , 2 at 2s. , 3 at 3s. , and so on. But every husbandpays altogether 63s. More than his wife, so we have to find in how manyways 63 may be the difference between two square numbers. These are thethree only possible ways: the square of 8 less the square of 1, thesquare of 12 less the square of 9, and the square of 32 less the squareof 31. Here 1, 9, and 31 represent the number of pigs bought and thenumber of shillings per pig paid by each woman, and 8, 12, and 32 thesame in the case of their respective husbands. From the furtherinformation given as to their purchases, we can now pair them off asfollows: Cornelius and Gurtrün bought 8 and 1; Elas and Katrün bought 12and 9; Hendrick and Anna bought 32 and 31. And these pairs representcorrectly the three married couples. 

The reader may here desire to know how we may determine the maximumnumber of ways in which a number may be expressed as the differencebetween two squares, and how we are to find the actual squares. Anyinteger except 1, 4, and twice any odd number, may be expressed as thedifference of two integral squares in as many ways as it can be split upinto pairs of factors, counting 1 as a factor. Suppose the number to be5, 940. The factors are 2². 3³. 5. 11. Here the exponents are 2, 3, 1, 1. Always deduct 1 from the exponents of 2 and add 1 to all the otherexponents; then we get 1, 4, 2, 2, and half the product of these fournumbers will be the required number of ways in which 5, 940 may be thedifference of two squares--that is, 8. To find these eight squares, asit is an _even_ number, we first divide by 4 and get 1485, the eightpairs of factors of which are 1 × 1485, 3 × 495, 5 × 297, 9 × 165, 11 ×135, 15 × 99, 27 × 55, and 33 × 45. The sum and difference of any one ofthese pairs will give the required numbers. Thus, the square of 1, 486less the square of 1, 484 is 5, 940, the square of 498 less the square of492 is the same, and so on. In the case of 63 above, the number is_odd_; so we factorize at once, 1 × 63, 3 × 21, 7 × 9. Then we find that_half_ the sum and difference will give us the numbers 32 and 31, 12 and9, and 8 and 1, as shown in the solution to the puzzle. 

The reverse problem, to find the factors of a number when you haveexpressed it as the difference of two squares, is obvious. For example, the sum and difference of any pair of numbers in the last sentence willgive us the factors of 63. Every prime number (except 1 and 2) may beexpressed as the difference of two squares in one way, and in one wayonly. If a number can be expressed as the difference of two squares inmore than one way, it is composite; and having so expressed it, we mayat once obtain the factors, as we have seen. Fermat showed in a letterto Mersenne or Frénicle, in 1643, how we may discover whether a numbermay be expressed as the difference of two squares in more than one way, or proved to be a prime. But the method, when dealing with largenumbers, is necessarily tedious, though in practice it may beconsiderably shortened. In many cases it is the shortest method knownfor factorizing large numbers, and I have always held the opinion thatFermat used it in performing a certain feat in factorizing that ishistorical and wrapped in mystery. 

140. --FIND ADA'S SURNAME. 

The girls' names were Ada Smith, Annie Brown, Emily Jones, MaryRobinson, and Bessie Evans. 

141. --SATURDAY MARKETING. 

As every person's purchase was of the value of an exact number ofshillings, and as the party possessed when they started out fortyshilling coins altogether, there was no necessity for any lady to haveany smaller change, or any evidence that they actually had such change. This being so, the only answer possible is that the women were namedrespectively Anne Jones, Mary Robinson, Jane Smith, and Kate Brown. Itwill now be found that there would be exactly eight shillings left, which may be divided equally among the eight persons in coin without anychange being required. 

142. --THE SILK PATCHWORK. 

[Illustration]

Our illustration will show how to cut the stitches of the patchwork soas to get the square F entire, and four equal pieces, G, H, I, K, thatwill form a perfect Greek cross. The reader will know how to assemblethese four pieces from Fig. 13 in the article. 

[Illustration: Fig. 1. ]

[Illustration: Fig. 2. ]

143. --TWO CROSSES FROM ONE. 

It will be seen that one cross is cut out entire, as A in Fig. 1, whilethe four pieces marked B, C, D and E form the second cross, as in Fig. 2, which will be of exactly the same size as the other. I will leave thereader the pleasant task of discovering for himself the best way offinding the direction of the cuts. Note that the Swastika again appears. 

The difficult question now presents itself: How are we to cut threeGreek crosses from one in the fewest possible pieces? As a matter offact, this problem may be solved in as few as thirteen pieces; but as Iknow many of my readers, advanced geometricians, will be glad to havesomething to work on of which they are not shown the solution, I leavethe mystery for the present undisclosed. 

144. --THE CROSS AND THE TRIANGLE. 

The line A B in the following diagram represents the side of a squarehaving the same area as the cross. I have shown elsewhere, as stated, how to make a square and equilateral triangle of equal area. I need notgo, therefore, into the preliminary question of finding the dimensionsof the triangle that is to equal our cross. We will assume that we havealready found this, and the question then becomes, How are we to cut upone of these into pieces that will form the other?

First draw the line A B where A and B are midway between the extremitiesof the two side arms. Next make the lines D C and E F equal in length tohalf the side of the triangle. Now from E and F describe with the sameradius the intersecting arcs at G and draw F G. Finally make I K equalto H C and L B equal to A D. If we now draw I L, it should be parallelto F G, and all the six pieces are marked out. These fit together andform a perfect equilateral triangle, as shown in the second diagram. Orwe might have first found the direction of the line M N in our triangle, then placed the point O over the point E in the cross and turned roundthe triangle over the cross until the line M N was parallel to A B. Thepiece 5 can then be marked off and the other pieces in succession. 

[Illustration]

I have seen many attempts at a solution involving the assumption thatthe height of the triangle is exactly the same as the height of thecross. This is a fallacy: the cross will always be higher than thetriangle of equal area. 

145. --THE FOLDED CROSS. 

[Illustration: FIG. 1. , FIG 2. ]

First fold the cross along the dotted line A B in Fig. 1. You then haveit in the form shown in Fig. 2. Next fold it along the dotted line C D(where D is, of course, the centre of the cross), and you get the formshown in Fig. 3. Now take your scissors and cut from G to F, and thefour pieces, all of the same size and shape, will fit together and forma square, as shown in Fig. 4. 

[Illustration: FIG. 3. , FIG. 4. ]

146. --AN EASY DISSECTION PUZZLE. 

[Illustration

 +===========+===========+- | · | · : \ | · | · : \ | · | · : \ | · | · : \ | · | · : \ +-----------+===========+===========+ | / : · | · : \ | / : · | · : \ | / : · | · : \ | / : · | · : \ | / : · | · : \ +===========+===========+===========+===========+

]

The solution to this puzzle is shown in the illustration. Divide thefigure up into twelve equal triangles, and it is easy to discover thedirections of the cuts, as indicated by the dark lines. 

147. --AN EASY SQUARE PUZZLE. 

[Illustration

 +-----------------------------------------+ | . /| | . / | | . / | | / / | | / . / | | / . / | | / . / | | / . / | | +--------------------+ | | / / | | / / | | / / | | / . / | | / . / | | / . / | | / . / | | / . | | / . | | / . | |/ . | +-----------------------------------------+

]

The diagram explains itself, one of the five pieces having been cut intwo to form a square. 

148. --THE BUN PUZZLE. 

[Illustration

 . . . . _ . . . |\ A . | \ . C | \ | | \ . | \ / . |______________________\/ | | . . . B . . . . . -

 _ . | . . | . . | . | | | D | E | | | . | . . | . . | . _

 _ . | . . -+- . . . . . - - . | G| F | |

 - - . . . . . - _ _ - . . | . -

 -+- . . - - . | H |

 - - . . - _ _ -

]

The secret of the bun puzzle lies in the fact that, with the relativedimensions of the circles as given, the three diameters will form aright-angled triangle, as shown by A, B, C. It follows that the twosmaller buns are exactly equal to the large bun. Therefore, if we giveDavid and Edgar the two halves marked D and E, they will have their fairshares--one quarter of the confectionery each. Then if we place thesmall bun, H, on the top of the remaining one and trace itscircumference in the manner shown, Fred's piece, F, will exactly equalHarry's small bun, H, with the addition of the piece marked G--half therim of the other. Thus each boy gets an exactly equal share, and thereare only five pieces necessary. 

149. --THE CHOCOLATE SQUARES. 

[Illustration]

Square A is left entire; the two pieces marked B fit together and make asecond square; the two pieces C make a third square; and the four piecesmarked D will form the fourth square. 

150. --DISSECTING A MITRE. 

The diagram on the next page shows how to cut into five pieces to form asquare. The dotted lines are intended to show how to find the points Cand F--the only difficulty. A B is half B D, and A E is parallel to B H. With the point of the compasses at B describe the arc H E, and A E willbe the distance of C from B. Then F G equals B C less A B. 

This puzzle--with the added condition that it shall be cut into fourparts of the same size and shape--I have not been able to trace to anearlier date than 1835. Strictly speaking, it is, in that form, impossible of solution; but I give the answer that is always presented, and that seems to satisfy most people. 

[Illustration]

We are asked to assume that the two portions containing the sameletter--AA, BB, CC, DD--are joined by "a mere hair, " and are, therefore, only one piece. To the geometrician this is absurd, and the four sharesare not equal in area unless they consist of two pieces each. If youmake them equal in area, they will not be exactly alike in shape. 

[Illustration]

151. --THE JOINER'S PROBLEM. 

[Illustration]

Nothing could be easier than the solution of this puzzle--when you knowhow to do it. And yet it is apt to perplex the novice a good deal if hewants to do it in the fewest possible pieces--three. All you have to dois to find the point A, midway between B and C, and then cut from A to Dand from A to E. The three pieces then form a square in the mannershown. Of course, the proportions of the original figure must becorrect; thus the triangle BEF is just a quarter of the square BCDF. Draw lines from B to D and from C to F and this will be clear. 

152. --ANOTHER JOINER'S PROBLEM. 

[Illustration]

THE point was to find a general rule for forming a perfect square out ofanother square combined with a "right-angled isosceles triangle. " Thetriangle to which geometricians give this high-sounding name is, ofcourse, nothing more or less than half a square that has been dividedfrom corner to corner. 

The precise relative proportions of the square and triangle are of noconsequence whatever. It is only necessary to cut the wood or materialinto five pieces. 

Suppose our original square to be ACLF in the above diagram and ourtriangle to be the shaded portion CED. Now, we first find half thelength of the long side of the triangle (CD) and measure off this lengthat AB. Then we place the triangle in its present position against thesquare and make two cuts--one from B to F, and the other from B to E. Strange as it may seem, that is all that is necessary! If we now removethe pieces G, H, and M to their new places, as shown in the diagram, weget the perfect square BEKF. 

Take any two square pieces of paper, of different sizes but perfectsquares, and cut the smaller one in half from corner to corner. Nowproceed in the manner shown, and you will find that the two pieces maybe combined to form a larger square by making these two simple cuts, andthat no piece will be required to be turned over. 

The remark that the triangle might be "a little larger or a good dealsmaller in proportion" was intended to bar cases where area of triangleis greater than area of square. In such cases six pieces are necessary, and if triangle and square are of equal area there is an obvioussolution in three pieces, by simply cutting the square in halfdiagonally. 

153. --A CUTTING-OUT PUZZLE. 

[Illustration]

The illustration shows how to cut the four pieces and form with them asquare. First find the side of the square (the mean proportional betweenthe length and height of the rectangle), and the method is obvious. Ifour strip is exactly in the proportions 9 × 1, or 16 × 1, or 25 × 1, wecan clearly cut it in 3, 4, or 5 rectangular pieces respectively to forma square. Excluding these special cases, the general law is that for astrip in length more than n² times the breadth, and not more than (n+1)²times the breadth, it may be cut in n+2 pieces to form a square, andthere will be n-1 rectangular pieces like piece 4 in the diagram. Thus, for example, with a strip 24 × 1, the length is more than 16 and lessthan 25 times the breadth. Therefore it can be done in 6 pieces (n herebeing 4), 3 of which will be rectangular. In the case where n equals 1, the rectangle disappears and we get a solution in three pieces. Withinthese limits, of course, the sides need not be rational: the solution ispurely geometrical. 

154. --MRS. HOBSON'S HEARTHRUG. 

[Illustration]

As I gave full measurements of the mutilated rug, it was quite an easymatter to find the precise dimensions for the square. The two pieces cutoff would, if placed together, make an oblong piece 12 × 6, giving anarea of 72 (inches or yards, as we please), and as the original completerug measured 36 × 27, it had an area of 972. If, therefore, we deductthe pieces that have been cut away, we find that our new rug willcontain 972 less 72, or 900; and as 900 is the square of 30, we knowthat the new rug must measure 30 × 30 to be a perfect square. This is agreat help towards the solution, because we may safely conclude that thetwo horizontal sides measuring 30 each may be left intact. 

There is a very easy way of solving the puzzle in four pieces, and alsoa way in three pieces that can scarcely be called difficult, but thecorrect answer is in only two pieces. 

It will be seen that if, after the cuts are made, we insert the teeth ofthe piece B one tooth lower down, the two portions will fit together andform a square. 

155. --THE PENTAGON AND SQUARE. 

A regular pentagon may be cut into as few as six pieces that will fittogether without any turning over and form a square, as I shall showbelow. Hitherto the best answer has been in seven pieces--the solutionproduced some years ago by a foreign mathematician, Paul Busschop. Wefirst form a parallelogram, and from that the square. The process willbe seen in the diagram on the next page. 

The pentagon is ABCDE. By the cut AC and the cut FM (F being the middlepoint between A and C, and M being the same distance from A as F) we gettwo pieces that may be placed in position at GHEA and form theparallelogram GHDC. We then find the mean proportional between thelength HD and the _height_ of the parallelogram. This distance we markoff from C at K, then draw CK, and from G drop the line GL, perpendicular to KC. The rest is easy and rather obvious. It will beseen that the six pieces will form either the pentagon or the square. 

I have received what purported to be a solution in five pieces, but themethod was based on the rather subtle fallacy that half the diagonalplus half the side of a pentagon equals the side of a square of the samearea. I say subtle, because it is an extremely close approximation thatwill deceive the eye, and is quite difficult to prove inexact. I am notaware that attention has before been drawn to this curiousapproximation. 

[Illustration]

Another correspondent made the side of his square 1¼ of the side ofthe pentagon. As a matter of fact, the ratio is irrational. I calculatethat if the side of the pentagon is 1--inch, foot, or anything else--theside of the square of equal area is 1. 3117 nearly, or say roughly1+3/10. So we can only hope to solve the puzzle by geometrical methods. 

156. --THE DISSECTED TRIANGLE. 

Diagram A is our original triangle. We will say it measures 5 inches (or5 feet) on each side. If we take off a slice at the bottom of anyequilateral triangle by a cut parallel with the base, the portion thatremains will always be an equilateral triangle; so we first cut offpiece 1 and get a triangle 3 inches on every side. The manner of findingdirections of the other cuts in A is obvious from the diagram. 

Now, if we want two triangles, 1 will be one of them, and 2, 3, 4, and 5will fit together, as in B, to form the other. If we want threeequilateral triangles, 1 will be one, 4 and 5 will form the second, asin C, and 2 and 3 will form the third, as in D. In B and C the piece 5is turned over; but there can be no objection to this, as it is notforbidden, and is in no way opposed to the nature of the puzzle. 

[Illustration]

157. --THE TABLE-TOP AND STOOLS. 

[Illustration]

One object that I had in view when presenting this little puzzle was topoint out the uncertainty of the meaning conveyed by the word "oval. "Though originally derived from the Latin word _ovum_, an egg, yet whatwe understand as the egg-shape (with one end smaller than the other) isonly one of many forms of the oval; while some eggs are spherical inshape, and a sphere or circle is most certainly not an oval. If we speakof an ellipse--a conical ellipse--we are on safer ground, but here wemust be careful of error. I recollect a Liverpool town councillor, manyyears ago, whose ignorance of the poultry-yard led him to substitute theword "hen" for "fowl, " remarking, "We must remember, gentlemen, thatalthough every cock is a hen, every hen is not a cock!" Similarly, wemust always note that although every ellipse is an oval, every oval isnot an ellipse. It is correct to say that an oval is an oblongcurvilinear figure, having two unequal diameters, and bounded by a curveline returning into itself; and this includes the ellipse, but all otherfigures which in any way approach towards the form of an oval withoutnecessarily having the properties above described are included in theterm "oval. " Thus the following solution that I give to our puzzleinvolves the pointed "oval, " known among architects as the "vesicapiscis. "

[Illustration: THE TWO STOOLS. ]

The dotted lines in the table are given for greater clearness, the cutsbeing made along the other lines. It will be seen that the eight piecesform two stools of exactly the same size and shape with similarhand-holes. These holes are a trifle longer than those in theschoolmaster's stools, but they are much narrower and of considerablysmaller area. Of course 5 and 6 can be cut out in one piece--also 7 and8--making only _six pieces_ in all. But I wished to keep the same numberas in the original story. 

When I first gave the above puzzle in a London newspaper, incompetition, no correct solution was received, but an ingenious andneatly executed attempt by a man lying in a London infirmary wasaccompanied by the following note: "Having no compasses here, I wascompelled to improvise a pair with the aid of a small penknife, a bit offirewood from a bundle, a piece of tin from a toy engine, a tin tack, and two portions of a hairpin, for points. They are a fairly serviceablepair of compasses, and I shall keep them as a memento of your puzzle. "

158. --THE GREAT MONAD. 

The areas of circles are to each other as the squares of theirdiameters. If you have a circle 2 in. In diameter and another 4 in. Indiameter, then one circle will be four times as great in area as theother, because the square of 4 is four times as great as the square of2. Now, if we refer to Diagram 1, we see how two equal squares may becut into four pieces that will form one larger square; from which it isself-evident that any square has just half the area of the square of itsdiagonal. In Diagram 2 I have introduced a square as it often occurs inancient drawings of the Monad; which was my reason for believing thatthe symbol had mathematical meanings, since it will be found todemonstrate the fact that the area of the outer ring or annulus isexactly equal to the area of the inner circle. Compare Diagram 2 withDiagram 1, and you will see that as the square of the diameter CD isdouble the square of the diameter of the inner circle, or CE, thereforethe area of the larger circle is double the area of the smaller one, andconsequently the area of the annulus is exactly equal to that of theinner circle. This answers our first question. 

[Illustration: 1 2 3 4]

In Diagram 3 I show the simple solution to the second question. It isobviously correct, and may be proved by the cutting and superposition ofparts. The dotted lines will also serve to make it evident. The thirdquestion is solved by the cut CD in Diagram 2, but it remains to beproved that the piece F is really one-half of the Yin or the Yan. Thiswe will do in Diagram 4. The circle K has one-quarter the area of thecircle containing Yin and Yan, because its diameter is just one-half thelength. Also L in Diagram 3 is, we know, one-quarter the area. It istherefore evident that G is exactly equal to H, and therefore half G isequal to half H. So that what F loses from L it gains from K, and F mustbe half of Yin or Yan. 

159. --THE SQUARE OF VENEER. 

[Illustration:

 +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | |: | || | | | | :| | |: | | :| | |: | || | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | |: | || | | | | :| | |: | | :| | |: | || | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | |: | || | |_ _|___|__:|___|___|:__|___|__:|___|___|:__|__||___| +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | |: | || | | | | :| | |: | | :| | |: | || | +---+---+---+---+---+---+---+---+---+---+===+===+---+ | | | :| | |: | | :| | ||: | | | | | | :| | |: | | :| | ||: | | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ |¯¯¯|¯¯¯|¯¯:|¯¯¯|¯¯¯|:¯¯|¯¯¯|¯¯:|¯¯¯|¯¯||:¯¯|¯¯¯|¯¯¯| | | | :| | |: | | :| | ||: | | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | ||: | | | | | | :| | |: | B | :| | ||: | C | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :| | ||: | | | |_ _|___|__:|___|___|:__|___|__:|___|__||:__|___|___| +---+---+---+---+---+---+---+---+===+===+===+===+===+ | | | :| | |: | | :|| | |: | | | | | | :| | |: | | :|| | |: | | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :|| | |: | | | | | | :| | |: | | :|| | |: | A | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ |¯¯¯|¯¯¯|¯¯:|¯¯¯|¯¯¯|:¯¯|¯¯¯|¯¯:||¯¯|¯¯¯|:¯¯|¯¯¯|¯¯¯| | | | :| | |: | | :|| | |: | | | +---+---+---+---+---+---+---+---+---+---+---+---+---+ | | | :| | |: | | :|| | |: | | | | | | :| | |: | | :|| | |: | | | +===+===+===+===+===+===+===+===+---+---+---+---+---+ | | | :| | |: | | :|| | |: | | | | | | :| | D |: | | :|| | |: | | | +---+---+---+---+---+---+---+---+===+===+===+===+===+

]

Any square number may be expressed as the sum of two squares in aninfinite number of different ways. The solution of the present puzzleforms a simple demonstration of this rule. It is a condition that wegive actual dimensions. 

In this puzzle I ignore the known dimensions of our square and work onthe assumption that it is 13n by 13n. The value of n we can afterwardsdetermine. Divide the square as shown (where the dotted lines indicatethe original markings) into 169 squares. As 169 is the sum of the twosquares 144 and 25, we will proceed to divide the veneer into twosquares, measuring respectively 12 × 12 and 5 × 5; and as we know thattwo squares may be formed from one square by dissection in four pieces, we seek a solution in this number. The dark lines in the diagram showwhere the cuts are to be made. The square 5 × 5 is cut out whole, andthe larger square is formed from the remaining three pieces, B, C, andD, which the reader can easily fit together. 

Now, n is clearly 5/13 of an inch. Consequently our larger square mustbe 60/13 in. × 60/13 in. , and our smaller square 25/13 in. × 25/13 in. The square of 60/13 added to the square of 25/13 is 25. The square isthus divided into as few as four pieces that form two squares of knowndimensions, and all the sixteen nails are avoided. 

Here is a general formula for finding two squares whose sum shall equala given square, say a². In the case of the solution of our puzzle p = 3, q = 2, and a = 5. 

 ________________________ 2pqa \/ a²( p² + q²)² - (2pqa)² --------- = x; --------------------------- = y p² + q² p² + q²

 Here x² + y² = a². 

160. --THE TWO HORSESHOES. 

The puzzle was to cut the two shoes (including the hoof contained withinthe outlines) into four pieces, two pieces each, that would fit togetherand form a perfect circle. It was also stipulated that all four piecesshould be different in shape. As a matter of fact, it is a puzzle basedon the principle contained in that curious Chinese symbol the Monad. (See No. 158. )

[Illustration]

The above diagrams give the correct solution to the problem. It will benoticed that 1 and 2 are cut into the required four pieces, alldifferent in shape, that fit together and form the perfect circle shownin Diagram 3. It will further be observed that the two pieces A and B ofone shoe and the two pieces C and D of the other form two exactlysimilar halves of the circle--the Yin and the Yan of the great Monad. Itwill be seen that the shape of the horseshoe is more easily determinedfrom the circle than the dimensions of the circle from the horseshoe, though the latter presents no difficulty when you know that the curve ofthe long side of the shoe is part of the circumference of your circle. The difference between B and D is instructive, and the idea is useful inall such cases where it is a condition that the pieces must be differentin shape. In forming D we simply add on a symmetrical piece, acurvilinear square, to the piece B. Therefore, in giving either B or D aquarter turn before placing in the new position, a precisely similareffect must be produced. 

161. --THE BETSY ROSS PUZZLE. 

Fold the circular piece of paper in half along the dotted line shown inFig. 1, and divide the upper half into five equal parts as indicated. Now fold the paper along the lines, and it will have the appearanceshown in Fig. 2. If you want a star like Fig. 3, cut from A to B; if youwish one like Fig. 4, cut from A to C. Thus, the nearer you cut to thepoint at the bottom the longer will be the points of the star, and thefarther off from the point that you cut the shorter will be the pointsof the star. 

[Illustration]

162. --THE CARDBOARD CHAIN. 

The reader will probably feel rewarded for any care and patience thathe may bestow on cutting out the cardboard chain. We will suppose thathe has a piece of cardboard measuring 8 in. By 2½ in. , though thedimensions are of no importance. Yet if you want a long chain youmust, of course, take a long strip of cardboard. First rule pencillines B B and C C, half an inch from the edges, and also the shortperpendicular lines half an inch apart. (See next page. ) Rule lines onthe other side in just the same way, and in order that they shallcoincide it is well to prick through the card with a needle the pointswhere the short lines end. Now take your penknife and split the cardfrom A A down to B B, and from D D up to C C. Then cut right throughthe card along all the short perpendicular lines, and half through thecard along the short portions of B B and C C that are not dotted. Nextturn the card over and cut half through along the short lines on B Band C C at the places that are immediately beneath the dotted lines onthe upper side. With a little careful separation of the parts with thepenknife, the cardboard may now be divided into two interlacingladder-like portions, as shown in Fig. 2; and if you cut away all theshaded parts you will get the chain, cut solidly out of the cardboard, without any join, as shown in the illustrations on page 40. 

It is an interesting variant of the puzzle to cut out two keys on aring--in the same manner without join. 

[Illustration]

164. --THE POTATO PUZZLE. 

As many as twenty-two pieces may be obtained by the six cuts. Theillustration shows a pretty symmetrical solution. The rule in such casesis that every cut shall intersect every other cut and no twointersections coincide; that is to say, every line passes through everyother line, but more than two lines do not cross at the same pointanywhere. There are other ways of making the cuts, but this rule mustalways be observed if we are to get the full number of pieces. 

The general formula is that with n cuts we can always produce (n(n +1) + 1)/2 . One of the problems proposed by the late Sam Loyd was toproduce the maximum number of pieces by n straight cuts through asolid cheese. Of course, again, the pieces cut off may not be moved orpiled. Here we have to deal with the intersection of planes (insteadof lines), and the general formula is that with n cuts we may produce((n - 1)n(n + 1))/6 + n + 1 pieces. It is extremely difficult to "see"the direction and effects of the successive cuts for more than a fewof the lowest values of n. 

165. --THE SEVEN PIGS. 

The illustration shows the direction for placing the three fences so asto enclose every pig in a separate sty. The greatest number of spacesthat can be enclosed with three straight lines in a square is seven, asshown in the last puzzle. Bearing this fact in mind, the puzzle must besolved by trial. 

[Illustration: THE SEVEN PIGS. ]

166. --THE LANDOWNER'S FENCES. 

Four fences only are necessary, as follows:--

[Illustration]

167. --THE WIZARD'S CATS. 

The illustration requires no explanation. It shows clearly how the threecircles may be drawn so that every cat has a separate enclosure, andcannot approach another cat without crossing a line. 

[Illustration: THE WIZARDS' CATS. ]

168. --THE CHRISTMAS PUDDING. 

The illustration shows how the pudding may be cut into two parts ofexactly the same size and shape. The lines must necessarily pass throughthe points A, B, C, D, and E. But, subject to this condition, they maybe varied in an infinite number of ways. For example, at a point midwaybetween A and the edge, the line may be completed in an unlimited numberof ways (straight or crooked), provided it be exactly reflected from Eto the opposite edge. And similar variations may be introduced at otherplaces. 

[Illustration]

169. --A TANGRAM PARADOX. 

