Chapter L (1)

LEL
LEVEL
LEVEVEL
LEVELEVEL
LEVEVEL
LEVEL
LEL
L ]

31.--_The Manciple's Puzzle._

The simple Ploughman, who was so ridiculed for his opinion, was perfectly correct: the Miller should receive seven pieces of money, and the Weaver only one. As all three ate equal shares of the bread, it should be evident that each ate 8/3 of a loaf. Therefore, as the Miller provided 15/3 and ate 8/3, he contributed 7/3 to the Manciple's meal; whereas the Weaver provided 9/3, ate 8/3, and contributed only 1/3. Therefore, since they contributed to the Manciple in the proportion of 7 to 1, they must divide the eight pieces of money in the same proportion.

PUZZLING TIMES AT SOLVAMHALL CASTLE

_SIR HUGH EXPLAINS HIS PROBLEMS_

The friends of Sir Hugh de Fortibus were so perplexed over many of his strange puzzles that at a gathering of his kinsmen and retainers he undertook to explain his posers.

"Of a truth," said he, "some of the riddles that I have put forth would greatly tax the wit of the unlettered knave to rede; yet will I try to show the manner thereof in such way that all may have understanding. For many there be who cannot of themselves do all these things, but will yet study them to their gain when they be given the answers, and will take pleasure therein."

32.--_The Game of Bandy-Ball._

Sir Hugh explained, in answer to this puzzle, that as the nine holes were 300, 250, 200, 325, 275, 350, 225, 375, and 400 yards apart, if a man could always strike the ball in a perfectly straight line and send it at will a distance of either 125 yards or 100 yards, he might go round the whole course in 26 strokes. This is clearly correct, for if we call the 125 stroke the "drive" and the 100 stroke the "approach," he could play as follows:--The first hole could be reached in 3 approaches, the second in 2 drives, the third in 2 approaches, the fourth in 2 approaches and 1 drive, the fifth in 3 drives and 1 backward approach, the sixth in 2 drives and 1 approach, the seventh in 1 drive and 1 approach, the eighth in 3 drives, and the ninth hole in 4 approaches. There are thus 26 strokes in all, and the feat cannot be performed in fewer.

33.--_Tilting at the Ring._

"By my halidame!" exclaimed Sir Hugh, "if some of yon varlets had been put in chains, which for their sins they do truly deserve, then would they well know, mayhap, that the length of any chain having like rings is equal to the inner width of a ring multiplied by the number of rings and added to twice the thickness of the iron whereof it is made. It may be shown that the inner width of the rings used in the tilting was one inch and two-thirds thereof, and the number of rings Stephen Malet did win was three, and those that fell to Henry de Gournay would be nine."

The knight was quite correct, for 1-2/3 in. x 3 + 1 in. = 6 in., and 1-2/3 in. x 9 + 1 in. = 16 in. Thus De Gournay beat Malet by six rings. The drawing showing the rings may assist the reader in verifying the answer and help him to see why the inner width of a link multiplied by the number of links and added to twice the thickness of the iron gives the exact length. It will be noticed that every link put on the chain loses a length equal to twice the thickness of the iron.

34.--_The Noble Demoiselle._

"Some here have asked me," continued Sir Hugh, "how they may find the cell in the Dungeon of the Death's-head wherein the noble maiden was cast. Beshrew me! but 'tis easy withal when you do but know how to do it. In attempting to pass through every door once, and never more, you must take heed that every cell hath two doors or four, which be even numbers, except two cells, which have but three. Now, certes, you cannot go in and out of any place, passing through all the doors once and no more, if the number of doors be an odd number. But as there be but two such odd cells, yet may we, by beginning at the one and ending at the other, so make our journey in many ways with success. I pray you, albeit, to mark that only one of these odd cells lieth on the outside of the dungeon, so we must perforce start therefrom. Marry, then, my masters, the noble demoiselle must needs have been wasting in the other."