The diagrams will show how the figures are constructed--each with theseven Tangrams. It will be noticed that in both cases the head, hat, andarm are precisely alike, and the width at the base of the body thesame. But this body contains four pieces in the first case, and in thesecond design only three. The first is larger than the second by exactlythat narrow strip indicated by the dotted line between A and B. Thisstrip is therefore exactly equal in area to the piece forming the footin the other design, though when thus distributed along the side of thebody the increased dimension is not easily apparent to the eye. 

[Illustration]

170. --THE CUSHION COVERS. 

[Illustration]

The two pieces of brocade marked A will fit together and form oneperfect square cushion top, and the two pieces marked B will form theother. 

171. --THE BANNER PUZZLE. 

The illustration explains itself. Divide the bunting into 25 squares(because this number is the sum of two other squares--16 and 9), andthen cut along the thick lines. The two pieces marked A form one square, and the two pieces marked B form the other. 

[Illustration]

172. --MRS. SMILEY'S CHRISTMAS PRESENT. 

[Illustration]

[Illustration]

The first step is to find six different square numbers that sum to 196. For example, 1 + 4 + 25 + 36 + 49 + 81 = 196; 1 + 4 + 9 + 25 + 36 + 121= 196; 1 + 9 + 16 + 25 + 64 + 81 = 196. The rest calls for individualjudgment and ingenuity, and no definite rules can be given forprocedure. The annexed diagrams will show solutions for the first twocases stated. Of course the three pieces marked A and those marked Bwill fit together and form a square in each case. The assembling of theparts may be slightly varied, and the reader may be interested infinding a solution for the third set of squares I have given. 

173. --MRS. PERKINS'S QUILT. 

The following diagram shows how the quilt should be constructed. 

[Illustration]

There is, I believe, practically only one solution to this puzzle. Thefewest separate squares must be eleven. The portions must be of thesizes given, the three largest pieces must be arranged as shown, and theremaining group of eight squares may be "reflected, " but cannot bedifferently arranged. 

174. --THE SQUARES OF BROCADE. 

[Illustration: Diagram 1]

So far as I have been able to discover, there is only one possiblesolution to fulfil the conditions. The pieces fit together as in Diagram1, Diagrams 2 and 3 showing how the two original squares are to be cut. It will be seen that the pieces A and C have each twenty chequers, andare therefore of equal area. Diagram 4 (built up with the dissectedsquare No. 5) solves the puzzle, except for the small conditioncontained in the words, "I cut the _two_ squares in the manner desired. "In this case the smaller square is preserved intact. Still I give it asan illustration of a feature of the puzzle. It is impossible in aproblem of this kind to give a _quarter-turn_ to any of the pieces ifthe pattern is to properly match, but (as in the case of F, in Diagram4) we may give a symmetrical piece a _half-turn_--that is, turn itupside down. Whether or not a piece may be given a quarter-turn, ahalf-turn, or no turn at all in these chequered problems, depends on thecharacter of the design, on the material employed, and also on the formof the piece itself. 

[Illustration: Diagram 2]

[Illustration: Diagram 3]

[Illustration: Diagram 4]

[Illustration: Diagram 5]

175. --ANOTHER PATCHWORK PUZZLE. 

The lady need only unpick the stitches along the dark lines in thelarger portion of patchwork, when the four pieces will fit together andform a square, as shown in our illustration. 

[Illustration]

176. --LINOLEUM CUTTING. 

There is only one solution that will enable us to retain the larger ofthe two pieces with as little as possible cut from it. Fig. 1 in thefollowing diagram shows how the smaller piece is to be cut, and Fig. 2how we should dissect the larger piece, while in Fig. 3 we have the newsquare 10 × 10 formed by the four pieces with all the chequers properlymatched. It will be seen that the piece D contains fifty-two chequers, and this is the largest piece that it is possible to preserve under theconditions. 

[Illustration]

177. --ANOTHER LINOLEUM PUZZLE. 

Cut along the thick lines, and the four pieces will fit together andform a perfect square in the manner shown in the smaller diagram. 

[Illustration: ANOTHER LINOLEUM PUZZLE. ]

178. --THE CARDBOARD BOX. 

The areas of the top and side multiplied together and divided by thearea of the end give the square of the length. Similarly, the product oftop and end divided by side gives the square of the breadth; and theproduct of side and end divided by the top gives the square of thedepth. But we only need one of these operations. Let us take the first. Thus, 120 × 96 divided by 80 equals 144, the square of 12. Therefore thelength is 12 inches, from which we can, of course, at once get thebreadth and depth--10 in. And 8 in. Respectively. 

179. --STEALING THE BELL-ROPES. 

Whenever we have one side (a) of a right-angled triangle, and know thedifference between the second side and the hypotenuse (which differencewe will call b), then the length of the hypotenuse will be

 a² b --- + -. 2b 2

In the case of our puzzle this will be

 48 × 48 ------- + 1½ in. = 32 ft. 1½ in. , 6

which is the length of the rope. 

180-- THE FOUR SONS. 

[Illustration]

The diagram shows the most equitable division of the land possible, "sothat each son shall receive land of exactly the same area and exactlysimilar in shape, " and so that each shall have access to the well inthe centre without trespass on another's land. The conditions do notrequire that each son's land shall be in one piece, but it is necessarythat the two portions assigned to an individual should be kept apart, ortwo adjoining portions might be held to be one piece, in which case thecondition as to shape would have to be broken. At present there is onlyone shape for each piece of land--half a square divided diagonally. AndA, B, C, and D can each reach their land from the outside, and have eachequal access to the well in the centre. 

181. --THE THREE RAILWAY STATIONS. 

The three stations form a triangle, with sides 13, 14, and 15 miles. Make the 14 side the base; then the height of the triangle is 12 and thearea 84. Multiply the three sides together and divide by four times thearea. The result is eight miles and one-eighth, the distance required. 

182. --THE GARDEN PUZZLE. 

Half the sum of the four sides is 144. From this deduct in turn the foursides, and we get 64, 99, 44, and 81. Multiply these together, and wehave as the result the square of 4, 752. Therefore the garden contained4, 752 square yards. Of course the tree being equidistant from the fourcorners shows that the garden is a quadrilateral that may be inscribedin a circle. 

183. --DRAWING A SPIRAL. 

Make a fold in the paper, as shown by the dotted line in theillustration. Then, taking any two points, as A and B, describesemicircles on the line alternately from the centres B and A, beingcareful to make the ends join, and the thing is done. Of course this isnot a _true_ spiral, but the puzzle was to produce the _particular_spiral that was shown, and that was drawn in this simple manner. 

[Illustration]

184. --HOW TO DRAW AN OVAL. 

If you place your sheet of paper round the surface of a cylindricalbottle or canister, the oval can be drawn with one sweep of thecompasses. 

185. --ST. GEORGE'S BANNER. 

As the flag measures 4 ft. By 3 ft. , the length of the diagonal (fromcorner to corner) is 5 ft. All you need do is to deduct half thelength of this diagonal (2½ ft. ) from a quarter of the distance allround the edge of the flag (3½ ft. )--a quarter of 14 ft. Thedifference (1 ft. ) is the required width of the arm of the red cross. The area of the cross will then be the same as that of the whiteground. 

186. --THE CLOTHES LINE PUZZLE. 

Multiply together, and also add together, the heights of the two polesand divide one result by the other. That is, if the two heights are aand b respectively, then ab/(a + b) will give the height of theintersection. In the particular case of our puzzle, the intersection wastherefore 2 ft. 11 in. From the ground. The distance that the poles areapart does not affect the answer. The reader who may have imagined thatthis was an accidental omission will perhaps be interested indiscovering the reason why the distance between the poles may beignored. 

187. --THE MILKMAID PUZZLE. 

[Illustration:

 A |\ | \ | \ | \ B RIVER +----+-------------- | / \ | / \ | / \ |/ DOOR STOOL

]

Draw a straight line, as shown in the diagram, from the milking-stoolperpendicular to the near bank of the river, and continue it to thepoint A, which is the same distance from that bank as the stool. If younow draw the straight line from A to the door of the dairy, it will cutthe river at B. Then the shortest route will be from the stool to B andthence to the door. Obviously the shortest distance from A to the dooris the straight line, and as the distance from the stool to any point ofthe river is the same as from A to that point, the correctness of thesolution will probably appeal to every reader without any acquaintancewith geometry. 

188. --THE BALL PROBLEM. 

If a round ball is placed on the level ground, six similar balls may beplaced round it (all on the ground), so that they shall all touch thecentral ball. 

As for the second question, the ratio of the diameter of a circle to itscircumference we call _pi_; and though we cannot express this ratio inexact numbers, we can get sufficiently near to it for all practicalpurposes. However, in this case it is not necessary to know the value of_pi_ at all. Because, to find the area of the surface of a sphere wemultiply the square of the diameter by _pi_; to find the volume of asphere we multiply the cube of the diameter by one-sixth of _pi_. Therefore we may ignore _pi_, and have merely to seek a number whosesquare shall equal one-sixth of its cube. This number is obviously 6. Therefore the ball was 6 ft. In diameter, for the area of its surfacewill be 36 times _pi_ in square feet, and its volume also 36 times _pi_in cubic feet. 

189. --THE YORKSHIRE ESTATES. 

The triangular piece of land that was not for sale contains exactlyeleven acres. Of course it is not difficult to find the answer if wefollow the eccentric and tricky tracks of intricate trigonometry; or Imight say that the application of a well-known formula reduces theproblem to finding one-quarter of the square root of (4 × 370 × 116)-(370 + 116 - 74)²--that is a quarter of the square root of 1936, whichis one-quarter of 44, or 11 acres. But all that the reader reallyrequires to know is the Pythagorean law on which many puzzles have beenbuilt, that in any right-angled triangle the square of the hypotenuse isequal to the sum of the squares of the other two sides. I shall dispensewith all "surds" and similar absurdities, notwithstanding the fact thatthe sides of our triangle are clearly incommensurate, since we cannotexactly extract the square roots of the three square areas. 

[Illustration:

 A |\ | \. | \ . |5 \ . | 7 \ . E +--------- +C . | | ` . . | | `. . |4 |4 ` . . | 7 | ` . . D----------+----------------- B F

]

In the above diagram ABC represents our triangle. ADB is a right-angledtriangle, AD measuring 9 and BD measuring 17, because the square of 9added to the square of 17 equals 370, the known area of the square onAB. Also AEC is a right-angled triangle, and the square of 5 added tothe square of 7 equals 74, the square estate on A C. Similarly, CFB is aright-angled triangle, for the square of 4 added to the square of 10equals 116, the square estate on BC. Now, although the sides of ourtriangular estate are incommensurate, we have in this diagram all theexact figures that we need to discover the area with precision. 

The area of our triangle ADB is clearly half of 9 × 17, or 76½ acres. The area of AEC is half of 5 × 7, or 17½ acres; the area of CFB is halfof 4 × 10, or 20 acres; and the area of the oblong EDFC is obviously 4 ×7, or 28 acres. Now, if we add together 17½, 20, and 28 = 65½, anddeduct this sum from the area of the large triangle ADB (which we havefound to be 76½ acres), what remains must clearly be the area of ABC. That is to say, the area we want must be 76½ - 65½ = 11 acres exactly. 

190. --FARMER WURZEL'S ESTATE. 

The area of the complete estate is exactly one hundred acres. To findthis answer I use the following little formula, __________________ \/4ab - (a + b + c)² -------------------- 4

where a, b, c represent the three square areas, in any order. Theexpression gives the area of the triangle A. This will be found to be 9acres. It can be easily proved that A, B, C, and D are all equal inarea; so the answer is 26 + 20 + 18 + 9 + 9 + 9 + 9 = 100 acres. 

[Illustration]

Here is the proof. If every little dotted square in the diagramrepresents an acre, this must be a correct plan of the estate, for thesquares of 5 and 1 together equal 26; the squares of 4 and 2 equal 20;and the squares of 3 and 3 added together equal 18. Now we see at oncethat the area of the triangle E is 2½, F is 4½, and G is 4. These addedtogether make 11 acres, which we deduct from the area of the rectangle, 20 acres, and we find that the field A contains exactly 9 acres. If youwant to prove that B, C, and D are equal in size to A, divide them intwo by a line from the middle of the longest side to the opposite angle, and you will find that the two pieces in every case, if cut out, willexactly fit together and form A. 

Or we can get our proof in a still easier way. The complete area of thesquared diagram is 12 × 12 = 144 acres, and the portions 1, 2, 3, 4, notincluded in the estate, have the respective areas of 12½, 17½, 9½, and4½. These added together make 44, which, deducted from 144, leaves 100as the required area of the complete estate. 

191. --THE CRESCENT PUZZLE. 

Referring to the original diagram, let AC be x, let CD be x - 9, and letEC be x - 5. Then x - 5 is a mean proportional between x - 9 and x, fromwhich we find that x equals 25. Therefore the diameters are 50 in. And41 in. Respectively. 

192. --THE PUZZLE WALL. 

[Illustration]

The answer given in all the old books is that shown in Fig. 1, where thecurved wall shuts out the cottages from access to the lake. But inseeking the direction for the "shortest possible" wall most readersto-day, remembering that the shortest distance between two points is astraight line, will adopt the method shown in Fig. 2. This is certainlyan improvement, yet the correct answer is really that indicated in Fig. 3. A measurement of the lines will show that there is a considerablesaving of length in this wall. 

193. --THE SHEEP-FOLD. 

This is the answer that is always given and accepted as correct: Twomore hurdles would be necessary, for the pen was twenty-four by one (asin Fig. A on next page), and by moving one of the sides and placing anextra hurdle at each end (as in Fig. B) the area would be doubled. Thediagrams are not to scale. Now there is no condition in the puzzle thatrequires the sheep-fold to be of any particular form. But even if weaccept the point that the pen was twenty-four by one, the answer utterlyfails, for two extra hurdles are certainly not at all necessary. Forexample, I arrange the fifty hurdles as in Fig. C, and as the area isincreased from twenty-four "square hurdles" to 156, there is nowaccommodation for 650 sheep. If it be held that the area must be exactlydouble that of the original pen, then I construct it (as in Fig. D) withtwenty-eight hurdles only, and have twenty-two in hand for otherpurposes on the farm. Even if it were insisted that all the originalhurdles must be used, then I should construct it as in Fig. E, where Ican get the area as exact as any farmer could possibly require, even ifwe have to allow for the fact that the sheep might not be able to grazeat the extreme ends. Thus we see that, from any point of view, theaccepted answer to this ancient little puzzle breaks down. And yetattention has never before been drawn to the absurdity. 

[Illustration

 A 24 +--------------------------------+ | 24 |1 +--------------------------------+

 B +--------------------------------+ | 48 |2 +--------------------------------+ 24 C +--------------------+ D | | +----------+ | | | | | |12 | 48 |6 | 156 | | | | | +----------+ | | 8 | | | | +--------------------+ 13

 12 . E 13 . ' ' . . ' ' . ' . . ' 12 ' . ' 13

]

194. --THE GARDEN WALLS. 

The puzzle was to divide the circular field into four equal parts bythree walls, each wall being of exactly the same length. There are twoessential difficulties in this problem. These are: (1) the thickness ofthe walls, and (2) the condition that these walls are three in number. As to the first point, since we are told that the walls are brick walls, we clearly cannot ignore their thickness, while we have to find asolution that will equally work, whether the walls be of a thickness ofone, two, three, or more bricks. 

[Illustration]

The second point requires a little more consideration. How are we todistinguish between a wall and walls? A straight wall without any bendin it, no matter how long, cannot ever become "walls, " if it is neitherbroken nor intersected in any way. Also our circular field is clearlyenclosed by one wall. But if it had happened to be a square or atriangular enclosure, would there be respectively four and three wallsor only one enclosing wall in each case? It is true that we speak of"the four walls" of a square building or garden, but this is only aconventional way of saying "the four sides. " If you were speaking of theactual brickwork, you would say, "I am going to enclose this squaregarden with a wall. " Angles clearly do not affect the question, for wemay have a zigzag wall just as well as a straight one, and the GreatWall of China is a good example of a wall with plenty of angles. Now, ifyou look at Diagrams 1, 2, and 3, you may be puzzled to declare whetherthere are in each case two or four new walls; but you cannot call themthree, as required in our puzzle. The intersection either affects thequestion or it does not affect it. 

If you tie two pieces of string firmly together, or splice them in anautical manner, they become "one piece of string. " If you simply letthem lie across one another or overlap, they remain "two pieces ofstring. " It is all a question of joining and welding. It may similarlybe held that if two walls be built into one another--I might almost say, if they be made homogeneous--they become one wall, in which caseDiagrams 1, 2, and 3 might each be said to show one wall or two, if itbe indicated that the four ends only touch, and are not really builtinto, the outer circular wall. 

The objection to Diagram 4 is that although it shows the three requiredwalls (assuming the ends are not built into the outer circular wall), yet it is only absolutely correct when we assume the walls to have nothickness. A brick has thickness, and therefore the fact throws thewhole method out and renders it only approximately correct. 

Diagram 5 shows, perhaps, the only correct and perfectly satisfactorysolution. It will be noticed that, in addition to the circular wall, there are three new walls, which touch (and so enclose) but are notbuilt into one another. This solution may be adapted to any desiredthickness of wall, and its correctness as to area and length of wallspace is so obvious that it is unnecessary to explain it. I will, however, just say that the semicircular piece of ground that each tenantgives to his neighbour is exactly equal to the semicircular piece thathis neighbour gives to him, while any section of wall space found in onegarden is precisely repeated in all the others. Of course there is aninfinite number of ways in which this solution may be correctly varied. 

195. --LADY BELINDA'S GARDEN. 

All that Lady Belinda need do was this: She should measure from A to B, fold her tape in four and mark off the point E, which is thus onequarter of the side. Then, in the same way, mark off the point F, one-fourth of the side AD Now, if she makes EG equal to AF, and GH equalto EF, then AH is the required width for the path in order that the bedshall be exactly half the area of the garden. An exact numericalmeasurement can only be obtained when the sum of the squares of the twosides is a square number. Thus, if the garden measured 12 poles by 5poles (where the squares of 12 and 5, 144 and 25, sum to 169, the squareof 13), then 12 added to 5, less 13, would equal four, and a quarter ofthis, 1 pole, would be the width of the path. 

196. --THE TETHERED GOAT. 

[Illustration]

This problem is quite simple if properly attacked. Let us suppose thetriangle ABC to represent our half-acre field, and the shaded portion tobe the quarter-acre over which the goat will graze when tethered to thecorner C. Now, as six equal equilateral triangles placed together willform a regular hexagon, as shown, it is evident that the shaded pastureis just one-sixth of the complete area of a circle. Therefore all werequire is the radius (CD) of a circle containing six quarter-acres or1½ acres, which is equal to 9, 408, 960 square inches. As we only wantour answer "to the nearest inch, " it is sufficiently exact for ourpurpose if we assume that as 1 is to 3. 1416, so is the diameter of acircle to its circumference. If, therefore, we divide the last number Igave by 3. 1416, and extract the square root, we find that 1, 731 inches, or 48 yards 3 inches, is the required length of the tether "to thenearest inch. "

197. --THE COMPASSES PUZZLE. 

Let AB in the following diagram be the given straight line. With thecentres A and B and radius AB describe the two circles. Mark off DE andEF equal to AD. With the centres A and F and radius DF describe arcsintersecting at G. With the centres A and B and distance BG describearcs GHK and N. Make HK equal to AB and HL equal to HB. Then withcentres K and L and radius AB describe arcs intersecting at I. Make BMequal to BI. Finally, with the centre M and radius MB cut the line in C, and the point C is the required middle of the line AB. For greaterexactitude you can mark off R from A (as you did M from B), and from Rdescribe another arc at C. This also solves the problem, to find a pointmidway between two given points without the straight line. 

[Illustration]

I will put the young geometer in the way of a rigid proof. First provethat twice the square of the line AB equals the square of the distanceBG, from which it follows that HABN are the four corners of a square. Toprove that I is the centre of this square, draw a line from H to Pthrough QIB and continue the arc HK to P. Then, conceiving the necessarylines to be drawn, the angle HKP, being in a semicircle, is a rightangle. Let fall the perpendicular KQ, and by similar triangles, and fromthe fact that HKI is an isosceles triangle by the construction, it canbe proved that HI is half of HB. We can similarly prove that C is thecentre of the square of which AIB are three corners. 

I am aware that this is not the simplest possible solution. 

198. --THE EIGHT STICKS. 

The first diagram is the answer that nearly every one will give to thispuzzle, and at first sight it seems quite satisfactory. But consider theconditions. We have to lay "every one of the sticks on the table. " Now, if a ladder be placed against a wall with only one end on the ground, itcan hardly be said that it is "laid on the ground. " And if we place thesticks in the above manner, it is only possible to make one end of twoof them touch the table: to say that every one lies on the table wouldnot be correct. To obtain a solution it is only necessary to have oursticks of proper dimensions. Say the long sticks are each 2 ft. Inlength and the short ones 1 ft. Then the sticks must be 3 in. Thick, when the three equal squares may be enclosed, as shown in the seconddiagram. If I had said "matches" instead of "sticks, " the puzzle wouldbe impossible, because an ordinary match is about twenty-one times aslong as it is broad, and the enclosed rectangles would not be squares. 

[Illustration]

199. --PAPA'S PUZZLE. 

I have found that a large number of people imagine that the following isa correct solution of the problem. Using the letters in the diagrambelow, they argue that if you make the distance BA one-third of BC, andtherefore the area of the rectangle ABE equal to that of the triangularremainder, the card must hang with the long side horizontal. Readerswill remember the jest of Charles II. , who induced the Royal Society tomeet and discuss the reason why the water in a vessel will not rise ifyou put a live fish in it; but in the middle of the proceedings one ofthe least distinguished among them quietly slipped out and made theexperiment, when he found that the water _did_ rise! If mycorrespondents had similarly made the experiment with a piece ofcardboard, they would have found at once their error. Area is one thing, but gravitation is quite another. The fact of that triangle sticking itsleg out to D has to be compensated for by additional area in therectangle. As a matter of fact, the ratio of BA to AC is as 1 is to thesquare root of 3, which latter cannot be given in an exact numericalmeasure, but is approximately 1. 732. Now let us look at the correctgeneral solution. There are many ways of arriving at the desired result, but the one I give is, I think, the simplest for beginners. 

[Illustration]

Fix your card on a piece of paper and draw the equilateral triangle BCF, BF and CF being equal to BC. Also mark off the point G so that DG shallequal DC. Draw the line CG and produce it until it cuts the line BF inH. If we now make HA parallel to BE, then A is the point from which ourcut must be made to the corner D, as indicated by the dotted line. 

A curious point in connection with this problem is the fact that theposition of the point A is independent of the side CD. The reason forthis is more obvious in the solution I have given than in any othermethod that I have seen, and (although the problem may be solved withall the working on the cardboard) that is partly why I have preferredit. It will be seen at once that however much you may reduce the widthof the card by bringing E nearer to B and D nearer to C, the line CG, being the diagonal of a square, will always lie in the same direction, and will cut BF in H. Finally, if you wish to get an approximate measurefor the distance BA, all you have to do is to multiply the length of thecard by the decimal . 366. Thus, if the card were 7 inches long, we get 7× . 366 = 2. 562, or a little more than 2½ inches, for the distance from Bto A. 

But the real joke of the puzzle is this: We have seen that the positionof the point A is independent of the width of the card, and dependsentirely on the length. Now, in the illustration it will be found thatboth cards have the same length; consequently all the little maid had todo was to lay the clipped card on top of the other one and mark off thepoint A at precisely the same distance from the top left-hand corner!So, after all, Pappus' puzzle, as he presented it to his little maid, was quite an infantile problem, when he was able to show her how toperform the feat without first introducing her to the elements ofstatics and geometry. 

200. --A KITE-FLYING PUZZLE. 

Solvers of this little puzzle, I have generally found, may be roughlydivided into two classes: those who get within a mile of the correctanswer by means of more or less complex calculations, involving "_pi_, "and those whose arithmetical kites fly hundreds and thousands of milesaway from the truth. The comparatively easy method that I shall showdoes not involve any consideration of the ratio that the diameter of acircle bears to its circumference. I call it the "hat-box method. "

[Illustration]

Supposing we place our ball of wire, A, in a cylindrical hat-box, B, that exactly fits it, so that it touches the side all round and exactlytouches the top and bottom, as shown in the illustration. Then, by aninvariable law that should be known by everybody, that box containsexactly half as much again as the ball. Therefore, as the ball is 24 in. In diameter, a hat-box of the same circumference but two-thirds of theheight (that is, 16 in. High) will have exactly the same contents as theball. 

Now let us consider that this reduced hat-box is a cylinder of metalmade up of an immense number of little wire cylinders close togetherlike the hairs in a painter's brush. By the conditions of the puzzle weare allowed to consider that there are no spaces between the wires. Howmany of these cylinders one one-hundredth of an inch thick are equal tothe large cylinder, which is 24 in. Thick? Circles are to one another asthe squares of their diameters. The square of 1/100 is 1/100000, and thesquare of 24 is 576; therefore the large cylinder contains 5, 760, 000 ofthe little wire cylinders. But we have seen that each of these wires is16 in. Long; hence 16 × 5, 760, 000 = 92, 160, 000 inches as the completelength of the wire. Reduce this to miles, and we get 1, 454 miles 2, 880ft. As the length of the wire attached to the professor's kite. 

Whether a kite would fly at such a height, or support such a weight, arequestions that do not enter into the problem. 

201. --HOW TO MAKE CISTERNS. 

Here is a general formula for solving this problem. Call the two sidesof the rectangle a and b. Then

 a + b - (a² + b² - ab)^½ --------------------------- 6

equals the side of the little square pieces to cut away. Themeasurements given were 8 ft. By 3 ft. , and the above rule gives 8 in. As the side of the square pieces that have to be cut away. Of course itwill not always come out exact, as in this case (on account of thatsquare root), but you can get as near as you like with decimals. 

202. --THE CONE PUZZLE. 

The simple rule is that the cone must be cut at one-third of itsaltitude. 

203. --CONCERNING WHEELS. 

If you mark a point A on the circumference of a wheel that runs on thesurface of a level road, like an ordinary cart-wheel, the curvedescribed by that point will be a common cycloid, as in Fig. 1. But ifyou mark a point B on the circumference of the flange of alocomotive-wheel, the curve will be a curtate cycloid, as in Fig. 2, terminating in nodes. Now, if we consider one of these nodes or loops, we shall see that "at any given moment" certain points at the bottom ofthe loop must be moving in the opposite direction to the train. As thereis an infinite number of such points on the flange's circumference, there must be an infinite number of these loops being described whilethe train is in motion. In fact, at any given moment certain points onthe flanges are always moving in a direction opposite to that in whichthe train is going. 

[Illustration: 1]

[Illustration: 2]

In the case of the two wheels, the wheel that runs round the stationaryone makes two revolutions round its own centre. As both wheels are ofthe same size, it is obvious that if at the start we mark a point on thecircumference of the upper wheel, at the very top, this point will be incontact with the lower wheel at its lowest part when half the journeyhas been made. Therefore this point is again at the top of the movingwheel, and one revolution has been made. Consequently there are two suchrevolutions in the complete journey. 

204. --A NEW MATCH PUZZLE. 

1. The easiest way is to arrange the eighteen matches as in Diagrams 1and 2, making the length of the perpendicular AB equal to a match and ahalf. Then, if the matches are an inch in length, Fig. 1 contains twosquare inches and Fig. 2 contains six square inches--4 × 1½. The secondcase (2) is a little more difficult to solve. The solution is given inFigs. 3 and 4. For the purpose of construction, place matchestemporarily on the dotted lines. Then it will be seen that as 3 containsfive equal equilateral triangles and 4 contains fifteen similartriangles, one figure is three times as large as the other, and exactlyeighteen matches are used. 

[Illustration: Figures 1, 2, 3, 4. ]

205. --THE SIX SHEEP-PENS. 

[Illustration] Place the twelve matches in the manner shown in theillustration, and you will have six pens of equal size. 

206. --THE KING AND THE CASTLES. 

There are various ways of building the ten castles so that they shallform five rows with four castles in every row, but the arrangement inthe next column is the only one that also provides that two castles (thegreatest number possible) shall not be approachable from the outside. Itwill be seen that you must cross the walls to reach these two. 

[Illustration: The King and the Castles]

207. --CHERRIES AND PLUMS. 

There are several ways in which this problem might be solved were it notfor the condition that as few cherries and plums as possible shall beplanted on the north and east sides of the orchard. The best possiblearrangement is that shown in the diagram, where the cherries, plums, and apples are indicated respectively by the letters C, P, and A. Thedotted lines connect the cherries, and the other lines the plums. Itwill be seen that the ten cherry trees and the ten plum trees are soplanted that each fruit forms five lines with four trees of its kind inline. This is the only arrangement that allows of so few as two cherriesor plums being planted on the north and east outside rows. 

[Illustration]

208. --A PLANTATION PUZZLE. 

The illustration shows the ten trees that must be left to form five rowswith four trees in every row. The dots represent the positions of thetrees that have been cut down. 

[Illustration]

209. --THE TWENTY-ONE TREES. 

I give two pleasing arrangements of the trees. In each case there aretwelve straight rows with five trees in every row. 

[Illustration: Figure 1, Figure 2. ]

210. --THE TEN COINS. 

The answer is that there are just 2, 400 different ways. Any three coinsmay be taken from one side to combine with one coin taken from the otherside. I give four examples on this and the next page. We may thus selectthree from the top in ten ways and one from the bottom in five ways, making fifty. But we may also select three from the bottom and one fromthe top in fifty ways. We may thus select the four coins in one hundredways, and the four removed may be arranged by permutation in twenty-fourways. Thus there are 24 × 100 = 2, 400 different solutions. 

[Illustration]

As all the points and lines puzzles that I have given so far, exceptingthe last, are variations of the case of ten points arranged to form fivelines of four, it will be well to consider this particular casegenerally. There are six fundamental solutions, and no more, as shown inthe six diagrams. These, for the sake of convenience, I named some yearsago the Star, the Dart, the Compasses, the Funnel, the Scissors, and theNail. (See next page. ) Readers will understand that any one of theseforms may be distorted in an infinite number of different ways withoutdestroying its real character. 

In "The King and the Castles" we have the Star, and its solution givesthe Compasses. In the "Cherries and Plums" solution we find that theCherries represent the Funnel and the Plums the Dart. The solution ofthe "Plantation Puzzle" is an example of the Dart distorted. Anysolution to the "Ten Coins" will represent the Scissors. Thus examplesof all have been given except the Nail. 

On a reduced chessboard, 7 by 7, we may place the ten pawns in justthree different ways, but they must all represent the Dart. The"Plantation" shows one way, the Plums show a second way, and the readermay like to find the third way for himself. On an ordinary chessboard, 8by 8, we can also get in a beautiful example of the Funnel--symmetricalin relation to the diagonal of the board. The smallest board that willtake a Star is one 9 by 7. The Nail requires a board 11 by 7, theScissors

[Illustration]

11 by 9, and the Compasses 17 by 12. At least these are the best resultsrecorded in my note-book. They may be beaten, but I do not think so. Ifyou divide a chessboard into two parts by a diagonal zigzag line, sothat the larger part contains 36 squares and the smaller part 28squares, you can place three separate schemes on the larger part and oneon the smaller part (all Darts) without their conflicting--that is, theyoccupy forty different squares. They can be placed in other ways withouta division of the board. The smallest square board that will contain sixdifferent schemes (not fundamentally different), without any line of onescheme crossing the line of another, is 14 by 14; and the smallest boardthat will contain one scheme entirely enclosed within the lines of asecond scheme, without any of the lines of the one, when drawn frompoint to point, crossing a line of the other, is 14 by 12. 

[Illustration: STAR DART COMPASSES FUNNEL SCISSORS NAIL]

211. --THE TWELVE MINCE-PIES. 

If you ignore the four black pies in our illustration, the remainingtwelve are in their original positions. Now remove the four detachedpies to the places occupied by the black ones, and you will have yourseven straight rows of four, as shown by the dotted lines. 

[Illustration: The Twelve Mince Pies. ]

212. --THE BURMESE PLANTATION. 

The arrangement on the next page is the most symmetrical answer that canprobably be found for twenty-one rows, which is, I believe, the greatestnumber of rows possible. There are several ways of doing it. 

213. --TURKS AND RUSSIANS. 

The main point is to discover the smallest possible number of Russiansthat there could have been. As the enemy opened fire from alldirections, it is clearly necessary to find what is the smallest numberof heads that could form sixteen lines with three heads in every line. Note that I say sixteen, and not thirty-two, because every line taken bya bullet may be also taken by another bullet fired in exactly theopposite direction. Now, as few as eleven points, or heads, may bearranged to form the required sixteen lines of three, but the discoveryof this arrangement is a hard nut. The diagram at the foot of this pagewill show exactly how the thing is to be done. 

[Illustration]

If, therefore, eleven Russians were in the positions shown by the stars, and the thirty-two Turks in the positions indicated by the black dots, it will be seen, by the lines shown, that each Turk may fire exactlyover the heads of three Russians. But as each bullet kills a man, it isessential that every Turk shall shoot one of his comrades and be shot byhim in turn; otherwise we should have to provide extra Russians to beshot, which would be destructive of the correct solution of our problem. As the firing was simultaneous, this point presents no difficulties. Theanswer we thus see is that there were at least eleven Russians amongstwhom there was no casualty, and that all the thirty-two Turks were shotby one another. It was not stated whether the Russians fired any shots, but it will be evident that even if they did their firing could not havebeen effective: for if one of their bullets killed a Turk, then we haveimmediately to provide another man for one of the Turkish bullets tokill; and as the Turks were known to be thirty-two in number, this wouldnecessitate our introducing another Russian soldier and, of course, destroying the solution. I repeat that the difficulty of the puzzleconsists in finding how to arrange eleven points so that they shall formsixteen lines of three. I am told that the possibility of doing this wasfirst discovered by the Rev. Mr. Wilkinson some twenty years ago. 

214. --THE SIX FROGS. 

Move the frogs in the following order: 2, 4, 6, 5, 3, 1 (repeat thesemoves in the same order twice more), 2, 4, 6. This is a solution intwenty-one moves--the fewest possible. 

If n, the number of frogs, be even, we require (n² + n)/2 moves, ofwhich (n² - n)/2 will be leaps and n simple moves. If n be odd, weshall need ((n² + 3n)/2) - 4 moves, of which (n² - n)/2 will be leapsand 2n - 4 simple moves. 

In the even cases write, for the moves, all the even numbers inascending order and the odd numbers in descending order. This seriesmust be repeated ½n times and followed by the even numbers inascending order once only. Thus the solution for 14 frogs will be (2, 4, 6, 8, 10, 12, 14, 13, 11, 9, 7, 5, 3, 1) repeated 7 times and followedby 2, 4, 6, 8, 10, 12, 14 = 105 moves. 

In the odd cases, write the even numbers in ascending order and the oddnumbers in descending order, repeat this series ½(n - 1) times, followwith the even numbers in ascending order (omitting n - 1), the oddnumbers in descending order (omitting 1), and conclude with all thenumbers (odd and even) in their natural order (omitting 1 and n). Thusfor 11 frogs: (2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 1) repeated 5 times, 2, 4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 7, 8, 9, 10 = 73 moves. 

This complete general solution is published here for the first time. 

215. --THE GRASSHOPPER PUZZLE. 

Move the counters in the following order. The moves in brackets are tobe made four times in succession. 12, 1, 3, 2, 12, 11, 1, 3, 2 (5, 7, 9, 10, 8, 6, 4), 3, 2, 12, 11, 2, 1, 2. The grasshoppers will then bereversed in forty-four moves. 

The general solution of this problem is very difficult. Of course it canalways be solved by the method given in the solution of the last puzzle, if we have no desire to use the fewest possible moves. But to employ afull economy of moves we have two main points to consider. There arealways what I call a lower movement (L) and an upper movement (U). Lconsists in exchanging certain of the highest numbers, such as 12, 11, 10 in our "Grasshopper Puzzle, " with certain of the lower numbers, 1, 2, 3; the former moving in a clockwise direction, the latter in anon-clockwise direction. U consists in reversing the intermediatecounters. In the above solution for 12, it will be seen that 12, 11, and1, 2, 3 are engaged in the L movement, and 4, 5, 6, 7, 8, 9, 10 in theU movement. The L movement needs 16 moves and U 28, making together 44. We might also involve 10 in the L movement, which would result in L 23, U 21, making also together 44 moves. These I call the first and secondmethods. But any other scheme will entail an increase of moves. Youalways get these two methods (of equal economy) for odd or evencounters, but the point is to determine just how many to involve in Land how many in U. Here is the solution in table form. But first note, in giving values to n, that 2, 3, and 4 counters are special cases, requiring respectively 3, 3, and 6 moves, and that 5 and 6 counters donot give a minimum solution by the second method--only by the first. 

 FIRST METHOD. +----------+---------------------------+-----------------------+-----------+ | Total No. | L MOVEMENT. | U MOVEMENT. | | | of +-------------+-------------+----------+------------+ Total No. | | Counters. | No. Of | No. Of | No. Of | No. Of | of Moves. | | | Counters. | Moves. |Counters. | Moves. | | +----------+-------------+-------------+----------+------------+-----------+ | 4n | n-1 and n |2(n-1)²+5n-7 | 2n+1 |2n²+3n+1 |4(n²+n-1) | | 4n-2 | n-1 " n |2(n-1)²+5n-7 | 2n-1 |2(n-1)²+3n-2|4n²-5 | | 4n+1 | n " n+1 |2n²+5n-2 | 2n |2n²+3n-4 |2(2n²+4n-3)| | 4n-1 | n-1 " n |2(n-1)²+5n-7 | 2n |2n²+3n-4 |4n²+4n-9 | +----------+-------------+-------------+----------+------------+-----------+

 SECOND METHOD. +---------+--------------------------+-------------------------+-----------+ |Total No. | L MOVEMENT. | U MOVEMENT. | | | of +-------------+------------+----------+--------------+ Total No. | |Counters. | No. Of | No. Of | No. Of | No. Of | of Moves. | | | Counters. | Moves. | Counters. | Moves. | | +---------+-------------+------------+----------+--------------+-----------+ | 4n | n and n |2n²+3n-4 | 2n | 2(n-1)²+5n-2 |4(n²+n-1) | | 4n-2 | n-1 " n-1 |2(n-1)²+3n-7| 2n | 2(n-1)²+5n-2 |4n²-5 | | 4n+1 | n " n |2n²+3n-4 | 2n+1 | 2n²+5n-2 |2(2n²+4n-3)| | 4n-1 | n " n |2n²+3n-4 | 2n-1 | 2(n-1)²+5n-7 |4n²+4n-9 | +---------+-------------+------------+----------+--------------+-----------+

More generally we may say that with m counters, where m is even andgreater than 4, we require (m² + 4m - 16)/4 moves; and where m is oddand greater than 3, (m² + 6m - 31)/4 moves. I have thus shown thereader how to find the minimum number of moves for any case, and thecharacter and direction of the moves. I will leave him to discover forhimself how the actual order of moves is to be determined. This is ahard nut, and requires careful adjustment of the L and the Umovements, so that they may be mutually accommodating. 

216. --THE EDUCATED FROGS. 

The following leaps solve the puzzle in ten moves: 2 to 1, 5 to 2, 3 to5, 6 to 3, 7 to 6, 4 to 7, 1 to 4, 3 to 1, 6 to 3, 7 to 6. 

217. --THE TWICKENHAM PUZZLE. 

Play the counters in the following order: K C E K W T C E H M K W T A NC E H M I K C E H M T, and there you are, at Twickenham. The positionitself will always determine whether you are to make a leap or a simplemove. 

218. --THE VICTORIA CROSS PUZZLE. 

In solving this puzzle there were two things to be achieved: first, soto manipulate the counters that the word VICTORIA should read round thecross in the same direction, only with the V on one of the dark arms;and secondly, to perform the feat in the fewest possible moves. Now, asa matter of fact, it would be impossible to perform the first part inany way whatever if all the letters of the word were different; but asthere are two I's, it can be done by making these letters changeplaces--that is, the first I changes from the 2nd place to the 7th, andthe second I from the 7th place to the 2nd. But the point I referred to, when introducing the puzzle, as a little remarkable is this: that asolution in twenty-two moves is obtainable by moving the letters in theorder of the following words: "A VICTOR! A VICTOR! A VICTOR I!"

There are, however, just six solutions in eighteen moves, and thefollowing is one of them: I (1), V, A, I (2), R, O, T, I (1), I (2), A, V, I (2), I (1), C, I (2), V, A, I (1). The first and second I in theword are distinguished by the numbers 1 and 2. 

It will be noticed that in the first solution given above one of the I'snever moves, though the movements of the other letters cause it tochange its relative position. There is another peculiarity I may pointout--that there is a solution in twenty-eight moves requiring no letterto move to the central division except the I's. I may also mention that, in each of the solutions in eighteen moves, the letters C, T, O, R moveonce only, while the second I always moves four times, the V alwaysbeing transferred to the right arm of the cross. 

219. --THE LETTER BLOCK PUZZLE. 

This puzzle can be solved in 23 moves--the fewest possible. Move theblocks in the following order: A, B, F, E, C, A, B, F, E, C, A, B, D, H, G, A, B, D, H, G, D, E, F. 

220. --A LODGING-HOUSE DIFFICULTY. 

The shortest possible way is to move the articles in the followingorder: Piano, bookcase, wardrobe, piano, cabinet, chest of drawers, piano, wardrobe, bookcase, cabinet, wardrobe, piano, chest of drawers, wardrobe, cabinet, bookcase, piano. Thus seventeen removals arenecessary. The landlady could then move chest of drawers, wardrobe, andcabinet. Mr. Dobson did not mind the wardrobe and chest of drawerschanging rooms so long as he secured the piano. 

221. --THE EIGHT ENGINES. 

The solution to the Eight Engines Puzzle is as follows: The engine thathas had its fire drawn and therefore cannot move is No. 5. Move theother engines in the following order: 7, 6, 3, 7, 6, 1, 2, 4, 1, 3, 8, 1, 3, 2, 4, 3, 2, seventeen moves in all, leaving the eight engines inthe required order. 

There are two other slightly different solutions. 

222. --A RAILWAY PUZZLE. 

This little puzzle may be solved in as few as nine moves. Play theengines as follows: From 9 to 10, from 6 to 9, from 5 to 6, from 2 to 5, from 1 to 2, from 7 to 1, from 8 to 7, from 9 to 8, and from 10 to 9. You will then have engines A, B, and C on each of the three circles andon each of the three straight lines. This is the shortest solution thatis possible. 

223. --A RAILWAY MUDDLE. 

[Illustration: 1]

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

[Illustration: 5]

[Illustration: 6]

Only six reversals are necessary. The white train (from A to D) isdivided into three sections, engine and 7 wagons, 8 wagons, and 1 wagon. The black train (D to A) never uncouples anything throughout. Fig. 1 isoriginal position with 8 and 1 uncoupled. The black train proceeds toposition in Fig. 2 (no reversal). The engine and 7 proceed towards D, and black train backs, leaves 8 on loop, and takes up position in Fig. 3(first reversal). Black train goes to position in Fig. 4 to fetch singlewagon (second reversal). Black train pushes 8 off loop and leaves singlewagon there, proceeding on its journey, as in Fig. 5 (third and fourthreversals). White train now backs on to loop to pick up single car andgoes right away to D (fifth and sixth reversals). 

224. --THE MOTOR-GARAGE PUZZLE. 

The exchange of cars can be made in forty-three moves, as follows: 6-G, 2-B, 1-E, 3-H, 4-I, 3-L, 6-K, 4-G, 1-I, 2-J, 5-H, 4-A, 7-F, 8-E, 4-D, 8-C, 7-A, 8-G, 5-C, 2-B, 1-E, 8-I, 1-G, 2-J, 7-H, 1-A, 7-G, 2-B, 6-E, 3-H, 8-L, 3-I, 7-K, 3-G, 6-I, 2-J, 5-H, 3-C, 5-G, 2-B, 6-E, 5-I, 6-J. Ofcourse, "6-G" means that the car numbered "6" moves to the point "G. "There are other ways in forty-three moves. 

225. --THE TEN PRISONERS. 

[Illustration]

It will be seen in the illustration how the prisoners may be arranged soas to produce as many as sixteen even rows. There are 4 such verticalrows, 4 horizontal rows, 5 diagonal rows in one direction, and 3diagonal rows in the other direction. The arrows here show the movementsof the four prisoners, and it will be seen that the infirm man in thebottom corner has not been moved. 

226. --ROUND THE COAST. 

In order to place words round the circle under the conditions, it isnecessary to select words in which letters are repeated in certainrelative positions. Thus, the word that solves our puzzle is "Swansea, "in which the first and fifth letters are the same, and the third andseventh the same. We make out jumps as follows, taking the letters ofthe word in their proper order: 2-5, 7-2, 4-7, 1-4, 6-1, 3-6, 8-3. Or wecould place a word like "Tarapur" (in which the second and fourthletters, and the third and seventh, are alike) with these moves: 6-1, 7-4, 2-7, 5--2, 8-5, 3-6, 8-3. But "Swansea" is the only word, apparently, that will fulfil the conditions of the puzzle. 

This puzzle should be compared with Sharp's Puzzle, referred to in mysolution to No. 341, "The Four Frogs. " The condition "touch and jumpover two" is identical with "touch and move along a line. "

227. --CENTRAL SOLITAIRE. 

Here is a solution in nineteen moves; the moves enclosed in bracketscount as one move only: 19-17, 16-18, (29-17, 17-19), 30-18, 27-25, (22-24, 24-26), 31-23, (4-16, 16-28), 7-9, 10-8, 12-10, 3-11, 18-6, (1-3, 3-11), (13-27, 27-25), (21-7, 7-9), (33-31, 31-23), (10-8, 8-22, 22-24, 24-26, 26-12, 12-10), 5-17. All the counters are now removedexcept one, which is left in the central hole. The solution needsjudgment, as one is tempted to make several jumps in one move, where itwould be the reverse of good play. For example, after playing the first3-11 above, one is inclined to increase the length of the move bycontinuing with 11-25, 25-27, or with 11-9, 9-7. 

I do not think the number of moves can be reduced. 

228. --THE TEN APPLES. 

Number the plates (1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), (13, 14, 15, 16) in successive rows from the top to the bottom. Then transfer theapple from 8 to 10 and play as follows, always removing the apple jumpedover: 9-11, 1-9, 13-5, 16-8, 4-12, 12-10, 3-1, 1-9, 9-11. 

229. --THE NINE ALMONDS. 

This puzzle may be solved in as few as four moves, in the followingmanner: Move 5 over 8, 9, 3, 1. Move 7 over 4. Move 6 over 2 and 7. Move5 over 6, and all the counters are removed except 5, which is left inthe central square that it originally occupied. 

230. --THE TWELVE PENNIES. 

Here is one of several solutions. Move 12 to 3, 7 to 4, 10 to 6, 8 to 1, 9 to 5, 11 to 2. 

231. --PLATES AND COINS. 

Number the plates from 1 to 12 in the order that the boy is seen to begoing in the illustration. Starting from 1, proceed as follows, where "1to 4" means that you take the coin from plate No. 1 and transfer it toplate No. 4: 1 to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, 11 to 2, andcomplete the last revolution to 1, making three revolutions in all. Oryou can proceed this way: 4 to 7, 8 to 11, 12 to 3, 2 to 5, 6 to 9, 10to 1. It is easy to solve in four revolutions, but the solutions inthree are more difficult to discover. 

This is "The Riddle of the Fishpond" (No. 41, _Canterbury Puzzles_) in adifferent dress. 

232. --CATCHING THE MICE. 

In order that the cat should eat every thirteenth mouse, and the whitemouse last of all, it is necessary that the count should begin at theseventh mouse (calling the white one the first)--that is, at the onenearest the tip of the cat's tail. In this case it is not at allnecessary to try starting at all the mice in turn until you come to theright one, for you can just start anywhere and note how far distant thelast one eaten is from the starting point. You will find it to be theeighth, and therefore must start at the eighth, counting backwards fromthe white mouse. This is the one I have indicated. 

In the case of the second puzzle, where you have to find the smallestnumber with which the cat may start at the white mouse and eat this onelast of all, unless you have mastered the general solution of theproblem, which is very difficult, there is no better course open to youthan to try every number in succession until you come to one that workscorrectly. The smallest number is twenty-one. If you have to proceed bytrial, you will shorten your labour a great deal by only counting outthe remainders when the number is divided successively by 13, 12, 11, 10, etc. Thus, in the case of 21, we have the remainders 8, 9, 10, 1, 3, 5, 7, 3, 1, 1, 3, 1, 1. Note that I do not give the remainders of 7, 3, and 1 as nought, but as 7, 3, and 1. Now, count round each of thesenumbers in turn, and you will find that the white mouse is killed lastof all. Of course, if we wanted simply any number, not the smallest, thesolution is very easy, for we merely take the least common multiple of13, 12, 11, 10, etc. Down to 2. This is 360360, and you will find thatthe first count kills the thirteenth mouse, the next the twelfth, thenext the eleventh, and so on down to the first. But the mostarithmetically inclined cat could not be expected to take such a bignumber when a small one like twenty-one would equally serve its purpose. 

In the third case, the smallest number is 100. The number 1, 000 wouldalso do, and there are just seventy-two other numbers between these thatthe cat might employ with equal success. 

233. --THE ECCENTRIC CHEESEMONGER. 

To leave the three piles at the extreme ends of the rows, the cheesesmay be moved as follows--the numbers refer to the cheeses and not totheir positions in the row: 7-2, 8-7, 9-8, 10-15, 6-10, 5-6, 14-16, 13-14, 12-13, 3-1, 4-3, 11-4. This is probably the easiest solution ofall to find. To get three of the piles on cheeses 13, 14, and 15, playthus: 9-4, 10-9, 11-10, 6-14, 5-6, 12-15, 8-12, 7-8, 16-5, 3-13, 2-3, 1-2. To leave the piles on cheeses 3, 5, 12, and 14, play thus: 8-3, 9-14, 16-12, 1-5, 10-9, 7-10, 11-8, 2-1, 4-16, 13-2, 6-11, 15-4. 

234. --THE EXCHANGE PUZZLE. 

Make the following exchanges of pairs: H-K, H-E, H-C, H-A, I-L, I-F, I-D, K-L, G-J, J-A, F-K, L-E, D-K, E-F, E-D, E-B, B-K. It will be foundthat, although the white counters can be moved to their proper places in11 moves, if we omit all consideration of exchanges, yet the blackcannot be so moved in fewer than 17 moves. So we have to introduce wastemoves with the white counters to equal the minimum required by theblack. Thus fewer than 17 moves must be impossible. Some of the movesare, of course, interchangeable. 

235. --TORPEDO PRACTICE. 

[Illustration:

 10 6 7 \ |/ 4 u u 2 \ u / 3-u u u u u u u u u u -----9--- / u 8 u u / \ 1 5

]

If the enemy's fleet be anchored in the formation shown in theillustration, it will be seen that as many as ten out of the sixteenships may be blown up by discharging the torpedoes in the orderindicated by the numbers and in the directions indicated by the arrows. As each torpedo in succession passes under three ships and sinks thefourth, strike out each vessel with the pencil as it is sunk. 

236. --THE HAT PUZZLE. 

[Illustration:

 1 2 3 4 5 6 7 8 9 10 11 12 +--+--+--+--+--+--+--+--+--+--+--+--+ | *| o| *| O| *| O| *| O| *| O| | | +--+--+--+--+--+--+--+--+--+--+--+--+ | *| | | O| *| O| *| O| *| O| O| *| +--+--+--+--+--+--+--+--+--+--+--+--+ | *| *| O| O| *| O| | | *| O| O| *| +--+--+--+--+--+--+--+--+--+--+--+--+ | *| *| O| | | O| O| *| *| O| O| *| +--+--+--+--+--+--+--+--+--+--+--+--+ | *| *| O| O| O| O| O| *| *| | | *| +--+--+--+--+--+--+--+--+--+--+--+--+ | | | O| O| O| O| O| *| *| *| *| *| +--+--+--+--+--+--+--+--+--+--+--+--+

]

I suggested that the reader should try this puzzle with counters, so Igive my solution in that form. The silk hats are represented by blackcounters and the felt hats by white counters. The first row shows thehats in their original positions, and then each successive row shows howthey appear after one of the five manipulations. It will thus be seenthat we first move hats 2 and 3, then 7 and 8, then 4 and 5, then 10 and11, and, finally, 1 and 2, leaving the four silk hats together, the fourfelt hats together, and the two vacant pegs at one end of the row. Thefirst three pairs moved are dissimilar hats, the last two pairs beingsimilar. There are other ways of solving the puzzle. 

237. --BOYS AND GIRLS. 

There are a good many different solutions to this puzzle. Any contiguouspair, except 7-8, may be moved first, and after the first move there arevariations. The following solution shows the position from the startright through each successive move to the end:--

 . . 1 2 3 4 5 6 7 8 4 3 1 2 . . 5 6 7 8 4 3 1 2 7 6 5 . . 8 4 3 1 2 7 . . 5 6 8 4 . . 2 7 1 3 5 6 8 4 8 6 2 7 1 3 5 . . 

238. --ARRANGING THE JAM POTS. 

Two of the pots, 13 and 19, were in their proper places. As everyinterchange may result in a pot being put in its place, it is clear thattwenty-two interchanges will get them all in order. But this number ofmoves is not the fewest possible, the correct answer being seventeen. Exchange the following pairs: (3-1, 2-3), (15-4, 16-15), (17-7, 20-17), (24-10, 11-24, 12-11), (8-5, 6-8, 21-6, 23-21, 22-23, 14-22, 9-14, 18-9). When you have made the interchanges within any pair of brackets, all numbers within those brackets are in their places. There are fivepairs of brackets, and 5 from 22 gives the number of changesrequired--17. 

239. --A JUVENILE PUZZLE. 

[Illustration:

 +-----------------+ | C E | | | | | | D F | +---------------B | G | A | | | H | +-----------------+

]

As the conditions are generally understood, this puzzle is incapable ofsolution. This can be demonstrated quite easily. So we have to look forsome catch or quibble in the statement of what we are asked to do. Nowif you fold the paper and then push the point of your pencil downbetween the fold, you can with one stroke make the two lines CD and EFin our diagram. Then start at A, and describe the line ending at B. Finally put in the last line GH, and the thing is done strictly withinthe conditions, since folding the paper is not actually forbidden. Ofcourse the lines are here left unjoined for the purpose of clearness. 