The drawing will make this quite clear to the reader. The two "odd cells" are indicated by the stars, and one of the many routes that will solve the puzzle is shown by the dotted line. It is perfectly certain that you must start at the lower star and end at the upper one; therefore the cell with the star situated over the left eye must be the one sought.

35.--_The Archery Butt._

"It hath been said that the proof of a pudding is ever in the eating thereof, and by the teeth of Saint George I know no better way of showing how this placing of the figures may be done than by the doing of it. Therefore have I in suchwise written the numbers that they do add up to twenty and three in all the twelve lines of three that are upon the butt."

I think it well here to supplement the solution of De Fortibus with a few remarks of my own. The nineteen numbers may be so arranged that the lines will add up to any number we may choose to select from 22 to 38 inclusive, excepting 30. In some cases there are several different solutions, but in the case of 23 there are only two. I give one of these. To obtain the second solution exchange respectively 7, 10, 5, 8, 9, in the illustration, with 13, 4, 17, 2, 15. Also exchange 18 with 12, and the other numbers may remain unmoved. In every instance there must be an even number in the central place, and any such number from 2 to 18 may occur. Every solution has its complementary. Thus, if for every number in the accompanying drawing we substitute the difference between it and 20, we get the solution in the case of 37. Similarly, from the arrangement in the original drawing, we may at once obtain a solution for the case of 38.

36.--_The Donjon Keep Window._

In this case Sir Hugh had greatly perplexed his chief builder by demanding that he should make a window measuring one foot on every side and divided by bars into eight lights, having all their sides equal. The illustration will show how this was to be done. It will be seen that if each side of the window measures one foot, then each of the eight triangular lights is six inches on every side.

"Of a truth, master builder," said De Fortibus slyly to the architect, "I did not tell thee that the window must be square, as it is most certain it never could be."

37.--_The Crescent and the Cross._

"By the toes of St. Moden," exclaimed Sir Hugh de Fortibus when this puzzle was brought up, "my poor wit hath never shaped a more cunning artifice or any more bewitching to look upon. It came to me as in a vision, and ofttimes have I marvelled at the thing, seeing its exceeding difficulty. My masters and kinsmen, it is done in this wise."

The worthy knight then pointed out that the crescent was of a particular and somewhat irregular form--the two distances _a_ to _b_ and _c_ to _d_ being straight lines, and the arcs _ac_ and _bd_ being precisely similar. He showed that if the cuts be made as in Figure 1, the four pieces will fit together and form a perfect square, as shown in Figure 2, if we there only regard the three curved lines. By now making the straight cuts also shown in Figure 2, we get the ten pieces that fit together, as in Figure 3, and form a perfectly symmetrical Greek cross. The proportions of the crescent and the cross in the original illustration were correct, and the solution can be demonstrated to be absolutely exact and not merely approximate.

I have a solution in considerably fewer pieces, but it is far more difficult to understand than the above method, in which the problem is simplified by introducing the intermediate square.

38.--_The Amulet._

The puzzle was to place your pencil on the A at the top of the amulet and count in how many different ways you could trace out the word "Abracadabra" downwards, always passing from a letter to an adjoining one.

A
B B
R R R
A A A A
C C C C C
A A A A A A
D D D D D D D
A A A A A A A A
B B B B B B B B B
R R R R R R R R R R
A A A A A A A A A A A

"Now, mark ye, fine fellows," said Sir Hugh to some who had besought him to explain, "that at the very first start there be two ways open: whichever B ye select, there will be two several ways of proceeding (twice times two are four); whichever R ye select, there be two ways of going on (twice times four are eight); and so on until the end. Each letter in order from A downwards may so be reached in 2, 4, 8, 16, 32, etc., ways. Therefore, as there be ten lines or steps in all from A to the bottom, all ye need do is to multiply ten 2's together, and truly the result, 1024, is the answer thou dost seek."