In the rubbing out form of the puzzle, first rub out A to B with asingle finger in one stroke. Then rub out the line GH with one finger. Finally, rub out the remaining two vertical lines with two fingers atonce! That is the old trick. 

240. --THE UNION JACK. 

[Illustration:

 +-------+ +----- A B | | / \ | | / |\ \ | | / /| | \ \ | | / / | | \ \| |/ / | | \ | / / | | \ |\ /| / | +-----\-|-\/-|-/-----+ \| /\ |/ |/ \/ |\ /\ /| \/ |\ +-----/-|-/\-|-\-----+ | / / \| \ | | / | \ \ | | / /| |\ \ | | / / | | \ \ | |/ / | | \ \| / | | \ / | | \ -----+ +-----

]

There are just sixteen points (all on the outside) where three roads maybe said to join. These are called by mathematicians "odd nodes. " Thereis a rule that tells us that in the case of a drawing like the presentone, where there are sixteen odd nodes, it requires eight separatestrokes or routes (that is, half as many as there are odd nodes) tocomplete it. As we have to produce as much as possible with only one ofthese eight strokes, it is clearly necessary to contrive that the sevenstrokes from odd node to odd node shall be as short as possible. Startat A and end at B, or go the reverse way. 

241. --THE DISSECTED CIRCLE. 

[Illustration:

 /---------------\ / \ / /------B \ / / | /^\ \ / / |\ | / \ \ / / | \ | / \ \ / / | \ | / A \ \ / / | \ | / | \ \ | / | \|/ | \ | | | -----+-----*-----+----- | | | | \ | /|\ | / | | | | \ | / | \ | / | | | | \ | / | \ | / | | | | \ | / | \ | / | | | | \|/ | \|/ | | D-+------*-----+-----*----E | | | /|\ | /|\ | | | / | \ | / | \ | | | / | \ | / | \ | | | / | \ | / | \ | | | / | \|/ | \ | | | -----+-----*-----+----- | | \ | /|\ | / | \ | / | \ | / / \ | / | \ | / / \ | / | \ | / / \ |/ | \| / / \ | / / \------+------/ / | / C-------/

]

It can be done in twelve continuous strokes, thus: Start at A in theillustration, and eight strokes, forming the star, will bring you backto A; then one stroke round the circle to B, one stroke to C, one roundthe circle to D, and one final stroke to E--twelve in all. Of course, inpractice the second circular stroke will be over the first one; it isseparated in the diagram, and the points of the star not joined to thecircle, to make the solution clear to the eye. 

242. --THE TUBE INSPECTOR'S PUZZLE. 

The inspector need only travel nineteen miles if he starts at B andtakes the following route: BADGDEFIFCBEHKLIHGJK. Thus the only portionsof line travelled over twice are the two sections D to G and F to I. Ofcourse, the route may be varied, but it cannot be shortened. 

243. --VISITING THE TOWNS. 

Note that there are six towns, from which only two roads issue. Thus 1must lie between 9 and 12 in the circular route. Mark these two roads assettled. Similarly mark 9, 5, 14, and 4, 8, 14, and 10, 6, 15, and 10, 2, 13, and 3, 7, 13. All these roads must be taken. Then you will findthat he must go from 4 to 15, as 13 is closed, and that he is compelledto take 3, 11, 16, and also 16, 12. Thus, there is only one route, asfollows: 1, 9, 5, 14, 8, 4, 15, 6, 10, 2, 13, 7, 3, 11, 16, 12, 1, orits reverse--reading the line the other way. Seven roads are not used. 

244. --THE FIFTEEN TURNINGS. 

[Illustration]

It will be seen from the illustration (where the roads not used areomitted) that the traveller can go as far as seventy miles in fifteenturnings. The turnings are all numbered in the order in which they aretaken. It will be seen that he never visits nineteen of the towns. Hemight visit them all in fifteen turnings, never entering any town twice, and end at the black town from which he starts (see "The Rook's Tour, "No. 320), but such a tour would only take him sixty-four miles. 

245. --THE FLY ON THE OCTAHEDRON. 

[Illustration]

Though we cannot really see all the sides of the octahedron at once, wecan make a projection of it that suits our purpose just as well. In thediagram the six points represent the six angles of the octahedron, andfour lines proceed from every point under exactly the same conditions asthe twelve edges of the solid. Therefore if we start at the point A andgo over all the lines once, we must always end our route at A. And thenumber of different routes is just 1, 488, counting the reverse way ofany route as different. It would take too much space to show how I makethe count. It can be done in about five minutes, but an explanation ofthe method is difficult. The reader is therefore asked to accept myanswer as correct. 

246. --THE ICOSAHEDRON PUZZLE. 

[Illustration]

There are thirty edges, of which eighteen were visible in the originalillustration, represented in the following diagram by the hexagonNAESGD. By this projection of the solid we get an imaginary view of theremaining twelve edges, and are able to see at once their direction andthe twelve points at which all the edges meet. The difference in thelength of the lines is of no importance; all we want is to present theirdirection in a graphic manner. But in case the novice should be puzzledat only finding nineteen triangles instead of the required twenty, Iwill point out that the apparently missing triangle is the outline HIK. 

In this case there are twelve odd nodes; therefore six distinct anddisconnected routes will be needful if we are not to go over any linestwice. Let us therefore find the greatest distance that we may so travelin one route. 

It will be noticed that I have struck out with little cross strokes fivelines or edges in the diagram. These five lines may be struck outanywhere so long as they do not join one another, and so long as one ofthem does not connect with N, the North Pole, from which we are tostart. It will be seen that the result of striking out these five linesis that all the nodes are now even except N and S. Consequently if webegin at N and stop at S we may go over all the lines, except the fivecrossed out, without traversing any line twice. There are many ways ofdoing this. Here is one route: N to H, I, K, S, I, E, S, G, K, D, H, A, N, B, A, E, F, B, C, G, D, N, C, F, S. By thus making five of the routesas short as is possible--simply from one node to the next--we are ableto get the greatest possible length for our sixth line. A greaterdistance in one route, without going over the same ground twice, it isnot possible to get. 

It is now readily seen that those five erased lines must be gone overtwice, and they may be "picked up, " so to speak, at any points of ourroute. Thus, whenever the traveller happens to be at I he can run up toA and back before proceeding on his route, or he may wait until he is atA and then run down to I and back to A. And so with the other lines thathave to be traced twice. It is, therefore, clear that he can go over 25of the lines once only (25 × 10, 000 miles = 250, 000 miles) and 5 of thelines twice (5 × 20, 000 miles = 100, 000 miles), the total, 350, 000 miles, being the length of his travels and the shortest distance that ispossible in visiting the whole body. 

It will be noticed that I have made him end his travels at S, the SouthPole, but this is not imperative. I might have made him finish at any ofthe other nodes, except the one from which he started. Suppose it hadbeen required to bring him home again to N at the end of his travels. Then instead of suppressing the line AI we might leave that open andclose IS. This would enable him to complete his 350, 000 miles tour at A, and another 10, 000 miles would take him to his own fireside. There are agreat many different routes, but as the lengths of the edges are allalike, one course is as good as another. To make the complete 350, 000miles tour from N to S absolutely clear to everybody, I will give itentire: N to H, I, A, I, K, H, K, S, I, E, S, G, F, G, K, D, C, D, H, A, N, B, E, B, A, E, F, B, C, G, D, N, C, F, S--that is, thirty-five linesof 10, 000 miles each. 

247. --INSPECTING A MINE. 

Starting from A, the inspector need only travel 36 furlongs if he takesthe following route: A to B, G, H, C, D, I, H, M, N, I, J, O, N, S, R, M, L, G, F, K, L, Q, R, S, T, O, J, E, D, C, B, A, F, K, P, Q. He thuspasses between A and B twice, between C and D twice, between F and Ktwice, between J and O twice, and between R and S twice--fiverepetitions. Therefore 31 passages plus 5 repeated equal 36 furlongs. The little pitfall in this puzzle lies in the fact that we start from aneven node. Otherwise we need only travel 35 furlongs. 

248. --THE CYCLIST'S TOUR. 

When Mr. Maggs replied, "No way, I'm sure, " he was not saying that thething was impossible, but was really giving the actual route by whichthe problem can be solved. Starting from the star, if you visit thetowns in the order, NO WAY, I'M SURE, you will visit every town once, and only once, and end at E. So both men were correct. This was thelittle joke of the puzzle, which is not by any means difficult. 

249. --THE SAILOR'S PUZZLE. 

[Illustration]

There are only four different routes (or eight, if we count the reverseways) by which the sailor can start at the island marked A, visit allthe islands once, and once only, and return again to A. Here they are:--

A I P T L O E H R Q D C F U G N S K M B A A I P T S N G L O E U F C D KM B Q R H A A B M K S N G L T P I O E U F C D Q R H A A I P T L O E U GN S K M B Q D C F R H A

Now, if the sailor takes the first route he will make C his 12th island(counting A as 1); by the second route he will make C his 13th island;by the third route, his 16th island; and by the fourth route, his 17thisland. If he goes the reverse way, C will be respectively his 10th, 9th, 6th, and 5th island. As these are the only possible routes, it isevident that if the sailor puts off his visit to C as long as possible, he must take the last route reading from left to right. This route Ishow by the dark lines in the diagram, and it is the correct answer tothe puzzle. 

The map may be greatly simplified by the "buttons and string" method, explained in the solution to No. 341, "The Four Frogs. "

250. --THE GRAND TOUR. 

The first thing to do in trying to solve a puzzle like this is toattempt to simplify it. If you look at Fig. 1, you will see that it is asimplified version of the map. Imagine the circular towns to be buttonsand the railways to be connecting strings. (See solution to No. 341. )Then, it will be seen, we have simply "straightened out" the previousdiagram without affecting the conditions. Now we can further simplify byconverting Fig. 1 into Fig. 2, which is a portion of a chessboard. Herethe directions of the railways will resemble the moves of a rook inchess--that is, we may move in any direction parallel to the sides ofthe diagram, but not diagonally. Therefore the first town (or square)visited must be a black one; the second must be a white; the third mustbe a black; and so on. Every odd square visited will thus be black andevery even one white. Now, we have 23 squares to visit (an odd number), so the last square visited must be black. But Z happens to be white, sothe puzzle would seem to be impossible of solution. 

[Illustration: Fig. 1. ]

[Illustration: Fig. 2. ]

As we were told that the man "succeeded" in carrying put his plan, wemust try to find some loophole in the conditions. He was to "enter everytown once and only once, " and we find no prohibition against hisentering once the town A after leaving it, especially as he has neverleft it since he was born, and would thus be "entering" it for the firsttime in his life. But he must return at once from the first town hevisits, and then he will have only 22 towns to visit, and as 22 is aneven number, there is no reason why he should not end on the whitesquare Z. A possible route for him is indicated by the dotted line fromA to Z. This route is repeated by the dark lines in Fig. 1, and thereader will now have no difficulty in applying; it to the original map. We have thus proved that the puzzle can only be solved by a return to Aimmediately after leaving it. 

251. --WATER, GAS, AND ELECTRICITY. 

[Illustration]

According to the conditions, in the strict sense in which one at firstunderstands them, there is no possible solution to this puzzle. In sucha dilemma one always has to look for some verbal quibble or trick. Ifthe owner of house A will allow the water company to run their pipe forhouse C through his property (and we are not bound to assume that hewould object), then the difficulty is got over, as shown in ourillustration. It will be seen that the dotted line from W to C passesthrough house A, but no pipe ever crosses another pipe. 

252. --A PUZZLE FOR MOTORISTS. 

[Illustration]

The routes taken by the eight drivers are shown in the illustration, where the dotted line roads are omitted to make the paths clearer to theeye. 

253. --A BANK HOLIDAY PUZZLE. 

The simplest way is to write in the number of routes to all the towns inthis manner. Put a 1 on all the towns in the top row and in the firstcolumn. Then the number of routes to any town will be the sum of theroutes to the town immediately above and to the town immediately to theleft. Thus the routes in the second row will be 1, 2, 3, 4, 5, 6, etc. , in the third row, 1, 3, 6, 10, 15, 21, etc. ; and so on with the otherrows. It will then be seen that the only town to which there are exactly1, 365 different routes is the twelfth town in the fifth row--the oneimmediately over the letter E. This town was therefore the cyclist'sdestination. 

The general formula for the number of routes from one corner to thecorner diagonally opposite on any such rectangular reticulatedarrangement, under the conditions as to direction, is (m+n)!/m!n!, where m is the number of towns on one side, less one, and n the numberon the other side, less one. Our solution involves the case wherethere are 12 towns by 5. Therefore m = 11 and n = 4. Then the formulagives us the answer 1, 365 as above. 

254. -- THE MOTOR-CAR TOUR. 

First of all I will ask the reader to compare the original squarediagram with the circular one shown in Figs. 1, 2, and 3 below. If forthe moment we ignore the shading (the purpose of which I shall proceedto explain), we find that the circular diagram in each case is merely asimplification of the original square one--that is, the roads from Alead to B, E, and M in both cases, the roads from L (London) lead to I, K, and S, and so on. The form below, being circular and symmetrical, answers my purpose better in applying a mechanical solution, and Itherefore adopt it without altering in any way the conditions of thepuzzle. If such a question as distances from town to town came into theproblem, the new diagrams might require the addition of numbers toindicate these distances, or they might conceivably not be at allpracticable. 

[Illustration: Figs. 1, 2, and 3]

Now, I draw the three circular diagrams, as shown, on a sheet of paperand then cut out three pieces of cardboard of the forms indicated by theshaded parts of these diagrams. It can be shown that every route, ifmarked out with a red pencil, will form one or other of the designsindicated by the edges of the cards, or a reflection thereof. Let usdirect our attention to Fig. 1. Here the card is so placed that the staris at the town T; it therefore gives us (by following the edge of thecard) one of the circular routes from London: L, S, R, T, M, A, E, P, O, J, D, C, B, G, N, Q, K, H, F, I, L. If we went the other way, we shouldget L, I, F, H, K, Q, etc. , but these reverse routes were not to becounted. When we have written out this first route we revolve the carduntil the star is at M, when we get another different route, at A athird route, at E a fourth route, and at P a fifth route. We have thusobtained five different routes by revolving the card as it lies. But itis evident that if we now take up the card and replace it with the otherside uppermost, we shall in the same manner get five other routes byrevolution. 

We therefore see how, by using the revolving card in Fig. 1, we may, without any difficulty, at once write out ten routes. And if we employthe cards in Figs. 2 and 3, we similarly obtain in each case ten otherroutes. These thirty routes are all that are possible. I do not give theactual proof that the three cards exhaust all the possible cases, butleave the reader to reason that out for himself. If he works out anyroute at haphazard, he will certainly find that it falls into one orother of the three categories. 

255. --THE LEVEL PUZZLE. 

Let us confine our attention to the L in the top left-hand corner. Suppose we go by way of the E on the right: we must then go straight onto the V, from which letter the word may be completed in four ways, forthere are four E's available through which we may reach an L. There aretherefore four ways of reading through the right-hand E. It is alsoclear that there must be the same number of ways through the E that isimmediately below our starting point. That makes eight. If, however, wetake the third route through the E on the diagonal, we then have theoption of any one of the three V's, by means of each of which we maycomplete the word in four ways. We can therefore spell LEVEL in twelveways through the diagonal E. Twelve added to eight gives twentyreadings, all emanating from the L in the top left-hand corner; and asthe four corners are equal, the answer must be four times twenty, oreighty different ways. 

256. --THE DIAMOND PUZZLE. 

There are 252 different ways. The general formula is that, for words ofn letters (not palindromes, as in the case of the next puzzle), whengrouped in this manner, there are always 2^(n+1) - 4 different readings. This does not allow diagonal readings, such as you would get if you usedinstead such a word as DIGGING, where it would be possible to pass fromone G to another G by a diagonal step. 

257. --THE DEIFIED PUZZLE. 

The correct answer is 1, 992 different ways. Every F is either a corner For a side F--standing next to a corner in its own square of F's. Now, FIED may be read _from_ a corner F in 16 ways; therefore DEIF may beread _into_ a corner F also in 16 ways; hence DEIFIED may be read_through_ a corner F in 16 × 16 = 256 ways. Consequently, the fourcorner F's give 4 × 256 = 1, 024 ways. Then FIED may be read from a sideF in 11 ways, and DEIFIED therefore in 121 ways. But there are eightside F's; consequently these give together 8 × 121 = 968 ways. Add 968to 1, 024 and we get the answer, 1, 992. 

In this form the solution will depend on whether the number of lettersin the palindrome be odd or even. For example, if you apply the word NUNin precisely the same manner, you will get 64 different readings; but ifyou use the word NOON, you will only get 56, because you cannot use thesame letter twice in immediate succession (since you must "always passfrom one letter to another") or diagonal readings, and every readingmust involve the use of the central N. 

The reader may like to find for himself the general formula in thiscase, which is complex and difficult. I will merely add that for such acase as MADAM, dealt with in the same way as DEIFIED, the number ofreadings is 400. 

258. -- THE VOTERS' PUZZLE. 

THE number of readings here is 63, 504, as in the case of "WAS IT A RAT ISAW" (No. 30, _Canterbury Puzzles_). The general formula is that forpalindromic sentences containing 2n + 1 letters there are (4(2^n -1))²readings. 

259. -- HANNAH'S PUZZLE. 

Starting from any one of the N's, there are 17 different readings ofNAH, or 68 (4 times 17) for the 4 N's. Therefore there are also 68 waysof spelling HAN. If we were allowed to use the same N twice in aspelling, the answer would be 68 times 68, or 4, 624 ways. But theconditions were, "always passing from one letter to another. " Therefore, for every one of the 17 ways of spelling HAN with a particular N, therewould be 51 ways (3 times 17) of completing the NAH, or 867 (17 times51) ways for the complete word. Hence, as there are four N's to use inHAN, the correct solution of the puzzle is 3, 468 (4 times 867) differentways. 

260. --THE HONEYCOMB PUZZLE. 

The required proverb is, "There is many a slip 'twixt the cup and thelip. " Start at the T on the outside at the bottom right-hand corner, pass to the H above it, and the rest is easy. 

261. -- THE MONK AND THE BRIDGES. 

[Illustration]

The problem of the Bridges may be reduced to the simple diagram shownin illustration. The point M represents the Monk, the point I theIsland, and the point Y the Monastery. Now the only direct ways from Mto I are by the bridges a and b; the only direct ways from I to Y areby the bridges c and d; and there is a direct way from M to Y by thebridge e. Now, what we have to do is to count all the routes that willlead from M to Y, passing over all the bridges, a, b, c, d, and e onceand once only. With the simple diagram under the eye it is quite easy, without any elaborate rule, to count these routes methodically. Thus, starting from a, b, we find there are only two ways of completing theroute; with _a, c_, there are only two routes; with a, d, only tworoutes; and so on. It will be found that there are sixteen such routesin all, as in the following list:--

 a b e c d b c d a e a b e d c b c e a d a c d b e b d c a e a c e b d b d e a c a d e b c e c a b d a d c b e e c b a d b a e c d e d a b c b a e d c e d b a c

If the reader will transfer the letters indicating the bridges from thediagram to the corresponding bridges in the original illustration, everything will be quite obvious. 

262. --THOSE FIFTEEN SHEEP. 

If we read the exact words of the writer in the cyclopædia, we find thatwe are not told that the pens were all necessarily empty! In fact, ifthe reader will refer back to the illustration, he will see that onesheep is already in one of the pens. It was just at this point that thewily farmer said to me, "_Now_ I'm going to start placing the fifteensheep. " He thereupon proceeded to drive three from his flock into thealready occupied pen, and then placed four sheep in each of the otherthree pens. "There, " says he, "you have seen me place fifteen sheep infour pens so that there shall be the same number of sheep in every pen. "I was, of course, forced to admit that he was perfectly correct, according to the exact wording of the question. 

263. --KING ARTHUR'S KNIGHTS. 

On the second evening King Arthur arranged the knights and himself inthe following order round the table: A, F, B, D, G, E, C. On the thirdevening they sat thus, A, E, B, G, C, F, D. He thus had B next but oneto him on both occasions (the nearest possible), and G was the thirdfrom him at both sittings (the furthest position possible). No other wayof sitting the knights would have been so satisfactory. 

264. --THE CITY LUNCHEONS. 

The men may be grouped as follows, where each line represents a day andeach column a table:--

 AB CD EF GH IJ KL AE DL GK FI CB HJ AG LJ FH KC DE IB AF JB KI HD LG CE AK BE HC IL JF DG AH EG ID CJ BK LF AI GF CL DB EH JK AC FK DJ LE GI BH AD KH LB JG FC EI AL HI JE BF KD GC AJ IC BG EK HL FD

Note that in every column (except in the case of the A's) all theletters descend cyclically in the same order, B, E, G, F, up to J, whichis followed by B. 

265. --A PUZZLE FOR CARD-PLAYERS. 

In the following solution each of the eleven lines represents a sitting, each column a table, and each pair of letters a pair of partners. 

 A B -- I L | E J -- G K | F H -- C D A C -- J B | F K -- H L | G I -- D E A D -- K C | G L -- I B | H J -- E F A E -- L D | H B -- J C | I K -- F G A F -- B E | I C -- K D | J L -- G H A G -- C F | J D -- L E | K B -- H I A H -- D G | K E -- B F | L C -- I J A I -- E H | L F -- C G | B D -- J K A J -- F I | B G -- D H | C E -- K L A K -- G J | C H -- E I | D F -- L B A L -- H K | D I -- F J | E G -- B C

It will be seen that the letters B, C, D . . . L descend cyclically. Thesolution given above is absolutely perfect in all respects. It will befound that every player has every other player once as his partner andtwice as his opponent. 

266. --A TENNIS TOURNAMENT. 

Call the men A, B, D, E, and their wives a, b, d, e. Then they may playas follows without any person ever playing twice with or against anyother person:--

 First Court. Second Court. 1st Day | A d against B e | D a against E b 2nd Day | A e " D b | E a " B d 3rd Day | A b " E d | B a " D e

It will be seen that no man ever plays with or against his own wife--anideal arrangement. If the reader wants a hard puzzle, let him try toarrange eight married couples (in four courts on seven days) underexactly similar conditions. It can be done, but I leave the reader inthis case the pleasure of seeking the answer and the general solution. 

267. --THE WRONG HATS. 

The number of different ways in which eight persons, with eight hats, can each take the wrong hat, is 14, 833. 

Here are the successive solutions for any number of persons from one toeight:--

 1 = 0 2 = 1 3 = 2 4 = 9 5 = 44 6 = 265 7 = 1, 854 8 = 14, 833

To get these numbers, multiply successively by 2, 3, 4, 5, etc. When themultiplier is even, add 1; when odd, deduct 1. Thus, 3 × 1 - 1 = 2; 4 ×2 + 1 = 9; 5 × 9 - 1 = 44; and so on. Or you can multiply the sum of thenumber of ways for n - 1 and n - 2 persons by n - 1, and so get thesolution for n persons. Thus, 4(2 + 9) = 44; 5(9 + 44) = 265; and so on. 

268. --THE PEAL OF BELLS. 

The bells should be rung as follows:--

 1 2 3 4 2 1 4 3 2 4 1 3 4 2 3 1 4 3 2 1 3 4 1 2 3 1 4 2 1 3 2 4 3 1 2 4 1 3 4 2 1 4 3 2 4 1 2 3 4 2 1 3 2 4 3 1 2 3 4 1 3 2 1 4 2 3 1 4 3 2 4 1 3 4 2 1 4 3 1 2 4 1 3 2 1 4 2 3 1 2 4 3 2 1 3 4

I have constructed peals for five and six bells respectively, and asolution is possible for any number of bells under the conditionspreviously stated. 

269. --THREE MEN IN A BOAT. 

If there were no conditions whatever, except that the men were all to goout together, in threes, they could row in an immense number ofdifferent ways. If the reader wishes to know how many, the number is455^7. And with the condition that no two may ever be together more thanonce, there are no fewer than 15, 567, 552, 000 different solutions--thatis, different ways of arranging the men. With one solution before him, the reader will realize why this must be, for although, as an example, Amust go out once with B and once with C, it does not necessarily followthat he must go out with C on the same occasion that he goes with B. Hemight take any other letter with him on that occasion, though the factof his taking other than B would have its effect on the arrangement ofthe other triplets. 

Of course only a certain number of all these arrangements are availablewhen we have that other condition of using the smallest possible numberof boats. As a matter of fact we need employ only ten different boats. Here is one the arrangements:--

 1 2 3 4 5 1st Day (ABC) (DBF) (GHI) (JKL) (MNO) 8 6 7 9 10 2nd Day (ADG) (BKN) (COL) (JEI) (MHF) 3 5 4 1 2 3rd Day (AJM) (BEH) (CFI) (DKO) (GNL) 7 6 8 9 1 4th Day (AEK) (CGM) (BOI) (DHL) (JNF) 4 5 3 10 2 5th Day (AHN) (CDJ) (BFL) (GEO) (MKI) 6 7 8 10 1 6th Day (AFO) (BGJ) (CKH) (DNI) (MEL) 5 4 3 9 2 7th Day (AIL) (BDM) (CEN) (GKF) (JHO)

It will be found that no two men ever go out twice together, and that noman ever goes out twice in the same boat. 

This is an extension of the well-known problem of the "FifteenSchoolgirls, " by Kirkman. The original conditions were simply thatfifteen girls walked out on seven days in triplets without any girl everwalking twice in a triplet with another girl. Attempts at a generalsolution of this puzzle had exercised the ingenuity of mathematicianssince 1850, when the question was first propounded, until recently. In1908 and the two following years I indicated (see _Educational TimesReprints_, Vols. XIV. , XV. , and XVII. ) that all our trouble had arisenfrom a failure to discover that 15 is a special case (too small to enterinto the general law for all higher numbers of girls of the form 6n+3), and showed what that general law is and how the groups should be posedfor any number of girls. I gave actual arrangements for numbers that hadpreviously baffled all attempts to manipulate, and the problem may nowbe considered generally solved. Readers will find an excellent fullaccount of the puzzle in W. W. Rouse Ball's _Mathematical Recreations_, 5th edition. 

270. --THE GLASS BALLS. 

There are, in all, sixteen balls to be broken, or sixteen places in theorder of breaking. Call the four strings A, B, C, and D--order is hereof no importance. The breaking of the balls on A may occupy any 4 out ofthese 16 places--that is, the combinations of 16 things, taken 4together, will be

 13 × 14 × 15 × 16 ----------------- = 1, 820 1 × 2 × 3 × 4

ways for A. In every one of these cases B may occupy any 4 out of theremaining 12 places, making

 9 × 10 × 11 × 12 ----------------- = 495 1 × 2 × 3 × 4

ways. Thus 1, 820 × 495 = 900, 900 different placings are open to A and B. But for every one of these cases C may occupy

 5 × 6 × 7 × 8 ------------- = 70 1 × 2 × 3 × 4

different places; so that 900, 900 × 70 = 63, 063, 000 different placingsare open to A, B, and C. In every one of these cases, D has no choicebut to take the four places that remain. Therefore the correct answer isthat the balls may be broken in 63, 063, 000 different ways under theconditions. Readers should compare this problem with No. 345, "The TwoPawns, " which they will then know how to solve for cases where there arethree, four, or more pawns on the board. 

271. --FIFTEEN LETTER PUZZLE. 

The following will be found to comply with the conditions of grouping:--

 ALE MET MOP BLM BAG CAP YOU CLT IRE OIL LUG LNR NAY BIT BUN BPR AIM BEY RUM GMY OAR GIN PLY CGR PEG ICY TRY CMN CUE COB TAU PNT ONE GOT PIU

The fifteen letters used are A, E, I, O, U, Y, and B, C, G, L, M, N, P, R, T. The number of words is 27, and these are all shown in the firstthree columns. The last word, PIU, is a musical term in common use; butalthough it has crept into some of our dictionaries, it is Italian, meaning "a little; slightly. " The remaining twenty-six are good words. Of course a TAU-cross is a T-shaped cross, also called the cross of St. Anthony, and borne on a badge in the Bishop's Palace at Exeter. It isalso a name for the toad-fish. 

We thus have twenty-six good words and one doubtful, obtained under therequired conditions, and I do not think it will be easy to improve onthis answer. Of course we are not bound by dictionaries but by commonusage. If we went by the dictionary only in a case of this kind, weshould find ourselves involved in prefixes, contractions, and suchabsurdities as I. O. U. , which Nuttall actually gives as a word. 

272. --THE NINE SCHOOLBOYS. 

The boys can walk out as follows:--

 1st Day. 2nd Day. 3rd Day. A B C B F H F A G D E F E I A I D B G H I C G D H C E

 4th Day. 5th Day. 6th Day. A D H G B I D C A B E G C F D E H B F I C H A E I G F

Every boy will then have walked by the side of every other boy once andonce only. 

Dealing with the problem generally, 12n+9 boys may walk out in tripletsunder the conditions on 9n+6 days, where n may be nought or any integer. Every possible pair will occur once. Call the number of boys m. Thenevery boy will pair m-1 times, of which (m-1)/4 times he will be in themiddle of a triplet and (m-1)/2 times on the outside. Thus, if we referto the solution above, we find that every boy is in the middle twice(making 4 pairs) and four times on the outside (making the remaining 4pairs of his 8). The reader may now like to try his hand at solving thetwo next cases of 21 boys on 15 days, and 33 boys on 24 days. It is, perhaps, interesting to note that a school of 489 boys could thus walkout daily in one leap year, but it would take 731 girls (referred to inthe solution to No. 269) to perform their particular feat by a dailywalk in a year of 365 days. 

273. --THE ROUND TABLE. 

The history of this problem will be found in _The Canterbury Puzzles_(No. 90). Since the publication of that book in 1907, so far as I know, nobody has succeeded in solving the case for that unlucky number ofpersons, 13, seated at a table on 66 occasions. A solution is possiblefor any number of persons, and I have recorded schedules for everynumber up to 25 persons inclusive and for 33. But as I know a good manymathematicians are still considering the case of 13, I will not at thisstage rob them of the pleasure of solving it by showing the answer. ButI will now display the solutions for all the cases up to 12 personsinclusive. Some of these solutions are now published for the first time, and they may afford useful clues to investigators. 

The solution for the case of 3 persons seated on 1 occasion needs noremark. 

A solution for the case of 4 persons on 3 occasions is as follows:--

 1 2 3 4 1 3 4 2 1 4 2 3

Each line represents the order for a sitting, and the person representedby the last number in a line must, of course, be regarded as sittingnext to the first person in the same line, when placed at the roundtable. 

The case of 5 persons on 6 occasions may be solved as follows:--

 1 2 3 4 5 1 2 4 5 3 1 2 5 3 4 --------- 1 3 2 5 4 1 4 2 3 5 1 5 2 4 3

The case for 6 persons on 10 occasions is solved thus:--

 1 2 3 6 4 5 1 3 4 2 5 6 1 4 5 3 6 2 1 5 6 4 2 3 1 6 2 5 3 4 ----------- 1 2 4 5 6 3 1 3 5 6 2 4 1 4 6 2 3 5 1 5 2 3 4 6 1 6 3 4 5 2

It will now no longer be necessary to give the solutions in full, forreasons that I will explain. It will be seen in the examples above thatthe 1 (and, in the case of 5 persons, also the 2) is repeated down thecolumn. Such a number I call a "repeater. " The other numbers descend incyclical order. Thus, for 6 persons we get the cycle, 2, 3, 4, 5, 6, 2, and so on, in every column. So it is only necessary to give the twolines 1 2 3 6 4 5 and 1 2 4 5 6 3, and denote the cycle and repeaters, to enable any one to write out the full solution straight away. Thereader may wonder why I do not start the last solution with the numbersin their natural order, 1 2 3 4 5 6. If I did so the numbers in thedescending cycle would not be in their natural order, and it is moreconvenient to have a regular cycle than to consider the order in thefirst line. 

The difficult case of 7 persons on 15 occasions is solved as follows, and was given by me in _The Canterbury Puzzles_:--

 1 2 3 4 5 7 6 1 6 2 7 5 3 4 1 3 5 2 6 7 4 1 5 7 4 3 6 2 1 5 2 7 3 4 6

In this case the 1 is a repeater, and there are _two_ separate cycles, 2, 3, 4, 2, and 5, 6, 7, 5. We thus get five groups of three lines each, for a fourth line in any group will merely repeat the first line. 

A solution for 8 persons on 21 occasions is as follows:--

 1 8 6 3 4 5 2 7 1 8 4 5 7 2 3 6 1 8 2 7 3 6 4 5

The 1 is here a repeater, and the cycle 2, 3, 4, 5, 6, 7, 8. Every oneof the 3 groups will give 7 lines. 

Here is my solution for 9 persons on 28 occasions:--

 2 1 9 7 4 5 6 3 8 2 9 5 1 6 8 3 4 7 2 9 3 1 8 4 7 5 6 2 9 1 5 6 4 7 8 3

There are here two repeaters, 1 and 2, and the cycle is 3, 4, 5, 6, 7, 8, 9. We thus get 4 groups of 7 lines each. 

The case of 10 persons on 36 occasions is solved as follows:--

 1 10 8 3 6 5 4 7 2 9 1 10 6 5 2 9 7 4 3 8 1 10 2 9 3 8 6 5 7 4 1 10 7 4 8 3 2 9 5 6

The repeater is 1, and the cycle, 2, 3, 4, 5, 6, 7, 8, 9, 10. We herehave 4 groups of 9 lines each. 

My solution for 11 persons on 45 occasions is as follows:--

 2 11 9 4 7 6 5 1 8 3 10 2 1 11 7 6 3 10 8 5 4 9 2 11 10 3 9 4 8 5 1 7 6 2 11 5 8 1 3 10 6 7 9 4 2 11 1 10 3 4 9 6 7 5 8

There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5, . . . 11. Wethus get 5 groups of 9 lines each. 

The case of 12 persons on 55 occasions is solved thus:--

 1 2 3 12 4 11 5 10 6 9 7 8 1 2 4 11 6 9 8 7 10 5 12 3 1 2 5 10 8 7 11 4 3 12 6 9 1 2 6 9 10 5 3 12 7 8 11 4 1 2 7 8 12 3 6 9 11 4 5 10

Here 1 is a repeater, and the cycle is 2, 3, 4, 5, . . . 12. We thus get 5groups of 11 lines each. 

274. --THE MOUSE-TRAP PUZZLE. 

If we interchange cards 6 and 13 and begin our count at 14, we may takeup all the twenty-one cards--that is, make twenty-one "catches"--in thefollowing order: 6, 8, 13, 2, 10, 1, 11, 4, 14, 3, 5, 7, 21, 12, 15, 20, 9, 16, 18, 17, 19. We may also exchange 10 and 14 and start at 16, orexchange 6 and 8 and start at 19. 

275. --THE SIXTEEN SHEEP. 

The six diagrams on next page show solutions for the cases where wereplace 2, 3, 4, 5, 6, and 7 hurdles. The dark lines indicate thehurdles that have been replaced. There are, of course, other ways ofmaking the removals. 

276. --THE EIGHT VILLAS. 

There are several ways of solving the puzzle, but there is very littledifference between them. The solver should, however, first of all bearin mind that in making his calculations he need only consider the fourvillas that stand at the corners, because the intermediate villas cannever vary when the corners are known. One way is to place the numbersnought to 9 one at a time in the top left-hand corner, and then considereach case in turn. 

Now, if we place 9 in the corner as shown in the Diagram A, two of thecorners cannot be occupied, while the corner that is diagonally oppositemay be filled by 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 persons. We thus seethat there are 10

[Illustration:

 +---+---+ +-+-----+ +---+---+ |O OHO O| |OHO O O| |O OHO O| | H | | + | | +=+ | |O OHO O| |OHO O O| |O OHOHO| +-+ +-+-+ +-+-----+ +---+ + | |O|O O|O| |O|O O O| |O O O|O| | +---+ | | +-+-+ | | +-+ | |O O O O| |O O OHO| |O O|O O| +-------+ +-------+ +-------+ 2 3 4

 +-----+-+ +-+-----+ +-------+ |O O OHO| |OHO O O| |O O O O| | +=+ | | +=+ | | +=+=+=+ |O OHO O| |OHOHO O| |OHOHO O| | +-+-+ + | + +-+ | + + + | |O|O O|O| |O|O O|O| |O|OHO O| +=+ +=+ | + +=+ +=+ + | |O O O O| |OHO O O| |O O|O O| +-------+ +-+-----+ +---+---+ 5 6 7 THE SIXTEEN SHEEP

]

solutions with a 9 in the corner. If, however, we substitute 8, the twocorners in the same row and column may contain 0, 0, or 1, 1, or 0, 1, or 1, 0. In the case of B, ten different selections may be made for thefourth corner; but in each of the cases C, D, and E, only nineselections are possible, because we cannot use the 9. Therefore with 8in the top left-hand corner there are 10 + (3 × 9) = 37 differentsolutions. If we then try 7 in the corner, the result will be 10 + 27 +40, or 77 solutions. With 6 we get 10 + 27 + 40 + 49 = 126; with 5, 10 +27 + 40 + 49 + 54 = 180; with 4, the same as with 5, + 55 = 235 ; with3, the same as with 4, + 52 = 287; with 2, the same as with 3, + 45 =332; with 1, the same as with 2, + 34 = 366, and with nought in the topleft-hand corner the number of solutions will be found to be 10 + 27 +40 + 49 + 54 + 55 + 52 + 45 + 34 + 19 = 385. As there is no other numberto be placed in the top left-hand corner, we have now only to add thesetotals together thus, 10 + 37 + 77 + 126 + 180 + 235 + 287 + 332 + 366 +385 = 2, 035. We therefore find that the total number of ways in whichtenants may occupy some or all of the eight villas so that there shallbe always nine persons living along each side of the square is 2, 035. Ofcourse, this method must obviously cover all the reversals andreflections, since each corner in turn is occupied by every number inall possible combinations with the other two corners that are in linewith it. 

[Illustration:

 A B C D E +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ |9| |0| |8| |0| |8| |1| |8| |0| |8| |1| +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ | |*| | | |*| | | |*| | | |*| | | |*| | +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ |0| | | |0| | | |1| | | |1| | | |0| | | +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+

]

Here is a general formula for solving the puzzle: (n² + 3n + 2)(n² +3n + 3)/6. Whatever may be the stipulated number of residents alongeach of the sides (which number is represented by n), the total numberof different arrangements may be thus ascertained. In our particularcase the number of residents was nine. Therefore (81 + 27 + 2) × (81 +27 + 3) and the product, divided by 6, gives 2, 035. If the number ofresidents had been 0, 1, 2, 3, 4, 5, 6, 7, or 8, the totalarrangements would be 1, 7, 26, 70, 155, 301, 532, 876, or 1, 365respectively. 

277. --COUNTER CROSSES. 

Let us first deal with the Greek Cross. There are just eighteen forms inwhich the numbers may be paired for the two arms. Here they are:--

 12978 13968 14958 34956 24957 23967

 23958 13769 14759 14967 24758 23768

 12589 23759 13579 34567 14768 24568

 14569 23569 14379 23578 14578 25368

 15369 24369 23189 24378 15378 45167

 24179 25169 34169 35168 34178 25178

Of course, the number in the middle is common to both arms. The firstpair is the one I gave as an example. I will suppose that we havewritten out all these crosses, always placing the first row of a pair inthe upright and the second row in the horizontal arm. Now, if we leavethe central figure fixed, there are 24 ways in which the numbers in theupright may be varied, for the four counters may be changed in 1 × 2 × 3× 4 = 24 ways. And as the four in the horizontal may also be changed in24 ways for every arrangement on the other arm, we find that there are24 × 24 = 576 variations for every form; therefore, as there are 18forms, we get 18 × 576 = 10, 368 ways. But this will include half thefour reversals and half the four reflections that we barred, so we mustdivide this by 4 to obtain the correct answer to the Greek Cross, whichis thus 2, 592 different ways. The division is by 4 and not by 8, becausewe provided against half the reversals and reflections by alwaysreserving one number for the upright and the other for the horizontal. 

In the case of the Latin Cross, it is obvious that we have to deal withthe same 18 forms of pairing. The total number of different ways in thiscase is the full number, 18 × 576. Owing to the fact that the upper andlower arms are unequal in length, permutations will repeat byreflection, but not by reversal, for we cannot reverse. Therefore thisfact only entails division by 2. But in every pair we may exchange thefigures in the upright with those in the horizontal (which we could notdo in the case of the Greek Cross, as the arms are there all alike);consequently we must multiply by 2. This multiplication by 2 anddivision by 2 cancel one another. Hence 10, 368 is here the correctanswer. 

278. --A DORMITORY PUZZLE. 

[Illustration:

 MON. TUES. WED. +---+---+---+ +---+---+---+ +---+---+---+ | 1 | 2 | 1 | | 1 | 3 | 1 | | 1 | 4 | 1 | +---+---+---+ +---+---+---+ +---+---+---+ | 2 | | 2 | | 1 | | 1 | | 1 | | 1 | +---+---+---+ +---+---+---+ +---+---+---+ | 1 | 22| 1 | | 3 | 19| 3 | | 4 | 16| 4 | +---+---+---+ +---+---+---+ +---+---+---+

 THURS. FRI. SAT. +---+---+---+ +---+---+---+ +---+---+---+ | 1 | 5 | 1 | | 2 | 6 | 2 | | 4 | 4 | 4 | +---+---+---+ +---+---+---+ +---+---+---+ | 2 | | 2 | | 1 | | 1 | | 4 | | 4 | +---+---+---+ +---+---+---+ +---+---+---+ | 4 | 13| 4 | | 7 | 6 | 7 | | 4 | 4 | 4 | +---+---+---+ +---+---+---+ +---+---+---+

]

Arrange the nuns from day to day as shown in the six diagrams. Thesmallest possible number of nuns would be thirty-two, and thearrangements on the last three days admit of variation. 

279. --THE BARRELS OF BALSAM. 

This is quite easy to solve for any number of barrels--if you know how. This is the way to do it. There are five barrels in each row Multiplythe numbers 1, 2, 3, 4, 5 together; and also multiply 6, 7, 8, 9, 10together. Divide one result by the other, and we get the number ofdifferent combinations or selections of ten things taken five at a time. This is here 252. Now, if we divide this by 6 (1 more than the number inthe row) we get 42, which is the correct answer to the puzzle, for thereare 42 different ways of arranging the barrels. Try this method ofsolution in the case of six barrels, three in each row, and you willfind the answer is 5 ways. If you check this by trial, you will discoverthe five arrangements with 123, 124, 125, 134, 135 respectively in thetop row, and you will find no others. 

The general solution to the problem is, in fact, this:

 n C 2n ----- n + 1

where 2n equals the number of barrels. The symbol C, of course, impliesthat we have to find how many combinations, or selections, we can makeof 2n things, taken n at a time. 

280. --BUILDING THE TETRAHEDRON. 

Take your constructed pyramid and hold it so that one stick only lies onthe table. Now, four sticks must branch off from it in differentdirections--two at each end. Any one of the five sticks may be left outof this connection; therefore the four may be selected in 5 differentways. But these four matches may be placed in 24 different orders. Andas any match may be joined at either of its ends, they may further bevaried (after their situations are settled for any particulararrangement) in 16 different ways. In every arrangement the sixth stickmay be added in 2 different ways. Now multiply these results together, and we get 5 × 24 × 16 × 2 = 3, 840 as the exact number of ways in whichthe pyramid may be constructed. This method excludes all possibility oferror. 

A common cause of error is this. If you calculate your combinations byworking upwards from a basic triangle lying on the table, you will gethalf the correct number of ways, because you overlook the fact that anequal number of pyramids may be built on that triangle downwards, so tospeak, through the table. They are, in fact, reflections of the others, and examples from the two sets of pyramids cannot be set up to resembleone another--except under fourth dimensional conditions!

281. --PAINTING A PYRAMID. 

It will be convenient to imagine that we are painting our pyramids onthe flat cardboard, as in the diagrams, before folding up. Now, if wetake any _four_ colours (say red, blue, green, and yellow), they may beapplied in only 2 distinctive ways, as shown in Figs, 1 and 2. Any otherway will only result in one of these when the pyramids are folded up. Ifwe take any _three_ colours, they may be applied in the 3 ways shown inFigs. 3, 4, and 5. If we take any _two_ colours, they may be applied inthe 3 ways shown in Figs. 6, 7, and 8. If we take any _single_ colour, it may obviously be applied in only 1 way. But four colours may beselected in 35 ways out of seven; three in 35 ways; two in 21 ways; andone colour in 7 ways. Therefore 35 applied in 2 ways = 70; 35 in 3 ways= 105; 21 in 3 ways = 63; and 7 in 1 way = 7. Consequently the pyramidmay be painted in 245 different ways (70 + 105 + 63 + 7), using theseven colours of the solar spectrum in accordance with the conditions ofthe puzzle. 

[Illustration:

 1 2 +---------------+ +---------------+ \ R / \ B / \ B / \ R / \ / \ / \ / \ / \ / G \ / \ / G \ / \-------/ \-------/ \ / \ / \ Y / \ Y / \ / \ / ' '

 3 4 5 +---------------+ +---------------+ +---------------+ \ R / \ R / \ R / \ G / \ Y / \ R / \ / \ / \ / \ / \ / \ / \ / G \ / \ / G \ / \ / G \ / \-------/ \-------/ \-------/ \ / \ / \ / \ Y / \ Y / \ Y / \ / \ / \ / ' ' '

 6 7 8 +---------------+ +---------------+ +---------------+ \ G / \ Y / \ Y / \ Y / \ G / \ G / \ / \ / \ / \ / \ / \ / \ / G \ / \ / G \ / \ / G \ / \-------/ \-------/ \-------/ \ / \ / \ / \ Y / \ Y / \ Y / \ / \ / \ / ' ' '

]

282. --THE ANTIQUARY'S CHAIN. 

[Illustration]

THE number of ways in which nine things may be arranged in a row withoutany restrictions is 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 362, 880. But weare told that the two circular rings must never be together; thereforewe must deduct the number of times that this would occur. The number is1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40, 320 × 2 = 80, 640, because if weconsider the two circular links to be inseparably joined together theybecome as one link, and eight links are capable of 40, 320 arrangements;but as these two links may always be put on in the orders AB or BA, wehave to double this number, it being a question of arrangement and notof design. The deduction required reduces our total to 282, 240. Then oneof our links is of a peculiar form, like an 8. We have therefore theoption of joining on either one end or the other on every occasion, sowe must double the last result. This brings up our total to 564, 480. 

We now come to the point to which I directed the reader'sattention--that every link may be put on in one of two ways. If we jointhe first finger and thumb of our left hand horizontally, and then linkthe first finger and thumb of the right hand, we see that the rightthumb may be either above or below. But in the case of our chain we mustremember that although that 8-shaped link has two independent _ends_ itis like every other link in having only two _sides_--that is, you cannotturn over one end without turning the other at the same time. 

We will, for convenience, assume that each link has a black side and aside painted white. Now, if it were stipulated that (with the chainlying on the table, and every successive link falling over itspredecessor in the same way, as in the diagram) only the white sidesshould be uppermost as in A, then the answer would be 564, 480, asabove--ignoring for the present all reversals of the completed chain. If, however, the first link were allowed to be placed either side up, then we could have either A or B, and the answer would be 2 × 564, 480 =1, 128, 960; if two links might be placed either way up, the answer wouldbe 4 × 564, 480; if three links, then 8 × 564, 480, and so on. Since, therefore, every link may be placed either side up, the number will be564, 480 multiplied by 2^9, or by 512. This raises our total to289, 013, 760. 

But there is still one more point to be considered. We have not yetallowed for the fact that with any given arrangement three of the otherarrangements may be obtained by simply turning the chain over throughits entire length and by reversing the ends. Thus C is really the sameas A, and if we turn this page upside down, then A and C give two otherarrangements that are still really identical. Thus to get the correctanswer to the puzzle we must divide our last total by 4, when we findthat there are just 72, 253, 440 different ways in which the smith mighthave put those links together. In other words, if the nine links hadoriginally formed a piece of chain, and it was known that the twocircular links were separated, then it would be 72, 253, 439 chances to 1that the smith would not have put the links together again precisely asthey were arranged before!

283. --THE FIFTEEN DOMINOES. 

The reader may have noticed that at each end of the line I give is afour, so that, if we like, we can form a ring instead of a line. It caneasily be proved that this must always be so. Every line arrangementwill make a circular arrangement if we like to join the ends. Now, curious as it may at first appear, the following diagram exactlyrepresents the conditions when we leave the doubles out of the questionand devote our attention to forming circular arrangements. Each number, or half domino, is in line with every other number, so that if we startat any one of the five numbers and go over all the lines of the pentagononce and once only we shall come back to the starting place, and theorder of our route will give us one of the circular arrangements for theten dominoes. Take your pencil and follow out the following route, starting at the 4: 41304210234. You have been over all the lines onceonly, and by repeating all these figures in this way, 41--13--30--04--42--21--10--02--23--34, you get an arrangement of thedominoes (without the doubles) which will be perfectly clear. Take otherroutes and you will get other arrangements. If, therefore, we canascertain just how many of these circular routes are obtainable fromthe pentagon, then the rest is very easy. 

Well, the number of different circular routes over the pentagon is 264. How I arrive at these figures I will not at present explain, because itwould take a lot of space. The dominoes may, therefore, be arranged in acircle in just 264 different ways, leaving out the doubles. Now, in anyone of these circles the five doubles may be inserted in 2^5 = 32different ways. Therefore when we include the doubles there are 264 × 32= 8, 448 different circular arrangements. But each of those circles maybe broken (so as to form our straight line) in any one of 15 differentplaces. Consequently, 8, 448 × 15 gives 126, 720 different ways as thecorrect answer to the puzzle. 

[Illustration:

 ----- | | / | | \ / ----- \ / . . \ ----- . . ----- | | . . | o o | | o | -. --------. --- | | | | . . . | o o | ----- . . . . ----- \ . . . . / ----- . . ----- | o | . . |o | | | --------- | o | | o |. . | o| ----- -----

]

I purposely refrained from asking the reader to discover in just howmany different ways the full set of twenty-eight dominoes may bearranged in a straight line in accordance with the ordinary rules of thegame, left to right and right to left of any arrangement counting asdifferent ways. It is an exceedingly difficult problem, but the correctanswer is 7, 959, 229, 931, 520 ways. The method of solving is very complex. 

284. --THE CROSS TARGET. 

[Illustration:

 -- -- (CD)( ) -- -- (AE)(A ) -- -- -- -- -- -- (CE)(E )(A )(AB)(C )(D ) -- -- -- -- -- -- (D )( )(B )(E )(EB)( ) -- -- -- -- -- -- (C )(B ) -- -- ( )(ED) -- --

]

Twenty-one different squares may be selected. Of these nine will be ofthe size shown by the four A's in the diagram, four of the size shown bythe B's, four of the size shown by the C's, two of the size shown by theD's, and two of the size indicated by the upper single A, the uppersingle E, the lower single C, and the EB. It is an interesting fact thatyou cannot form any one of these twenty-one squares without using atleast one of the six circles marked E. 

285. --THE FOUR POSTAGE STAMPS. 

Referring to the original diagram, the four stamps may be given in theshape 1, 2, 3, 4, in three ways; in the shape 1, 2, 5, 6, in six ways;in the shape 1, 2, 3, 5, or 1, 2, 3, 7, or 1, 5, 6, 7, or 3, 5, 6, 7, intwenty-eight ways; in shape 1, 2, 3, 6, or 2, 5, 6, 7, in fourteen ways;in shape 1, 2, 6, 7, or 2, 3, 5, 6, or 1, 5, 6, 10, or 2, 5, 6, 9, infourteen ways. Thus there are sixty-five ways in all. 

286. --PAINTING THE DIE. 

The 1 can be marked on any one of six different sides. For every sideoccupied by 1 we have a selection of four sides for the 2. For everysituation of the 2 we have two places for the 3. (The 6, 5, and 4 neednot be considered, as their positions are determined by the 1, 2, and3. ) Therefore 6, 4, and 2 multiplied together make 48 differentways--the correct answer. 

287. --AN ACROSTIC PUZZLE. 

There are twenty-six letters in the alphabet, giving 325 differentpairs. Every one of these pairs may be reversed, making 650 ways. Butevery initial letter may be repeated as the final, producing 26 otherways. The total is therefore 676 different pairs. In other words, theanswer is the square of the number of letters in the alphabet. 

288. --CHEQUERED BOARD DIVISIONS. 

There are 255 different ways of cutting the board into two pieces ofexactly the same size and shape. Every way must involve one of the fivecuts shown in Diagrams A, B, C, D, and E. To avoid repetitions byreversal and reflection, we need only consider cuts that enter at thepoints a, b, and c. But the exit must always be at a point in a straightline from the entry through the centre. This is the most importantcondition to remember. In case B you cannot enter at a, or you will getthe cut provided for in E. Similarly in C or D, you must not enter thekey-line in the same direction as itself, or you will get A or B. If youare working on A or C and entering at a, you must consider joins at oneend only of the key-line, or you will get repetitions. In other casesyou must consider joins at both ends of the key; but after leaving a incase D, turn always either to right or left--use one direction only. Figs. 1 and 2 are examples under A; 3 and 4 are examples under B; 5 and6 come under C;

[Illustration]

and 7 is a pretty example of D. Of course, E is a peculiar type, andobviously admits of only one way of cutting, for you clearly cannotenter at b or c. 

Here is a table of the results:--

 a b c Ways. A = 8 + 17 + 21 = 46 B = 0 + 17 + 21 = 38 C = 15 + 31 + 39 = 85 D = 17 + 29 + 39 = 85 E = 1 + 0 + 0 = 1 -- -- -- --- 41 94 120 255

I have not attempted the task of enumerating the ways of dividing aboard 8 × 8--that is, an ordinary chessboard. Whatever the methodadopted, the solution would entail considerable labour. 

289. --LIONS AND CROWNS. 

[Illustration]

Here is the solution. It will be seen that each of the four pieces(after making the cuts along the thick lines) is of exactly the samesize and shape, and that each piece contains a lion and a crown. Two ofthe pieces are shaded so as to make the solution quite clear to the eye. 

290. --BOARDS WITH AN ODD NUMBER OF SQUARES. 

There are fifteen different ways of cutting the 5 × 5 board (with thecentral square removed) into two pieces of the same size and shape. Limitations of space will not allow me to give diagrams of all these, but I will enable the reader to draw them all out for himself withoutthe slightest difficulty. At whatever point on the edge your cut enters, it must always end at a point on the edge, exactly opposite in a linethrough the centre of the square. Thus, if you enter at point 1 (seeFig. 1) at the top, you must leave at point 1 at the bottom. Now, 1 and2 are the only two really different points of entry; if we use anyothers they will simply produce similar solutions. The directions of thecuts in the following fifteen

[Illustration: Fig. 1. Fig. 2. ]

solutions are indicated by the numbers on the diagram. The duplicationof the numbers can lead to no confusion, since every successive numberis contiguous to the previous one. But whichever direction you take fromthe top downwards you must repeat from the bottom upwards, one directionbeing an exact reflection of the other. 

 1, 4, 8. 1, 4, 3, 7, 8. 1, 4, 3, 7, 10, 9. 1, 4, 3, 7, 10, 6, 5, 9. 1, 4, 5, 9. 1, 4, 5, 6, 10, 9. 1, 4, 5, 6, 10, 7, 8. 2, 3, 4, 8. 2, 3, 4, 5, 9. 2, 3, 4, 5, 6, 10, 9. 2, 3, 4, 5, 6, 10, 7, 8. 2, 3, 7, 8. 2, 3, 7, 10, 9. 2, 3, 7, 10, 6, 5, 9. 2, 3, 7, 10, 6, 5, 4, 8. 

It will be seen that the fourth direction (1, 4, 3, 7, 10, 6, 5, 9)produces the solution shown in Fig. 2. The thirteenth produces thesolution given in propounding the puzzle, where the cut entered at theside instead of at the top. The pieces, however, will be of the sameshape if turned over, which, as it was stated in the conditions, wouldnot constitute a different solution. 

291. --THE GRAND LAMA'S PROBLEM. 

The method of dividing the chessboard so that each of the four partsshall be of exactly the same size and shape, and contain one of thegems, is shown in the diagram. The method of shading the squares isadopted to make the shape of the pieces clear to the eye. Two of thepieces are shaded and two left white. 

The reader may find it interesting to compare this puzzle with that ofthe "Weaver" (No. 14, _Canterbury Puzzles_). 

[Illustration: THE GRAND LAMA'S PROBLEM. 

 +===+===+===+===+===+===+===+===+ |:o:| : : : : : : : I. . . I. . . +===+===+===+===+===+===+ |:::| o |:::::::::::::::::::::::| I. . . I. . . I. . . +===+===+===+===+. . . I |:::| |:o:| : : : |:::| I. . . I. . . I. . . I. . . I===+===+. . . I. . . I |:::| |:::| o |:::::::| |:::| I. . . I. . . I. . . +===I===+. . . I. . . I. . . I |:::| |:::::::| |:::| |:::| I. . . I. . . +===+===+. . . +. . . I. . . I. . . I |:::| : : : |:::| |:::| I. . . +===+===+===+===I. . . I. . . I. . . I |:::::::::::::::::::::::| |:::| +===+===+===+===+===+===+. . . I. . . I | : : : : : : |:::| +===+===+===+===+===+===+===+===+

]

292. --THE ABBOT'S WINDOW. 

THE man who was "learned in strange mysteries" pointed out to FatherJohn that the orders of the Lord Abbot of St. Edmondsbury might beeasily carried out by blocking up twelve of the lights in the window asshown by the dark squares in the following sketch:--

[Illustration:

 +===+===+===+===+===+===+===+===+ | : : : : : : : | I. . . +===+. . . +. . . +. . . +. . . +===+. . . I | IIIII : : : IIIII | I. . . +===+===+. . . +. . . +===+===+. . . I | : IIIII : IIIII : | I. . . +. . . +===+===+===+===+. . . +. . . I | : : IIIIIIIII : : | I. . . +. . . +. . . +===+===+. . . +. . . +. . . I | : : IIIIIIIII : : | I. . . +. . . +===+===+===+===+. . . +. . . I | : IIIII : IIIII : | I. . . +===+===+. . . +. . . +===+===+. . . I | IIIII : : : IIIII | I. . . +===+. . . +. . . +. . . +. . . +===+. . . I | : : : : : : : | +===+===+===+===+===+===+===+===+

]

Father John held that the four corners should also be darkened, but thesage explained that it was desired to obstruct no more light than wasabsolutely necessary, and he said, anticipating Lord Dundreary, "Asingle pane can no more be in a _line_ with itself than one bird can gointo a corner and flock in solitude. The Abbot's condition was that nodiagonal _lines_ should contain an odd number of lights. "

Now, when the holy man saw what had been done he was well pleased, andsaid, "Truly, Father John, thou art a man of deep wisdom, in that thouhast done that which seemed impossible, and yet withal adorned ourwindow with a device of the cross of St. Andrew, whose name I receivedfrom my godfathers and godmothers. " Thereafter he slept well and aroserefreshed. The window might be seen intact to-day in the monastery ofSt. Edmondsbury, if it existed, which, alas! the window does not. 

293. --THE CHINESE CHESSBOARD. 

 +===I===+===+===+===I===+===+===+ | |:::: 2 ::::| 3 |:::| 5 |:6:| I. . . +===+. . . +===+. . . I. . . I. . . +===I |:::: 1 |:::| ::::| 4 |:::| 7 | I. . . +===+===+. . . I===I. . . I===+===I | |:::: |:::| ::::| 9 |:::| I===I. . . I===============I. . . I. . . I |:::: 11|:::: ::::: 10|:::| 8 | I=======I===I===========I. . . I. . . I | ::::: 12|:::: 13::::| |:::| I=======+. . . I. . . +===+===|===+===I |:::: 14|:::| |:::| 16::::| 17| I. . . +. . . I===I===+. . . +. . . +===+. . . I | ::::| ::::: 15|:::| ::::| I=======+===========+===+=======I |:::: ::::: 18::::: ::::: | +===+===+===+===+===+===+===+===+

 +===+===I===I===+===I===+===+===+ | ::::| |:::: |:::| ::::| I. . . +===I. . . I=======I. . . I===+. . . I |:::| |:::: |:::: |:::| | I. . . I===I===============I===I. . . I | |:::: ::::| ::::: |:::| I===I=======I=======I=======I===I |:::| ::::| ::::| ::::| | I. . . I===+. . . I. . . +. . . I. . . +===+. . . I | ::::| |:::: |:::| ::::| I. . . +===I. . . +===I===+. . . I===+. . . I |:::| |:::: |:::: |:::| | I===I. . . +=======I=======+. . . I===I | |:::: ::::| ::::: |:::| I. . . +=======+. . . I. . . +=======+. . . I |:::: ::::| |:::| ::::: | +===+===+===+===+===+===+===+===+

Eighteen is the maximum number of pieces. I give two solutions. Thenumbered diagram is so cut that the eighteenth piece has the largestarea--eight squares--that is possible under the conditions. The seconddiagram was prepared under the added condition that no piece shouldcontain more than five squares. 

No. 74 in _The Canterbury Puzzles_ shows how to cut the board intotwelve pieces, all different, each containing five squares, with onesquare piece of four squares. 

294. --THE CHESSBOARD SENTENCE. 

 +===I===I===I===I=======I=======+ | |:::| |:::| ::::| ::::| I===I. . . I===I. . . I. . . +===I. . . +===I |:::| ::::: |:::| ::::: | |. . . |. . . +===I. . . I. . . +===+. . . +===I | |:::| |:::| ::::| ::::| |. . . +===+. . . +===I===I===I=======I |:::: ::::: |:::| ::::: | I===========I===I. . . I===I===+. . . | | ::::: |:::| |:::| |:::| |. . . +===+. . . |. . . |. . . |. . . I===+. . . | |:::| |:::| |:::| |:::: | |. . . |. . . |. . . |. . . I===+. . . +===+. . . | | |:::| |:::| ::::: |:::| I===+. . . +===I. . . +=======I===+. . . | |:::: ::::| ::::: |:::: | +===========I===================+

The pieces may be fitted together, as shown in the illustration, to forma perfect chessboard. 

295. --THE EIGHT ROOKS. 

Obviously there must be a rook in every row and every column. Startingwith the top row, it is clear that we may put our first rook on any oneof eight different squares. Wherever it is placed, we have the option ofseven squares for the second rook in the second row. Then we have sixsquares from which to select the third row, five in the fourth, and soon. Therefore the number of our different ways must be 8 × 7 × 6 × 5 × 4× 3 × 2 × 1 = 40, 320 (that is 8!), which is the correct answer. 

How many ways there are if mere reversals and reflections are notcounted as different has not yet been determined; it is a difficultproblem. But this point, on a smaller square, is considered in the nextpuzzle. 

296. --THE FOUR LIONS. 

There are only seven different ways under the conditions. They are asfollows: 1 2 3 4, 1 2 4 3, 1 3 2 4, 1 3 4 2, 1 4 3 2, 2 1 4 3, 2 4 1 3. Taking the last example, this notation means that we place a lion in thesecond square of first row, fourth square of second row, first square ofthird row, and third square of fourth row. The first example is, ofcourse, the one we gave when setting the puzzle. 

297. --BISHOPS--UNGUARDED. 

 +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + : ::::: ::::: ::::: ::::: +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + ::::: ::::: ::::: ::::: : +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + : ::::: ::::: ::::: ::::: +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + ::B:: B ::B:: B ::B:: B ::B:: B : +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + : ::::: ::::: ::::: ::::: +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + ::::: ::::: ::::: ::::: : +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + ::::: ::::: ::::: ::::: : +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + : ::::: ::::: ::::: ::::: +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . +

This cannot be done with fewer bishops than eight, and the simplestsolution is to place the bishops in line along the fourth or fifth rowof the board (see diagram). But it will be noticed that no bishop ishere guarded by another, so we consider that point in the next puzzle. 

298. --BISHOPS--GUARDED. 

 +. . . +. . . +. . . +. . . +. . . . . . . +. . . . . . . + : ::::: ::::: ::::: ::::: +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + ::::: ::::: ::::: ::::: : +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + : ::::: ::::: ::::: ::::: +. . . +. . . +. . . +. . . +. . . +. . . +. . . . . . . + ::::: B ::B:: B ::::: B ::B:: : +. . . . . . . . . . . +. . . +. . . +. . . +. . . +. . . + : ::B:: B ::B:: ::B:: B ::::: +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + ::::: ::::: ::::: ::::: : +. . . +. . . +. . . +. . . +. . . +. . . +. . . +. . . + : ::::: ::::: ::::: ::::: +. . . +. . . +. . . +. . . +. . . . . . . +. . . +. . . + ::::: ::::: ::::: ::::: : +. . . +. . . +. . . +. . . +. . . . . . . +. . . +. . . +

This puzzle is quite easy if you first of all give it a little thought. You need only consider squares of one colour, for whatever can be donein the case of the white squares can always be repeated on the black, and they are here quite independent of one another. This equality, ofcourse, is in consequence of the fact that the number of squares on anordinary chessboard, sixty-four, is an even number. If a squarechequered board has an odd number of squares, then there will always beone more square of one colour than of the other. 

Ten bishops are necessary in order that every square shall be attackedand every bishop guarded by another bishop. I give one way of arrangingthem in the diagram. It will be noticed that the two central bishops inthe group of six on the left-hand side of the board serve no purpose, except to protect those bishops that are on adjoining squares. Anothersolution would therefore be obtained by simply raising the upper one ofthese one square and placing the other a square lower down. 

299. --BISHOPS IN CONVOCATION. 

The fourteen bishops may be placed in 256 different ways. But everybishop must always be placed on one of the sides of the board--thatis, somewhere on a row or file on the extreme edge. The puzzle, therefore, consists in counting the number of different ways that wecan arrange the fourteen round the edge of the board without attack. This is not a difficult matter. On a chessboard of n² squares 2n - 2bishops (the maximum number) may always be placed in 2^n ways withoutattacking. On an ordinary chessboard n would be 8; therefore 14bishops may be placed in 256 different ways. It is rather curious thatthe general result should come out in so simple a form. 

[Illustration]

300. --THE EIGHT QUEENS. 

[Illustration]

The solution to this puzzle is shown in the diagram. It will be foundthat no queen attacks another, and also that no three queens are in astraight line in any oblique direction. This is the only arrangement outof the twelve fundamentally different ways of placing eight queenswithout attack that fulfils the last condition. 

301. --THE EIGHT STARS. 

The solution of this puzzle is shown in the first diagram. It is theonly possible solution within the conditions stated. But if one of theeight stars had not already been placed as shown, there would then havebeen eight ways of arranging the stars according to this scheme, if wecount reversals and reflections as different. If you turn this pageround so that each side is in turn at the bottom, you will get the fourreversals; and if you reflect each of these in a mirror, you will getthe four reflections. These are, therefore, merely eight aspects of one"fundamental solution. " But without that first star being so placed, there is another fundamental solution, as shown in the second diagram. But this arrangement being in a way symmetrical, only produces fourdifferent aspects by reversal and reflection. 

[Illustration]

302. --A PROBLEM IN MOSAICS. 

[Illustration]

The diagram shows how the tiles may be rearranged. As before, one yellowand one purple tile are dispensed with. I will here point out that inthe previous arrangement the yellow and purple tiles in the seventh rowmight have changed places, but no other arrangement was possible. 

303. --UNDER THE VEIL. 

Some schemes give more diagonal readings of four letters than others, and we are at first tempted to favour these; but this is a false scent, because what you appear to gain in this direction you lose in others. Ofcourse it immediately occurs to the solver that every LIVE or EVIL isworth twice as much as any other word, since it reads both ways andalways counts as 2. This is an important consideration, though sometimesthose arrangements that contain most readings of these two words arefruitless in other words, and we lose in the general count. 

[Illustration:

 _ _ I V E L _ _ E V L _ _ I _ _ L _ _ I _ _ V E I _ V E _ _ _ L _ E _ _ L V _ I _ L I _ _ I _ E V /V _ E L _ _ I _ _ I _ _ V E L _\

]

The above diagram is in accordance with the conditions requiring noletter to be in line with another similar letter, and it gives twentyreadings of the five words--six horizontally, six vertically, four inthe diagonals indicated by the arrows on the left, and four in thediagonals indicated by the arrows on the right. This is the maximum. 

Four sets of eight letters may be placed on the board of sixty-foursquares in as many as 604 different ways, without any letter ever beingin line with a similar one. This does not count reversals andreflections as different, and it does not take into consideration theactual permutations of the letters among themselves; that is, forexample, making the L's change places with the E's. Now it is a singularfact that not only do the twenty word-readings that I have given proveto be the real maximum, but there is actually only that one arrangementfrom which this maximum may be obtained. But if you make the V's changeplaces with the I's, and the L's with the E's, in the solution given, you still get twenty readings--the same number as before in everydirection. Therefore there are two ways of getting the maximum from thesame arrangement. The minimum number of readings is zero--that is, theletters can be so arranged that no word can be read in any of thedirections. 

304. --BACHET'S SQUARE. 

[Illustration: 1]

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

Let us use the letters A, K, Q, J, to denote ace, king, queen, jack; andD, S, H, C, to denote diamonds, spades, hearts, clubs. In Diagrams 1and 2 we have the two available ways of arranging either group ofletters so that no two similar letters shall be in line--though aquarter-turn of 1 will give us the arrangement in 2. If we superimposeor combine these two squares, we get the arrangement of Diagram 3, whichis one solution. But in each square we may put the letters in the topline in twenty-four different ways without altering the scheme ofarrangement. Thus, in Diagram 4 the S's are similarly placed to the D'sin 2, the H's to the S's, the C's to the H's, and the D's to the C's. Itclearly follows that there must be 24×24 = 576 ways of combining the twoprimitive arrangements. But the error that Labosne fell into was that ofassuming that the A, K, Q, J must be arranged in the form 1, and the D, S, H, C in the form 2. He thus included reflections and half-turns, butnot quarter-turns. They may obviously be interchanged. So that thecorrect answer is 2 × 576 = 1, 152, counting reflections and reversals asdifferent. Put in another manner, the pairs in the top row may bewritten in 16 × 9 × 4 × 1 = 576 different ways, and the square thencompleted in 2 ways, making 1, 152 ways in all. 

305. --THE THIRTY-SIX LETTER BLOCKS. 

I pointed out that it was impossible to get all the letters into the boxunder the conditions, but the puzzle was to place as many as possible. 

This requires a little judgment and careful investigation, or we areliable to jump to the hasty conclusion that the proper way to solve thepuzzle must be first to place all six of one letter, then all six ofanother letter, and so on. As there is only one scheme (with itsreversals) for placing six similar letters so that no two shall be in aline in any direction, the reader will find that after he has placedfour different kinds of letters, six times each, every place is occupiedexcept those twelve that form the two long diagonals. He is, therefore, unable to place more than two each of his last two letters, and thereare eight blanks left. I give such an arrangement in Diagram 1. 

[Illustration: 1]

[Illustration: 2]

The secret, however, consists in not trying thus to place all six ofeach letter. It will be found that if we content ourselves with placingonly five of each letter, this number (thirty in all) may be got intothe box, and there will be only six blanks. But the correct solution isto place six of each of two letters and five of each of the remainingfour. An examination of Diagram 2 will show that there are six each of Cand D, and five each of A, B, E, and F. There are, therefore, only fourblanks left, and no letter is in line with a similar letter in anydirection. 

306. --THE CROWDED CHESSBOARD. 

[Illustration]

Here is the solution. Only 8 queens or 8 rooks can be placed on theboard without attack, while the greatest number of bishops is 14, and ofknights 32. But as all these knights must be placed on squares of thesame colour, while the queens occupy four of each colour and the bishops7 of each colour, it follows that only 21 knights can be placed on thesame colour in this puzzle. More than 21 knights can be placed alone onthe board if we use both colours, but I have not succeeded in placingmore than 21 on the "crowded chessboard. " I believe the above solutioncontains the maximum number of pieces, but possibly some ingeniousreader may succeed in getting in another knight. 

307. --THE COLOURED COUNTERS. 

The counters may be arranged in this order:--

 R1, B2, Y3, O4, GS. Y4, O5, G1, R2, B3. G2, R3, B4, Y5, O1. B5, Y1, O2, G3, R4. O3, G4, R5, B1, Y2. 

308. --THE GENTLE ART OF STAMP-LICKING. 

The following arrangement shows how sixteen stamps may be stuck on thecard, under the conditions, of a total value of fifty pence, or 4s. 2d. :--

[Illustration]

If, after placing the four 5d. Stamps, the reader is tempted to placefour 4d. Stamps also, he can afterwards only place two of each of thethree other denominations, thus losing two spaces and counting no morethan forty-eight pence, or 4s. This is the pitfall that was hinted at. (Compare with No. 43, _Canterbury Puzzles_. )

309. --THE FORTY-NINE COUNTERS. 

The counters may be arranged in this order:--

 A1, B2, C3, D4, E5, F6, G7. F4, G5, A6, B7, C1, D2, E3. D7, E1, F2, G3, A4, B5, C6. B3, C4, D5, E6, F7, G1, A2. G6, A7, B1, C2, D3, E4, F5. E2, F3, G4, A5, B6, C7, D1. C5, D6, E7, F1, G2, A3, B4. 

310. --THE THREE SHEEP. 

The number of different ways in which the three sheep may be placed sothat every pen shall always be either occupied or in line with at leastone sheep is forty-seven. 

The following table, if used with the key in Diagram 1, will enable thereader to place them in all these ways:--

 +------------+---------------------------+----------+ | | | No. Of | | Two Sheep. | Third Sheep. | Ways. | +------------+---------------------------+----------+ | A and B | C, E, G, K, L, N, or P | 7 | | A and C | I, J, K, or O | 4 | | A and D | M, N, or J | 3 | | A and F | J, K, L, or P | 4 | | A and G | H, J, K, N, O, or P | 6 | | A and H | K, L, N, or O | 4 | | A and O | K or L | 2 | | B and C | N | 1 | | B and E | F, H, K, or L | 4 | | B and F | G, J, N, or O | 4 | | B and G | K, L, or N | 3 | | B and H | J or N | 2 | | B and J | K or L | 2 | | F and G | J | 1 | | | | ---- | | | | 47 | +------------+---------------------------+----------+

This, of course, means that if you place sheep in the pens marked A andB, then there are seven different pens in which you may place the thirdsheep, giving seven different solutions. It was understood thatreversals and reflections do not count as different. 

If one pen at least is to be _not_ in line with a sheep, there would bethirty solutions to that problem. If we counted all the reversals andreflections of these 47 and 30 cases respectively as different, theirtotal would be 560, which is the number of different ways in which thesheep may be placed in three pens without any conditions. I will remarkthat there are three ways in which two sheep may be placed so that everypen is occupied or in line, as in Diagrams 2, 3, and 4, but in everycase each sheep is in line with its companion. There are only two waysin which three sheep may be so placed that every pen shall be occupiedor in line, but no sheep in line with another. These I show in Diagrams5 and 6. Finally, there is only one way in which three sheep may beplaced so that at least one pen shall not be in line with a sheep andyet no sheep in line with another. Place the sheep in C, E, L. This ispractically all there is to be said on this pleasant pastoral subject. 

[Illustration]

311. --THE FIVE DOGS PUZZLE. 

The diagrams show four fundamentally different solutions. In the case ofA we can reverse the order, so that the single dog is in the bottom rowand the other four shifted up two squares. Also we may use the nextcolumn to the right and both of the two central horizontal rows. Thus Agives 8 solutions. Then B may be reversed and placed in either diagonal, giving 4 solutions. Similarly C will give 4 solutions. The line in Dbeing symmetrical, its reversal will not be different, but it may bedisposed in 4 different directions. We thus have in all 20 differentsolutions. 

[Illustration]

312. --THE FIVE CRESCENTS OF BYZANTIUM. 

[Illustration]

If that ancient architect had arranged his five crescent tiles in themanner shown in the following diagram, every tile would have beenwatched over by, or in a line with, at least one crescent, and spacewould have been reserved for a perfectly square carpet equal in area toexactly half of the pavement. It is a very curious fact that, althoughthere are two or three solutions allowing a carpet to be laid downwithin the conditions so as to cover an area of nearly twenty-nine ofthe tiles, this is the only possible solution giving exactly half thearea of the pavement, which is the largest space obtainable. 

313. --QUEENS AND BISHOP PUZZLE. 

[Illustration: FIG. 1. ]

[Illustration: FIG. 2. ]

The bishop is on the square originally occupied by the rook, and thefour queens are so placed that every square is either occupied orattacked by a piece. (Fig. 1. )

I pointed out in 1899 that if four queens are placed as shown in thediagram (Fig. 2), then the fifth queen may be placed on any one of thetwelve squares marked a, b, c, d, and e; or a rook on the two squares, c; or a bishop on the eight squares, a, b, and e; or a pawn on thesquare b; or a king on the four squares, b, c, and e. The only knownarrangement for four queens and a knight is that given by Mr. J. Wallisin _The Strand Magazine_ for August 1908, here reproduced. (Fig. 3. )

[Illustration: FIG. 3. ]

I have recorded a large number of solutions with four queens and a rook, or bishop, but the only arrangement, I believe, with three queens andtwo rooks in which all the pieces are guarded is that of which I give anillustration (Fig. 4), first published by Dr. C. Planck. But I havesince found the accompanying solution with three queens, a rook, and abishop, though the pieces do not protect one another. (Fig. 5. )

[Illustration: FIG. 4. ]

[Illustration: FIG. 5. ]

314. --THE SOUTHERN CROSS. 

My readers have been so familiarized with the fact that it requires atleast five planets to attack every one of a square arrangement ofsixty-four stars that many of them have, perhaps, got to believe that alarger square arrangement of stars must need an increase of planets. Itwas to correct this possible error of reasoning, and so warn readersagainst another of those numerous little pitfalls in the world ofpuzzledom, that I devised this new stellar problem. Let me then state atonce that, in the case of a square arrangement of eighty one stars, there are several ways of placing five planets so that every star shallbe in line with at least one planet vertically, horizontally, ordiagonally. Here is the solution to the "Southern Cross": --

It will be remembered that I said that the five planets in their newpositions "will, of course, obscure five other stars in place of thoseat present covered. " This was to exclude an easier solution in whichonly four planets need be moved. 

315. --THE HAT-PEG PUZZLE. 

The moves will be made quite clear by a reference to the diagrams, whichshow the position on the board after each of the four moves. The dartsindicate the successive removals that have been made. It will be seenthat at every stage all the squares are either attacked or occupied, andthat after the fourth move no queen attacks any other. In the case ofthe last move the queen in the top row might also have been moved onesquare farther to the left. This is, I believe, the only solution to thepuzzle. 

[Illustration: 1]

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

316. --THE AMAZONS. 

It will be seen that only three queens have been removed from theirpositions on the edge of the board, and that, as a consequence, elevensquares (indicated by the black dots) are left unattacked by any queen. I will hazard the statement that eight queens cannot be placed on thechessboard so as to leave more than eleven squares unattacked. It istrue that we have no rigid proof of this yet, but I have entirelyconvinced myself of the truth of the statement. There are at least fivedifferent ways of arranging the queens so as to leave eleven squaresunattacked. 

[Illustration]

317. --A PUZZLE WITH PAWNS. 

Sixteen pawns may be placed so that no three shall be in a straight linein any possible direction, as in the diagram. We regard, as theconditions required, the pawns as mere points on a plane. 

[Illustration]

318. --LION-HUNTING. 

There are 6, 480 ways of placing the man and the lion, if there are norestrictions whatever except that they must be on different spots. Thisis obvious, because the man may be placed on any one of the 81 spots, and in every case there are 80 spots remaining for the lion; therefore81 × 80 = 6, 480. Now, if we deduct the number of ways in which the lionand the man may be placed on the same path, the result must be thenumber of ways in which they will not be on the same path. The number ofways in which they may be in line is found without much difficulty to be816. Consequently, 6, 480 - 816 = 5, 664, the required answer. 

The general solution is this: 1/3n(n - 1)(3n² - n + 2). This is, ofcourse, equivalent to saying that if we call the number of squares onthe side of a "chessboard" n, then the formula shows the number ofways in which two bishops may be placed without attacking one another. Only in this case we must divide by two, because the two bishops have nodistinct individuality, and cannot produce a different solution by mereexchange of places. 

319. --THE KNIGHT-GUARDS. 

[Illustration: DIAGRAM 1. ]

[Illustration: DIAGRAM 2. ]

The smallest possible number of knights with which this puzzle can besolved is fourteen. 

It has sometimes been assumed that there are a great many differentsolutions. As a matter of fact, there are only three arrangements--notcounting mere reversals and reflections as different. Curiously enough, nobody seems ever to have hit on the following simple proof, or to havethought of dealing with the black and the white squares separately. 

[Illustration: DIAGRAM 3. ]

[Illustration: DIAGRAM 4. ]

[Illustration: DIAGRAM 5. ]

Seven knights can be placed on the board on white squares so as toattack every black square in two ways only. These are shown in Diagrams1 and 2. Note that three knights occupy the same position in botharrangements. It is therefore clear that if we turn the board so that ablack square shall be in the top left-hand corner instead of a white, and place the knights in exactly the same positions, we shall have twosimilar ways of attacking all the white squares. I will assume thereader has made the two last described diagrams on transparent paper, and marked them _1a_ and _2a_. Now, by placing the transparent Diagram_1a_ over 1 you will be able to obtain the solution in Diagram 3, byplacing _2a_ over 2 you will get Diagram 4, and by placing _2a_ over 1you will get Diagram 5. You may now try all possible combinations ofthose two pairs of diagrams, but you will only get the threearrangements I have given, or their reversals and reflections. Thereforethese three solutions are all that exist. 

320. --THE ROOK'S TOUR. 

[Illustration]

The only possible minimum solutions are shown in the two diagrams, whereit will be seen that only sixteen moves are required to perform thefeat. Most people find it difficult to reduce the number of moves belowseventeen*. 

[Illustration: THE ROOK'S TOUR. ]

321. --THE ROOK'S JOURNEY. 

[Illustration]

I show the route in the diagram. It will be seen that the tenth movelands us at the square marked "10, " and that the last move, thetwenty-first, brings us to a halt on square "21. "

322. --THE LANGUISHING MAIDEN. 

The dotted line shows the route in twenty-two straight paths by whichthe knight may rescue the maiden. It is necessary, after entering thefirst cell, immediately to return before entering another. Otherwise asolution would not be possible. (See "The Grand Tour, " p. 200. )

323. --A DUNGEON PUZZLE. 

If the prisoner takes the route shown in the diagram--where forclearness the doorways are omitted--he will succeed in visiting everycell once, and only once, in as many as fifty-seven straight lines. Norook's path over the chessboard can exceed this number of moves. 

[Illustration: THE LANGUISHING MAIDEN]

[Illustration: A DUNGEON PUZZLE. ]

324. --THE LION AND THE MAN. 

First of all, the fewest possible straight lines in each case aretwenty-two, and in order that no cell may be visited twice it isabsolutely necessary that each should pass into one cell and thenimmediately "visit" the one from which he started, afterwards proceedingby way of the second available cell. In the following diagram the man'sroute is indicated by the unbroken lines, and the lion's by the dottedlines. It will be found, if the two routes are followed cell by cellwith two pencil points, that the lion and the man never meet. But therewas one little point that ought not to be overlooked--"they occasionallygot glimpses of one another. " Now, if we take one route for the man andmerely reverse it for the lion, we invariably find that, going at thesame speed, they never get a glimpse of one another. But in our diagramit will be found that the man and the lion are in the cells marked A atthe same moment, and may see one another through the open doorways;while the same happens when they are in the two cells marked B, theupper letters indicating the man and the lower the lion. In the firstcase the lion goes straight for the man, while the man appears toattempt to get in the rear of the lion; in the second case it lookssuspiciously like running away from one another!

[Illustration]

325. --AN EPISCOPAL VISITATION. 

[Illustration]

In the diagram I show how the bishop may be made to visit every one ofhis white parishes in seventeen moves. It is obvious that we must startfrom one corner square and end at the one that is diagonally opposite toit. The puzzle cannot be solved in fewer than seventeen moves. 

326. --A NEW COUNTER PUZZLE. 

Play as follows: 2--3, 9--4, 10--7, 3--8, 4--2, 7--5, 8--6, 5--10, 6--9, 2--5, 1--6, 6--4, 5--3, 10--8, 4--7, 3--2, 8--1, 7--10. The whitecounters have now changed places with the red ones, in eighteen moves, without breaking the conditions. 

327. --A NEW BISHOP'S PUZZLE. 

[Illustration: A]

[Illustration: B]

Play as follows, using the notation indicated by the numbered squares inDiagram A:--

 White. | Black. | White. | Black. 1. 18--15 | 1. 3--6 | 10. 20--10 | 10. 1--11 2. 17--8 | 2. 4--13 | 11. 3--9 | 11. 18--12 3. 19--14 | 3. 2--7 | 12. 10--13 | 12. 11--8 4. 15--5 | 4. 6--16 | 13. 19--16 | 13. 2--5 5. 8--3 | 5. 13-18 | 14. 16--1 | 14. 5--20 6. 14--9 | 6. 7--12 | 15. 9--6 | 15. 12--15 7. 5--10 | 7. 16-11 | 16. 13-7 | 16. 8--14 8. 9--19 | 8. 12--2 | 17. 6--3 | 17. 15-18 9. 10--4 | 9. 11-17 | 18. 7--2 | 18. 14--19

Diagram B shows the position after the ninth move. Bishops at 1 and 20have not yet moved, but 2 and 19 have sallied forth and returned. In theend, 1 and 19, 2 and 20, 3 and 17, and 4 and 18 will have exchangedplaces. Note the position after the thirteenth move. 

328. --THE QUEEN'S TOUR. 

[Illustration]

The annexed diagram shows a second way of performing the Queen's Tour. If you break the line at the point J and erase the shorter portion ofthat line, you will have the required path solution for any J square. Ifyou break the line at I, you will have a non-re-entrant solutionstarting from any I square. And if you break the line at G, you willhave a solution for any G square. The Queen's Tour previously given maybe similarly broken at three different places, but I seized theopportunity of exhibiting a second tour. 

329. --THE STAR PUZZLE. 

The illustration explains itself. The stars are all struck out infourteen straight strokes, starting and ending at a white star. 

[Illustration]

330. --THE YACHT RACE. 

The diagram explains itself. The numbers will show the direction of thelines in their proper order, and it will be seen that the seventh courseends at the flag-buoy, as stipulated. 

[Illustration]

331. --THE SCIENTIFIC SKATER. 

In this case we go beyond the boundary of the square. Apart from that, the moves are all queen moves. There are three or four ways in which itcan be done. 

Here is one way of performing the feat:--

[Illustration]

It will be seen that the skater strikes out all the stars in onecontinuous journey of fourteen straight lines, returning to the pointfrom which he started. To follow the skater's course in the diagram itis necessary always to go as far as we can in a straight line beforeturning. 

332. --THE FORTY-NINE STARS. 

The illustration shows how all the stars may be struck out in twelvestraight strokes, beginning and ending at a black star. 

[Illustration]

333. --THE QUEEN'S JOURNEY. 

The correct solution to this puzzle is shown in the diagram by the darkline. The five moves indicated will take the queen the greatest distancethat it is possible for her to go in five moves, within the conditions. The dotted line shows the route that most people suggest, but it is notquite so long as the other. Let us assume that the distance from thecentre of any square to the centre of the next in the same horizontal orvertical line is 2 inches, and that the queen travels from the centre ofher original square to the centre of the one at which she rests. Thenthe first route will be found to exceed 67. 9 inches, while the dottedroute is less than 67. 8 inches. The difference is small, but it issufficient to settle the point as to the longer route. All other routesare shorter still than these two. 

[Illustration]

334. --ST. GEORGE AND THE DRAGON. 

We select for the solution of this puzzle one of the prettiest designsthat can be formed by representing the moves of the knight by lines fromsquare to square. The chequering of the squares is omitted to givegreater clearness. St. George thus slays the Dragon in strict accordancewith the conditions and in the elegant manner we should expect of him. 

[Illustration: St. George and the Dragon. ]

335. --FARMER LAWRENCE'S CORNFIELDS. 

There are numerous solutions to this little agricultural problem. Theversion I give in the next column is rather curious on account of thelong parallel straight lines formed by some of the moves. 

[Illustration: Farmer Lawrence's Cornfields. ]

336. --THE GREYHOUND PUZZLE. 

There are several interesting points involved in this question. In thefirst place, if we had made no stipulation as to the positions of thetwo ends of the string, it is quite impossible to form any such stringunless we begin and end in the top and bottom row of kennels. We maybegin in the top row and end in the bottom (or, of course, the reverse), or we may begin in one of these rows and end in the same. But we cannever begin or end in one of the two central rows. Our places ofstarting and ending, however, were fixed for us. Yet the first half ofour route must be confined entirely to those squares that aredistinguished in the following diagram by circles, and the second halfwill therefore be confined to the squares that are not circled. Thesquares reserved for the two half-strings will be seen to be symmetricaland similar. 

The next point is that the first half-string must end in one of thecentral rows, and the second half-string must begin in one of theserows. This is now obvious, because they have to link together to formthe complete string, and every square on an outside row is connected bya knight's move with similar squares only--that is, circled ornon-circled as the case may be. The half-strings can, therefore, only belinked in the two central rows. 

[Illustration]

Now, there are just eight different first half-strings, and consequentlyalso eight second half-strings. We shall see that these combine to formtwelve complete strings, which is the total number that exist and thecorrect solution of our puzzle. I do not propose to give all the routesat length, but I will so far indicate them that if the reader hasdropped any he will be able to discover which they are and work them outfor himself without any difficulty. The following numbers apply to thosein the above diagram. 

The eight first half-strings are: 1 to 6 (2 routes); 1 to 8 (1 route);1 to 10 (3 routes); 1 to 12 (1 route); and 1 to 14 (1 route). The eightsecond half-strings are: 7 to 20 (1 route); 9 to 20 (1 route); 11 to 20(3 routes); 13 to 20 (1 route); and 15 to 20 (2 routes). Every differentway in which you can link one half-string to another gives a differentsolution. These linkings will be found to be as follows: 6 to 13 (2cases); 10 to 13 (3 cases); 8 to 11 (3 cases); 8 to 15 (2 cases); 12 to9 (1 case); and 14 to 7 (1 case). There are, therefore, twelve differentlinkings and twelve different answers to the puzzle. The route given inthe illustration with the greyhound will be found to consist of one ofthe three half-strings 1 to 10, linked to the half-string 13 to 20. Itshould be noted that ten of the solutions are produced by fivedistinctive routes and their reversals--that is, if you indicate thesefive routes by lines and then turn the diagrams upside down you will getthe five other routes. The remaining two solutions are symmetrical(these are the cases where 12 to 9 and 14 to 7 are the links), andconsequently they do not produce new solutions by reversal. 

337. --THE FOUR KANGAROOS. 

[Illustration]

A pretty symmetrical solution to this puzzle is shown in the diagram. Each of the four kangaroos makes his little excursion and returns to hiscorner, without ever entering a square that has been visited by anotherkangaroo and without crossing the central line. It will at once occur tothe reader, as a possible improvement of the puzzle, to divide the boardby a central vertical line and make the condition that this also shallnot be crossed. This would mean that each kangaroo had to confinehimself to a square 4 by 4, but it would be quite impossible, as I shallexplain in the next two puzzles. 

338. --THE BOARD IN COMPARTMENTS. 

[Illustration]

In attempting to solve this problem it is first necessary to take thetwo distinctive compartments of twenty and twelve squares respectivelyand analyse them with a view to determining where the necessary pointsof entry and exit lie. In the case of the larger compartment it will befound that to complete a tour of it we must begin and end on two of theoutside squares on the long sides. But though you may start at any oneof these ten squares, you are restricted as to those at which you canend, or (which is the same thing) you may end at whichever of these youlike, provided you begin your tour at certain particular squares. In thecase of the smaller compartment you are compelled to begin and end atone of the six squares lying at the two narrow ends of the compartments, but similar restrictions apply as in the other instance. A very littlethought will show that in the case of the two small compartments youmust begin and finish at the ends that lie together, and it thenfollows that the tours in the larger compartments must also start andend on the contiguous sides. 

In the diagram given of one of the possible solutions it will be seenthat there are eight places at which we may start this particular tour;but there is only one route in each case, because we must complete thecompartment in which we find ourself before passing into another. In anysolution we shall find that the squares distinguished by stars must beentering or exit points, but the law of reversals leaves us the optionof making the other connections either at the diamonds or at thecircles. In the solution worked out the diamonds are used, but othervariations occur in which the circle squares are employed instead. Ithink these remarks explain all the essential points in the puzzle, which is distinctly instructive and interesting. 

339. --THE FOUR KNIGHTS' TOURS. 

[Illustration]

It will be seen in the illustration how a chessboard may be divided intofour parts, each of the same size and shape, so that a completere-entrant knight's tour may be made on each portion. There is only onepossible route for each knight and its reversal. 

340. --THE CUBIC KNIGHT'S TOUR. 

[Illustration]

If the reader should cut out the above diagram, fold it in the form of acube, and stick it together by the strips left for that purpose at theedges, he would have an interesting little curiosity. Or he can make oneon a larger scale for himself. It will be found that if we imagine thecube to have a complete chessboard on each of its sides, we may startwith the knight on any one of the 384 squares, and make a complete tourof the cube, always returning to the starting-point. The method ofpassing from one side of the cube to another is easily understood, but, of course, the difficulty consisted in finding the proper points ofentry and exit on each board, the order in which the different boardsshould be taken, and in getting arrangements that would comply with therequired conditions. 

341. --THE FOUR FROGS. 

The fewest possible moves, counting every move separately, are sixteen. But the puzzle may be solved in seven plays, as follows, if any numberof successive moves by one frog count as a single play. All the movescontained within a bracket are a single play; the numbers refer to thetoadstools: (1--5), (3--7, 7--1), (8--4, 4--3, 3--7), (6--2, 2--8, 8--4, 4--3), (5--6, 6--2, 2--8), (1--5, 5--6), (7--1). 

This is the familiar old puzzle by Guarini, propounded in 1512, and Igive it here in order to explain my "buttons and string" method ofsolving this class of moving-counter problem. 

Diagram A shows the old way of presenting Guarini's puzzle, the pointbeing to make the white knights change places with the black ones. In"The Four Frogs" presentation of the idea the possible directions of themoves are indicated by lines, to obviate the necessity of the reader'sunderstanding the nature of the knight's move in chess. But it will atonce be seen that the two problems are identical. The central squarecan, of course, be ignored, since no knight can ever enter it. Now, regard the toadstools as buttons and the connecting lines as strings, asin Diagram B. Then by disentangling these strings we can clearly presentthe diagram in the form shown in Diagram C, where the relationshipbetween the buttons is precisely the same as in B. Any solution on Cwill be applicable to B, and to A. Place your white knights on 1 and 3and your black knights on 6 and 8 in the C diagram, and the simplicityof the solution will be very evident. You have simply to move theknights round the circle in one direction or the other. Play over themoves given above, and you will find that every little difficulty hasdisappeared. 

[Illustrations: A B C D E]

In Diagram D I give another familiar puzzle that first appeared in abook published in Brussels in 1789, _Les Petites Aventures de JeromeSharp_. Place seven counters on seven of the eight points in thefollowing manner. You must always touch a point that is vacant with acounter, and then move it along a straight line leading from that pointto the next vacant point (in either direction), where you deposit thecounter. You proceed in the same way until all the counters are placed. Remember you always touch a vacant place and slide the counter from itto the next place, which must be also vacant. Now, by the "buttons andstring" method of simplification we can transform the diagram into E. Then the solution becomes obvious. "Always move _to_ the point that youlast moved _from_. " This is not, of course, the only way of placing thecounters, but it is the simplest solution to carry in the mind. 

There are several puzzles in this book that the reader will find lendthemselves readily to this method. 

342. --THE MANDARIN'S PUZZLE. 

The rather perplexing point that the solver has to decide for himself inattacking this puzzle is whether the shaded numbers (those that areshown in their right places) are mere dummies or not. Ninety-ninepersons out of a hundred might form the opinion that there can be noadvantage in moving any of them, but if so they would be wrong. 

The shortest solution without moving any shaded number is in thirty-twomoves. But the puzzle can be solved in thirty moves. The trick lies inmoving the 6, or the 15, on the second move and replacing it on thenineteenth move. Here is the solution: 2, 6, 13, 4, 1, 21, 4, 1, 10, 2, 21, 10, 2, 5, 22, 16, 1, 13, 6, 19, 11, 2, 5, 22, 16, 5, 13, 4, 10, 21. Thirty moves. 

343. --EXERCISE FOR PRISONERS. 

There are eighty different arrangements of the numbers in the form of aperfect knight's path, but only forty of these can be reached withouttwo men ever being in a cell at the same time. Two is the greatestnumber of men that can be given a complete rest, and though the knight'spath can be arranged so as to leave either 7 and 13, 8 and 13, 5 and 7, or 5 and 13 in their original positions, the following fourarrangements, in which 7 and 13 are unmoved, are the only ones that canbe reached under the moving conditions. It therefore resolves itselfinto finding the fewest possible moves that will lead up to one of thesepositions. This is certainly no easy matter, and no rigid rules can belaid down for arriving at the correct answer. It is largely a matter forindividual judgment, patient experiment, and a sharp eye for revolutionsand position. 

 A +--+--+--+--+ | 6| 1|10|15| +--+--+--+--+ | 9|12| 7| 4| +--+--+--+--+ | 2| 5|14|11| +--+--+--+--+ |13| 8| 3|**| +--+--+--+--+

 B +--+--+--+--+ | 6| 1|10|15| +--+--+--+--+ |11|14| 7| 4| +--+--+--+--+ | 2| 5|12| 9| +--+--+--+--+ |13| 8| 3|**| +--+--+--+--+

 C +--+--+--+--+ | 6| 9| 4|15| +--+--+--+--+ | 1|12| 7|10| +--+--+--+--+ | 8| 5|14| 3| +--+--+--+--+ |13| 2|11|**| +--+--+--+--+

 D +--+--+--+--+ | 6|11| 4|15| +--+--+--+--+ | 1|14| 7|10| +--+--+--+--+ | 8| 5|12| 3| +--+--+--+--+ |13| 2| 9|**| +--+--+--+--+

[Illustration: A, B, C, D]

As a matter of fact, the position C can be reached in as few assixty-six moves in the following manner: 12, 11, 15, 12, 11, 8, 4, 3, 2, 6, 5, 1, 6, 5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8, 4, 12, 11, 3, 2, 5, 10, 15, 6, 1, 8, 4, 9, 8, 1, 6, 4, 9, 12, 2, 5, 10, 15, 4, 9, 12, 2, 5, 3, 11, 14, 2, 5, 14, 11 = 66 moves. Though this isthe shortest that I know of, and I do not think it can be beaten, Icannot state positively that there is not a shorter way yet to bediscovered. The most tempting arrangement is certainly A; but thingsare not what they seem, and C is really the easiest to reach. 

If the bottom left-hand corner cell might be left vacant, the followingis a solution in forty-five moves by Mr. R. Elrick: 15, 11, 10, 9, 13, 14, 11, 10, 7, 8, 4, 3, 8, 6, 9, 7, 12, 4, 6, 9, 5, 13, 7, 5, 13, 1, 2, 13, 5, 7, 1, 2, 13, 8, 3, 6, 9, 12, 7, 11, 14, 1, 11, 14, 1. But everyman has moved. 

344. --THE KENNEL PUZZLE. 

The first point is to make a choice of the most promising knight'sstring and then consider the question of reaching the arrangement in thefewest moves. I am strongly of opinion that the best string is the onerepresented in the following diagram, in which it will be seen that eachsuccessive number is a knight's move from the preceding one, and thatfive of the dogs (1, 5, 10, 15, and 20) never leave their originalkennels. 

 +-----+------+------+------+------+ |1 |2 |3 |4 |5 | | | | | | | | 1 | 18 | 9 | 14 | 5 | | | | | | | +-----+------+------+------+------+ |6 |7 |8 |9 |10 | | | | | | | | 8 | 13 | 4 | 19 | 10 | | | | | | | +-----+------+------+------+------+ |11 |12 |13 |14 |15 | | | | | | | | 17 | 2 | 11 | 6 | 15 | | | | | | | +-----+------+------+------+------+ |16 |17 |18 |19 |20 | | | | | | | | 12 | 7 | 16 | 3 | 20 | | | | | | | +-----+------+------+------+------+ |21 |22 |23 |24 |25 | | | | | | | | | | | | | | | | | | | +-----+------+------+------+------+

[Illustration]

This position may be arrived at in as few as forty-six moves, asfollows: 16--21, 16--22, 16--23, 17--16, 12--17, 12--22, 12--21, 7--12, 7--17, 7--22, 11--12, 11--17, 2--7, 2--12, 6--11, 8--7, 8--6, 13--8, 18--13, 11--18, 2--17, 18--12, 18--7, 18--2, 13--7, 3--8, 3--13, 4--3, 4--8, 9--4, 9--3, 14--9, 14--4, 19--14, 19--9, 3--14, 3--19, 6--12, 6--13, 6--14, 17--11, 12--16, 2--12, 7--17, 11--13, 16--18 = 46 moves. Iam, of course, not able to say positively that a solution cannot bediscovered in fewer moves, but I believe it will be found a very hardtask to reduce the number. 

345. --THE TWO PAWNS. 

Call one pawn A and the other B. Now, owing to that optional first move, either pawn may make either 5 or 6 moves in reaching the eighth square. There are, therefore, four cases to be considered: (1) A 6 moves and B 6moves; (2) A 6 moves and B 5 moves; (3) A 5 moves and B 6 moves; (4) A 5moves and B 5 moves. In case (1) there are 12 moves, and we may selectany 6 of these for A. Therefore 7×8×9×10×11×12 divided by 1×2×3×4×5×6gives us the number of variations for this case--that is, 924. Similarlyfor case (2), 6 selections out of 11 will be 462; in case (3), 5selections out of 11 will also be 462; and in case (4), 5 selections outof 10 will be 252. Add these four numbers together and we get 2, 100, which is the correct number of different ways in which the pawns mayadvance under the conditions. (See No. 270, on p. 204. )

346. --SETTING THE BOARD. 

The White pawns may be arranged in 40, 320 ways, the White rooks in 2ways, the bishops in 2 ways, and the knights in 2 ways. Multiply thesenumbers together, and we find that the White pieces may be placed in322, 560 different ways. The Black pieces may, of course, be placed inthe same number of ways. Therefore the men may be set up in 322, 560 ×322, 560 = 104, 044, 953, 600 ways. But the point that nearly everybodyoverlooks is that the board may be placed in two different ways forevery arrangement. Therefore the answer is doubled, and is208, 089, 907, 200 different ways. 

347. --COUNTING THE RECTANGLES. 

There are 1, 296 different rectangles in all, 204 of which are squares, counting the square board itself as one, and 1, 092 rectangles that arenot squares. The general formula is that a board of n² squarescontains ((n² + n)²)/4 rectangles, of which (2n³ + 3n² + n)/6 aresquares and (3n^4 + 2n³ - 3n² - 2n)/12 are rectangles that are notsquares. It is curious and interesting that the total number ofrectangles is always the square of the triangular number whose side isn. 

348. --THE ROOKERY. 

The answer involves the little point that in the final position thenumbered rooks must be in numerical order in the direction contrary tothat in which they appear in the original diagram, otherwise it cannotbe solved. Play the rooks in the following order of their numbers. Asthere is never more than one square to which a rook can move (except onthe final move), the notation is obvious--5, 6, 7, 5, 6, 4, 3, 6, 4, 7, 5, 4, 7, 3, 6, 7, 3, 5, 4, 3, 1, 8, 3, 4, 5, 6, 7, 1, 8, 2, 1, and rooktakes bishop, checkmate. These are the fewest possiblemoves--thirty-two. The Black king's moves are all forced, and need notbe given. 

349. --STALEMATE. 

Working independently, the same position was arrived at by Messrs. S. Loyd, E. N. Frankenstein, W. H. Thompson, and myself. So the following maybe accepted as the best solution possible to this curious problem :--

 White. Black. 1. P--Q4 1. P--K4 2. Q--Q3 2. Q--R5 3. Q--KKt3 3. B--Kt5 ch 4. Kt--Q2 4. P--QR4 5. P--R4 5. P--Q3 6. P--R3 6. B--K3 7. R--R3 7. P--KB4 8. Q--R2 8. P--B4 9. R--KKt3 9. B--Kt6 10. P--QB4 10. P--B5 11. P--B3 11. P--K5 12. P--Q5 12. P--K6

And White is stalemated. 

We give a diagram of the curious position arrived at. It will be seenthat not one of White's pieces may be moved. 

[Illustration]

 +-+-+-+-+-+-+-+-+ |r|n| | |k| |n|r| +-+-+-+-+-+-+-+-+ | |p| | | | |p|p| +-+-+-+-+-+-+-+-+ | | | |p| | | | | +-+-+-+-+-+-+-+-+ |p| |p|P| | | | | +-+-+-+-+-+-+-+-+ |P|b|P| | |p| |q| +-+-+-+-+-+-+-+-+ | |b| | |p|P|R|P| +-+-+-+-+-+-+-+-+ | |P| |N|P| |P|Q| +-+-+-+-+-+-+-+-+ | | |B| |K|B|N|R| +-+-+-+-+-+-+-+-+

350. --THE FORSAKEN KING. 

Play as follows:--

 White. Black. 1. P to K 4th 1. Any move 2. Q to Kt 4th 2. Any move except on KB file (a) 3. Q to Kt 7th 3. K moves to royal row 4. B to Kt 5th 4. Any move 5. Mate in two moves If 3. K other than to royal row 4. P to Q 4th 4. Any move 5. Mate in two moves (a) If 2. Any move on KB file 3. Q to Q 7th 3. K moves to royal row 4. P to Q Kt 3rd 4. Any move 5. Mate in two moves If 3. K other than to royal row 4. P to Q 4th 4. Any move 5. Mate in two moves

Of course, by "royal row" is meant the row on which the king originallystands at the beginning of a game. Though, if Black plays badly, he may, in certain positions, be mated in fewer moves, the above provides forevery variation he can possibly bring about. 

351. --THE CRUSADER. 

 White. Black. 1. Kt to QB 3rd 1. P to Q 4th 2. Kt takes QP 2. Kt to QB 3rd 3. Kt takes KP 3. P to KKt 4th 4. Kt takes B 4. Kt to KB 3rd 5. Kt takes P 5. Kt to K 5th 6. Kt takes Kt 6. Kt to B 6th 7. Kt takes Q 7. R to KKt sq 8. Kt takes BP 8. R to KKt 3rd 9. Kt takes P 9. R to K 3rd 10. Kt takes P 10. Kt to Kt 8th 11. Kt takes B 11. R to R 6th 12. Kt takes R 12. P to Kt 4th 13. Kt takes P (ch) 13. K to B 2nd 14. Kt takes P 14. K to Kt 3rd 15. Kt takes R 15. K to R 4th 16. Kt takes Kt 16. K to R 5th White now mates in three moves. 17. P to Q 4th 17. K to R 4th 18. Q to Q 3rd 18. K moves 19. Q to KR 3rd (mate) If 17. K to Kt 5th 18. P to K 4th (dis. Ch) 18. K moves 19. P to KKt 3rd (mate)

The position after the sixteenth move, with the mate in three moves, wasfirst given by S. Loyd in _Chess Nuts_. 

352. --IMMOVABLE PAWNS. 

 1. Kt to KB 3 2. Kt to KR 4 3. Kt to Kt 6 4. Kt takes R 5. Kt to Kt 6 6. Kt takes B 7. K takes Kt 8. Kt to QB 3 9. Kt to R 4 10. Kt to Kt 6 11. Kt takes R 12. Kt to Kt 6 13. Kt takes B 14. Kt to Q 6 15. Q to K sq 16. Kt takes Q 17. K takes Kt, and the position is reached. 

Black plays precisely the same moves as White, and therefore we give oneset of moves only. The above seventeen moves are the fewest possible. 

353. --THIRTY-SIX MATES. 

Place the remaining eight White pieces thus: K at KB 4th, Q at QKt 6th, R at Q 6th, R at KKt 7th, B at Q 5th, B at KR 8th, Kt at QR 5th, and Ktat QB 5th. The following mates can then be given:--

 By discovery from Q 8 By discovery from R at Q 6th 13 By discovery from B at R 8th 11 Given by Kt at R 5th 2 Given by pawns 2 -- Total 36

Is it possible to construct a position in which more than thirty-sixdifferent mates on the move can be given? So far as I know, nobody hasyet beaten my arrangement. 

354. --AN AMAZING DILEMMA. 

Mr Black left his king on his queen's knight's 7th, and no matter whatpiece White chooses for his pawn, Black cannot be checkmated. As wesaid, the Black king takes no notice of checks and never moves. Whitemay queen his pawn, capture the Black rook, and bring his three piecesup to the attack, but mate is quite impossible. The Black king cannot beleft on any other square without a checkmate being possible. 

The late Sam Loyd first pointed out the peculiarity on which this puzzleis based. 

355. --CHECKMATE!

Remove the White pawn from B 6th to K 4th and place a Black pawn onBlack's KB 2nd. Now, White plays P to K 5th, check, and Black must playP to B 4th. Then White plays P takes P _en passant_, checkmate. This wastherefore White's last move, and leaves the position given. It is theonly possible solution. 

356. --QUEER CHESS. 

 +-+-+-+-+-+-+-+-+ | | | | | | | | | +-+-+-+-+-+-+-+-+ | | |R|k|R|N| | | +-+-+-+-+-+-+-+-+ | | | | | | | | | +-+-+-+-+-+-+-+-+

If you place the pieces as follows (where only a portion of the board isgiven, to save space), the Black king is in check, with no possible moveopen to him. The reader will now see why I avoided the term "checkmate, "apart from the fact that there is no White king. The position isimpossible in the game of chess, because Black could not be given checkby both rooks at the same time, nor could he have moved into check onhis last move. 

I believe the position was first published by the late S. Loyd. 

357. --ANCIENT CHINESE PUZZLE. 

Play as follows:--

 1. R--Q 6 2. K--R 7 3. R (R 6)--B 6 (mate). 

Black's moves are forced, so need not be given. 

358. --THE SIX PAWNS. 

The general formula for six pawns on all squares greater than 2² isthis: Six times the square of the number of combinations of n thingstaken three at a time, where n represents the number of squares on theside of the board. Of course, where n is even the unoccupied squaresin the rows and columns will be even, and where n is odd the number ofsquares will be odd. Here n is 8, so the answer is 18, 816 differentways. This is "The Dyer's Puzzle" (_Canterbury Puzzles_, No. 27) inanother form. I repeat it here in order to explain a method of solvingthat will be readily grasped by the novice. First of all, it is evidentthat if we put a pawn on any line, we must put a second one in that linein order that the remainder may be even in number. We cannot put four orsix in any row without making it impossible to get an even number in allthe columns interfered with. We have, therefore, to put two pawns ineach of three rows and in each of three columns. Now, there are just sixschemes or arrangements that fulfil these conditions, and these areshown in Diagrams A to F, inclusive, on next page. 

[Illustration]

I will just remark in passing that A and B are the only distinctivearrangements, because, if you give A a quarter-turn, you get F; and ifyou give B three quarter-turns in the direction that a clock handmoves, you will get successively C, D, and E. No matter how you mayplace your six pawns, if you have complied with the conditions of thepuzzle they will fall under one of these arrangements. Of course itwill be understood that mere expansions do not destroy the essentialcharacter of the arrangements. Thus G is only an expansion of form A. The solution therefore consists in finding the number of theseexpansions. Supposing we confine our operations to the first threerows, as in G, then with the pairs a and b placed in the first andsecond columns the pair c may be disposed in any one of the remainingsix columns, and so give six solutions. Now slide pair b into thethird column, and there are five possible positions for c. Slide binto the fourth column, and c may produce four new solutions. And soon, until (still leaving a in the first column) you have b in theseventh column, and there is only one place for c--in the eighthcolumn. Then you may put a in the second column, b in the third, and cin the fourth, and start sliding c and b as before for another seriesof solutions. 

We find thus that, by using form A alone and confining our operations tothe three top rows, we get as many answers as there are combinations of8 things taken 3 at a time. This is (8 × 7 × 6)/(1 × 2 × 3) = 56. And itwill at once strike the reader that if there are 56 different ways ofelecting the columns, there must be for each of these ways just 56 waysof selecting the rows, for we may simultaneously work that "sliding"process downwards to the very bottom in exactly the same way as we haveworked from left to right. Therefore the total number of ways in whichform A may be applied is 56 × 6 = 3, 136. But there are, as we have seen, six arrangements, and we have only dealt with one of these, A. We must, therefore, multiply this result by 6, which gives us 3, 136 × 6 = 18, 816, which is the total number of ways, as we have already stated. 

359. --COUNTER SOLITAIRE. 

Play as follows: 3--11, 9--10, 1--2, 7--15, 8--16, 8--7, 5--13, 1--4, 8--5, 6--14, 3--8, 6--3, 6--12, 1--6, 1--9, and all the counters willhave been removed, with the exception of No. 1, as required by theconditions. 

360. --CHESSBOARD SOLITAIRE. 

Play as follows: 7--15, 8--16, 8--7, 2--10, 1--9, 1--2, 5--13, 3--4, 6--3, 11--1, 14--8, 6--12, 5--6, 5--11, 31--23, 32--24, 32--31, 26--18, 25--17, 25--26, 22--32, 14--22, 29--21, 14--29, 27--28, 30--27, 25--14, 30--20, 25--30, 25--5. The two counters left on the board are 25 and19--both belonging to the same group, as stipulated--and 19 has neverbeen moved from its original place. 

I do not think any solution is possible in which only one counter isleft on the board. 

361. --THE MONSTROSITY. 

 White Black, 1. P to KB 4 P to QB 3 2. K to B 2 Q to R 4 3. K to K 3 K to Q sq 4. P to B 5 K to B 2 5. Q to K sq K to Kt 3 6. Q to Kt 3 Kt to QR 3 7. Q to Kt 8 P to KR 4 8. Kt to KB 3 R to R 3 9. Kt to K 5 R to Kt 3 10. Q takes B R to Kt 6, ch 11. P takes R K to Kt 4 12. R to R 4 P to B 3 13. R to Q 4 P takes Kt 14. P to QKt 4 P takes R, ch 15. K to B 4 P to R 5 16. Q to K 8 P to R 6 17. Kt to B 3, ch P takes Kt 18. B to R 3 P to R 7 19. R to Kt sq P to R 8 (Q) 20. R to Kt 2 P takes R 21. K to Kt 5 Q to KKt 8 22. Q to R 5 K to R 5 23. P to Kt 5 R to B sq 24. P to Kt 6 R to B 2 25. P takes R P to Kt 8 (B) 26. P to B 8 (R) Q to B 2 27. B to Q 6 Kt to Kt 5 28. K to Kt 6 K to R 6 29. R to R 8 K to Kt 7 30. P to R 4 Q (Kt 8) to Kt 3 31. P to R 5 K to B 8 32. P takes Q K to Q 8 33. P takes Q K to K 8 34. K to B 7 Kt to KR 3, ch 35. K to K 8 B to R 7 36. P to B 6 B to Kt sq 37. P to B 7 K takes B 38. P to B 8 (B) Kt to Q 4 39. B to Kt 8 Kt to B 3, ch 40. K to Q 8 Kt to K sq 41. P takes Kt (R) Kt to B 2, ch 42. K to B 7 Kt to Q sq 43. Q to B 7, ch K to Kt 8

And the position is reached. 

The order of the moves is immaterial, and this order may be greatlyvaried. But, although many attempts have been made, nobody has succeededin reducing the number of my moves. 

362. --THE WASSAIL BOWL. 

The division of the twelve pints of ale can be made in elevenmanipulations, as below. The six columns show at a glance the quantityof ale in the barrel, the five-pint jug, the three-pint jug, and thetramps X, Y, and Z respectively after each manipulation. 

 Barrel. 5-pint. 3-pint. X. Y. Z. 

 7 . . 5 . . 0 . . 0 . . 0 . . 0 7 . . 2 . . 3 . . 0 . . 0 . . 0 7 . . 0 . . 3 . . 2 . . 0 . . 0 7 . . 3 . . 0 . . 2 . . 0 . . 0 4 . . 3 . . 3 . . 2 . . 0 . . 0 0 . . 3 . . 3 . . 2 . . 4 . . 0 0 . . 5 . . 1 . . 2 . . 4 . . 0 0 . . 5 . . 0 . . 2 . . 4 . . 1 0 . . 2 . . 3 . . 2 . . 4 . . 1 0 . . 0 . . 3 . . 4 . . 4 . . 1 0 . . 0 . . 0 . . 4 . . 4 . . 4

And each man has received his four pints of ale. 

363. --THE DOCTOR'S QUERY. 

The mixture of spirits of wine and water is in the proportion of 40 to1, just as in the other bottle it was in the proportion of 1 to 40. 

364. --THE BARREL PUZZLE. 

[Illustration: Figs. 1, 2, and 3]

All that is necessary is to tilt the barrel as in Fig. 1, and if theedge of the surface of the water exactly touches the lip a at the sametime that it touches the edge of the bottom b, it will be just halffull. To be more exact, if the bottom is an inch or so from the ground, then we can allow for that, and the thickness of the bottom, at the top. If when the surface of the water reached the lip a it had risen to thepoint c in Fig. 2, then it would be more than half full. If, as inFig. 3, some portion of the bottom were visible and the level of thewater fell to the point d, then it would be less than half full. 

This method applies to all symmetrically constructed vessels. 

365. --NEW MEASURING PUZZLE. 

The following solution in eleven manipulations shows the contents ofevery vessel at the start and after every manipulation:--

 10-quart. 10-quart. 5-quart. 4-quart. 

 10 . . 10 . . 0 . . 0 5 . . 10 . . 5 . . 0 5 . . 10 . . 1 . . 4 9 . . 10 . . 1 . . 0 9 . . 6 . . 1 . . 4 9 . . 7 . . 0 . . 4 9 . . 7 . . 4 . . 0 9 . . 3 . . 4 . . 4 9 . . 3 . . 5 . . 3 9 . . 8 . . 0 . . 3 4 . . 8 . . 5 . . 3 4 . . 10 . . 3 . . 3

366. --THE HONEST DAIRYMAN. 

Whatever the respective quantities of milk and water, the relativeproportion sent to London would always be three parts of water to one ofmilk. But there are one or two points to be observed. There mustoriginally be more water than milk, or there will be no water in A todouble in the second transaction. And the water must not be more thanthree times the quantity of milk, or there will not be enough liquid inB to effect the second transaction. The third transaction has no effecton A, as the relative proportions in it must be the same as after thesecond transaction. It was introduced to prevent a quibble if thequantity of milk and water were originally the same; for though double"nothing" would be "nothing, " yet the third transaction in such a casecould not take place. 

367. --WINE AND WATER. 

The wine in small glass was one-sixth of the total liquid, and the winein large glass two-ninths of total. Add these together, and we find thatthe wine was seven-eighteenths of total fluid, and therefore the watereleven-eighteenths. 

368. --THE KEG OF WINE. 

The capacity of the jug must have been a little less than three gallons. To be more exact, it was 2. 93 gallons. 

369. --MIXING THE TEA. 

There are three ways of mixing the teas. Taking them in the order ofquality, 2s. 6d. , 2s. 3d. , 1s. 9p. , mix 16 lbs. , 1 lb. , 3 lbs. ; or 14lbs. , 4 lbs. , 2 lbs. ; or 12 lbs. , 7 lbs. , 1 lb. In every case thetwenty pounds mixture should be worth 2s. 4½d. Per pound; but the lastcase requires the smallest quantity of the best tea, therefore it isthe correct answer. 

370. --A PACKING PUZZLE. 

On the side of the box, 14 by 22+4/5, we can arrange 13 rows containingalternately 7 and 6 balls, or 85 in all. Above this we can place anotherlayer consisting of 12 rows of 7 and 6 alternately, or a total of 78. Inthe length of 24+9/10 inches 15 such layers may be packed, the alternatelayers containing 85 and 78 balls. Thus 8 times 85 added to 7 times 78gives us 1, 226 for the full contents of the box. 

371. --GOLD PACKING IN RUSSIA. 

The box should be 100 inches by 100 inches by 11 inches deep, internaldimensions. We can lay flat at the bottom a row of eight slabs, lengthways, end to end, which will just fill one side, and nine of theserows will dispose of seventy-two slabs (all on the bottom), with a spaceleft over on the bottom measuring 100 inches by 1 inch by 1 inch. Nowmake eleven depths of such seventy-two slabs, and we have packed 792, and have a space 100 inches by 1 inch by 11 inches deep. In this we mayexactly pack the remaining eight slabs on edge, end to end. 

372. --THE BARRELS OF HONEY. 

The only way in which the barrels could be equally divided among thethree brothers, so that each should receive his 3½ barrels of honeyand his 7 barrels, is as follows:--

 Full. Half-full. Empty. A 3 1 3 B 2 3 2 C 2 3 2

There is one other way in which the division could be made, were it notfor the objection that all the brothers made to taking more than fourbarrels of the same description. Except for this difficulty, they mighthave given B his quantity in exactly the same way as A above, and thenhave left C one full barrel, five half-full barrels, and one emptybarrel. It will thus be seen that in any case two brothers would have toreceive their allowance in the same way. 

373. --CROSSING THE STREAM. 

First, the two sons cross, and one returns Then the man crosses and theother son returns. Then both sons cross and one returns. Then the ladycrosses and the other son returns Then the two sons cross and one ofthem returns for the dog. Eleven crossings in all. 

It would appear that no general rule can be given for solving theseriver-crossing puzzles. A formula can be found for a particular case(say on No. 375 or 376) that would apply to any number of individualsunder the restricted conditions; but it is not of much use, for somelittle added stipulation will entirely upset it. As in the case of themeasuring puzzles, we generally have to rely on individual ingenuity. 

374. --CROSSING THE RIVER AXE. 

Here is the solution:--

 | {J 5) | G T8 3 5 | ( J } | G T8 3 5 | {G 3) | JT8 53 | ( G } | JT8 53 | {J T) | G 8 J 5 | (T 3} | G 8 J 5 | {G 8) | T 3 G 8 | (J 5} | T G 8 | {J T) | 53 JT8 | ( G } | 53 JT8 | {G 3) | 5 G T8 3 | ( J } | 5 G T8 3 | {J 5) |

G, J, and T stand for Giles, Jasper, and Timothy; and 8, 5, 3, for £800, £500, and £300 respectively. The two side columns represent the leftbank and the right bank, and the middle column the river. Thirteencrossings are necessary, and each line shows the position when the boatis in mid-stream during a crossing, the point of the bracket indicatingthe direction. 

It will be found that not only is no person left alone on the land or inthe boat with more than his share of the spoil, but that also no twopersons are left with more than their joint shares, though this lastpoint was not insisted upon in the conditions. 

375. --FIVE JEALOUS HUSBANDS. 

It is obvious that there must be an odd number of crossings, and that ifthe five husbands had not been jealous of one another the party mighthave all got over in nine crossings. But no wife was to be in thecompany of a man or men unless her husband was present. This entails twomore crossings, eleven in all. 

The following shows how it might have been done. The capital lettersstand for the husbands, and the small letters for their respectivewives. The position of affairs is shown at the start, and after eachcrossing between the left bank and the right, and the boat isrepresented by the asterisk. So you can see at a glance that a, b, and cwent over at the first crossing, that b and c returned at the secondcrossing, and so on. 

 ABCDE abcde *|. . | | | 1. ABCDE de |. . |* abc 2. ABCDE bcde *|. . | a 3. ABCDE e |. . |* abcd 4. ABCDE de *|. . | abc 5. DE de |, ,|* ABC abc 6. CDE cde *|. . | AB ab 7. Cde |. . |* ABCDE ab 8. Bcde *|. . | ABCDE a 9. E |. . |* ABCDE abcd 10. Bc e *|. . | ABCDE a d 11. |. . |* ABCDE abcde

There is a little subtlety concealed in the words "show the _quickest_way. "

Everybody correctly assumes that, as we are told nothing of the rowingcapabilities of the party, we must take it that they all row equallywell. But it is obvious that two such persons should row more quicklythan one. 

Therefore in the second and third crossings two of the ladies shouldtake back the boat to fetch d, not one of them only. This does notaffect the number of landings, so no time is lost on that account. Asimilar opportunity occurs in crossings 10 and 11, where the party againhad the option of sending over two ladies or one only. 

To those who think they have solved the puzzle in nine crossings I wouldsay that in every case they will find that they are wrong. No suchjealous husband would, in the circumstances, send his wife over to theother bank to a man or men, even if she assured him that she was comingback next time in the boat. If readers will have this fact in mind, theywill at once discover their errors. 

376. --THE FOUR ELOPEMENTS. 

If there had been only three couples, the island might have beendispensed with, but with four or more couples it is absolutely necessaryin order to cross under the conditions laid down. It can be done inseventeen passages from land to land (though French mathematicians havedeclared in their books that in such circumstances twenty-four areneeded), and it cannot be done in fewer. I will give one way. A, B, C, and D are the young men, and a, b, c, and d are the girls to whom theyare respectively engaged. The three columns show the positions of thedifferent individuals on the lawn, the island, and the opposite shorebefore starting and after each passage, while the asterisk indicates theposition of the boat on every occasion. 

 Lawn. | Island. | Shore. | | ABCDabcd * | | ABCD cd | | ab * ABCD bcd * | | a ABCD d | bc * | a ABCD cd * | b | a CD cd | b | AB a * BCD cd * | b | A a BCD | bcd * | A a BCD d * | bc | A a D d | bc | ABC a * D d | abc * | ABC D d | b | ABC a c * B D d * | b | A C a c d | b | ABCD a c * d | bc * | ABCD a d | | ABCD abc * cd * | | ABCD ab | | ABCD abcd *

Having found the fewest possible passages, we should consider two otherpoints in deciding on the "quickest method": Which persons were the mostexpert in handling the oars, and which method entails the fewestpossible delays in getting in and out of the boat? We have no data uponwhich to decide the first point, though it is probable that, as the boatbelonged to the girls' household, they would be capable oarswomen. Theother point, however, is important, and in the solution I have given(where the girls do 8-13ths of the rowing and A and D need not row atall) there are only sixteen gettings-in and sixteen gettings-out. A manand a girl are never in the boat together, and no man ever lands on theisland. There are other methods that require several more exchanges ofplaces. 

377. --STEALING THE CASTLE TREASURE. 

Here is the best answer, in eleven manipulations:--

 Treasure down. Boy down--treasure up. Youth down--boy up. Treasure down. Man down--youth and treasure up. Treasure down. Boy down--treasure up. Treasure down. Youth down--boy up. Boy down--treasure up. Treasure down. 

378. --DOMINOES IN PROGRESSION. 

There are twenty-three different ways. You may start with any domino, except the 4--4 and those that bear a 5 or 6, though only certaininitial dominoes may be played either way round. If you are given thecommon difference and the first domino is played, you have no option asto the other dominoes. Therefore all I need do is to give the initialdomino for all the twenty-three ways, and state the common difference. This I will do as follows:--

With a common difference of 1, the first domino may be either of these:0--0, 0--1, 1--0, 0--2, 1--1, 2--0, 0--3, 1--2, 2--1, 3--0, 0--4, 1--3, 2--2, 3--1, 1--4, 2--3, 3--2, 2--4, 3--3, 3--4. With a difference of 2, the first domino may be 0--0, 0--2, or 0--1. Take the last case of allas an example. Having played the 0--1, and the difference being 2, weare compelled to continue with 1--2, 2--3, 3--4. 4--5, 5--6. There arethree dominoes that can never be used at all. These are 0--5, 0--6, and1--6. If we used a box of dominoes extending to 9--9, there would beforty different ways. 

379. --THE FIVE DOMINOES. 

There are just ten different ways of arranging the dominoes. Here is oneof them:--

(2--0) (0--0) (0--1) (1--4) (4--0). 

I will leave my readers to find the remaining nine for themselves. 

380. --THE DOMINO FRAME PUZZLE. 

[Illustration:

 +---+-------+-------+-------+-------+-------+-------+-------+ | 2 | 2 | 5 | 5 | 6 | 6 | 6 | 6 | 1 | 1 | | | | | 4 | | - +-------+-------+-------+-------+-------+-------+---+---+ | 2 | | 4 | +---+ | - | | 2 | | 3 | | - | +---+ | 6 | | 3 | +---+ T H E | - | | 6 | | 3 | | - | +---+ | 3 | | 3 | +---+ | - | | 3 | | 1 | | - | D O M I N O F R A M E +---+ | | | 1 | +---+ | - | | | | 1 | | - | +---+ | 5 | | 1 | +---+ -S-O-L-U-T-I-O-N- | - | | 5 | | 4 | | - | +---+ | 3 | | 4 | +---+ | - | | 3 | | 6 | | - | +---+ | 2 | | 6 | +---+---+-------+-------+-------+-------+-------+-------+ - | | 2 | 1 | 1 | 5 | 5 | 5 | 5 | 4 | 4 | 4 | 4 | 2 | 2 | | | +-------+-------+-------+-------+-------+-------+-------+---+

]

The illustration is a solution. It will be found that all four sides ofthe frame add up 44. The sum of the pips on all the dominoes is 168, andif we wish to make the sides sum to 44, we must take care that the fourcorners sum to 8, because these corners are counted twice, and 168 addedto 8 will equal 4 times 44, which is necessary. There are many differentsolutions. Even in the example given certain interchanges are possibleto produce different arrangements. For example, on the left-hand sidethe string of dominoes from 2--2 down to 3--2 may be reversed, or from2--6 to 3--2, or from 3--0 to 5--3. Also, on the right-hand side we mayreverse from 4--3 to 1--4. These changes will not affect the correctnessof the solution. 

381. --THE CARD FRAME PUZZLE. 

The sum of all the pips on the ten cards is 55. Suppose we are trying toget 14 pips on every side. Then 4 times 14 is 56. But each of the fourcorner cards is added in twice, so that 55 deducted from 56, or 1, mustrepresent the sum of the four corner cards. This is clearly impossible;therefore 14 is also impossible. But suppose we came to trying 18. Then4 times 18 is 72, and if we deduct 55 we get 17 as the sum of thecorners. We need then only try different arrangements with the fourcorners always summing to 17, and we soon discover the followingsolution:--

[Illustration:

 +-------+-------+-------+ | 2 | 10 | 6 | +---+---+------ +---+---+ | | | | | 3 | | 7 | | | | | +---+ +---+ | | | | | 8 | | 1 | | | | | +---+---+-------+--+----+ | 5 | 9 | 4 | +-------+-------+-------+

]

The final trials are very limited in number, and must with a littlejudgment either bring us to a correct solution or satisfy us that asolution is impossible under the conditions we are attempting. The twocentre cards on the upright sides can, of course, always beinterchanged, but I do not call these different solutions. If youreflect in a mirror you get another arrangement, which also is notconsidered different. In the answer given, however, we may exchange the5 with the 8 and the 4 with the 1. This is a different solution. Thereare two solutions with 18, four with 19, two with 20, and two with22--ten arrangements in all. Readers may like to find all these forthemselves. 

382. --THE CROSS OF CARDS. 

There are eighteen fundamental arrangements, as follows, where I onlygive the numbers in the horizontal bar, since the remainder mustnaturally fall into their places. 

 5 6 1 7 4 2 4 5 6 8 3 5 1 6 8 3 4 5 6 7 3 4 1 7 8 1 4 7 6 8 2 5 1 7 8 2 3 7 6 8 2 5 3 6 8 2 4 7 5 8 1 5 3 7 8 3 4 9 5 6 2 4 3 7 8 2 4 9 5 7 1 4 5 7 8 1 4 9 6 7 2 3 5 7 8 2 3 9 6 7

It will be noticed that there must always be an odd number in thecentre, that there are four ways each of adding up 23, 25, and 27, butonly three ways each of summing to 24 and 26. 

383. --THE "T" CARD PUZZLE. 

If we remove the ace, the remaining cards may he divided into two groups(each adding up alike) in four ways; if we remove 3, there are threeways; if 5, there are four ways; if 7, there are three ways; and if weremove 9, there are four ways of making two equal groups. There are thuseighteen different ways of grouping, and if we take any one of these andkeep the odd card (that I have called "removed") at the head of thecolumn, then one set of numbers can be varied in order in twenty-fourways in the column and the other four twenty-four ways in thehorizontal, or together they may be varied in 24 × 24 = 576 ways. And asthere are eighteen such cases, we multiply this number by 18 and get10, 368, the correct number of ways of placing the cards. As this numberincludes the reflections, we must divide by 2, but we have also toremember that every horizontal row can change places with a verticalrow, necessitating our multiplying by 2; so one operation cancels theother. 

384. --CARD TRIANGLES. 

The following arrangements of the cards show (1) the smallest possiblesum, 17; and (2) the largest possible, 23. 

 1 7 9 6 4 2 4 8 3 6 3 7 5 2 9 5 1 8

It will be seen that the two cards in the middle of any side may alwaysbe interchanged without affecting the conditions. Thus there are eightways of presenting every fundamental arrangement. The number offundamentals is eighteen, as follows: two summing to 17, four summing to19, six summing to 20, four summing to 21, and two summing to 23. Theseeighteen fundamentals, multiplied by eight (for the reason statedabove), give 144 as the total number of different ways of placing thecards. 

385. --"STRAND" PATIENCE. 

The reader may find a solution quite easy in a little over 200 moves, but, surprising as it may at first appear, not more than 62 moves arerequired. Here is the play: By "4 C up" I mean a transfer of the 4 ofclubs with all the cards that rest on it. 1 D on space, 2 S on space, 3D on space, 2 S on 3 D, 1 H on 2 S, 2 C on space, 1 D on 2 C, 4 S onspace, 3 H on 4 S (9 moves so far), 2 S up on 3 H (3 moves), 5 H and 5 Dexchanged, and 4 C on 5 D (6 moves), 3 D on 4 C (1), 6 S (with 5 H) onspace (3), 4 C up on 5 H (3), 2 C up on 3 D (3), 7 D on space (1), 6 Cup on 7 D (3), 8 S on space (1), 7 H on 8 S (1), 8 C on 9 D (1), 7 H on8 C (1), 8 S on 9 H (1), 7 H on 8 S (1), 7 D up on 8 C (5), 4 C up on 5D (9), 6 S up on 7 H (3), 4 S up on 5 H (7) = 62 moves in all. This ismy record; perhaps the reader can beat it. 

386. --A TRICK WITH DICE. 

All you have to do is to deduct 250 from the result given, and the threefigures in the answer will be the three points thrown with the dice. Thus, in the throw we gave, the number given would be 386; and when wededuct 250 we get 136, from which we know that the throws were 1, 3, and6. 

The process merely consists in giving 100a + 10b + c + 250, where a, b, and c represent the three throws. The result is obvious. 

387. --THE VILLAGE CRICKET MATCH. 

[Illustration:

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