39.--_The Snail on the Flagstaff._

Though there was no need to take down and measure the staff, it is undoubtedly necessary to find its height before the answer can be given. It was well known among the friends and retainers of Sir Hugh de Fortibus that he was exactly six feet in height. It will be seen in the original picture that Sir Hugh's height is just twice the length of his shadow. Therefore we all know that the flagstaff will, at the same place and time of day, be also just twice as long as its shadow. The shadow of the staff is the same length as Sir Hugh's height; therefore this shadow is six feet long, and the flagstaff must be twelve feet high. Now, the snail, by climbing up three feet in the daytime and slipping back two feet by night, really advances one foot in a day of twenty-four hours. At the end of nine days it is three feet from the top, so that it reaches its journey's end on the tenth day.

The reader will doubtless here exclaim, "This is all very well; but how were we to know the height of Sir Hugh? It was never stated how tall he was!" No, it was not stated in so many words, but it was none the less clearly indicated to the reader who is sharp in these matters. In the original illustration to the donjon keep window Sir Hugh is shown standing against a wall, the window in which is stated to be one foot square on the inside. Therefore, as his height will be found by measurement to be just six times the inside height of the window, he evidently stands just six feet in his boots!

40.--_Lady Isabel's Casket._

The last puzzle was undoubtedly a hard nut, but perhaps difficulty does not make a good puzzle any the less interesting when we are shown the solution. The accompanying diagram indicates exactly how the top of Lady Isabel de Fitzarnulph's casket was inlaid with square pieces of rare wood (no two squares alike) and the strip of gold 10 inches by a quarter of an inch. This is the only possible solution, and it is a singular fact (though I cannot here show the subtle method of working) that the number, sizes, and order of those squares are determined by the given dimensions of the strip of gold, and the casket can have no other dimensions than 20 inches square. The number in a square indicates the length in inches of the side of that square, so the accuracy of the answer can be checked almost at a glance.

Sir Hugh de Fortibus made some general concluding remarks on the occasion that are not altogether uninteresting to-day.

"Friends and retainers," he said, "if the strange offspring of my poor wit about which we have held pleasant counsel to-night hath mayhap had some small interest for ye, let these matters serve to call to mind the lesson that our fleeting life is rounded and beset with enigmas. Whence we came and whither we go be riddles, and albeit such as these we may never bring within our understanding, yet there be many others with which we and they that do come after us will ever strive for the answer. Whether success do attend or do not attend our labour, it is well that we make the attempt; for 'tis truly good and honourable to train the mind, and the wit, and the fancy of man, for out of such doth issue all manner of good in ways unforeseen for them that do come after us."

THE MERRY MONKS OF RIDDLEWELL

41.--_The Riddle of the Fish-pond._

Number the fish baskets in the illustration from 1 to 12 in the direction that Brother Jonathan is seen to be going. Starting from 1, proceed as follows, where "1 to 4" means, take the fish from basket No. 1 and transfer it to basket No. 4:--

1 to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, 11 to 2, and complete the last revolution to 1, making three revolutions in all. Or you can proceed this way:--

4 to 7, 8 to 11, 12 to 3, 2 to 5, 6 to 9, 10 to 1.

It is easy to solve in four revolutions, but the solutions in three are more difficult to discover.

42.--_The Riddle of the Pilgrims._

If it were not for the Abbot's conditions that the number of guests in any room may not exceed three, and that every room must be occupied, it would have been possible to accommodate either 24, 27, 30, 33, 36, 39, or 42 pilgrims. But to accommodate 24 pilgrims so that there shall be twice as many sleeping on the upper floor as on the lower floor, and eleven persons on each side of the building, it will be found necessary to leave some of the rooms empty. If, on the other hand, we try to put up 33, 36, 39 or 42 pilgrims, we shall find that in every case we are obliged to place more than three persons in some of the rooms. Thus we know that the number of pilgrims originally announced (whom, it will be remembered, it was possible to accommodate under the conditions of the Abbot) must have been 27, and that, since three more than this number were actually provided with beds, the total number of pilgrims was 30. The accompanying diagram shows how they might be arranged, and if in each instance we regard the upper floor as placed above the lower one, it will be seen that there are eleven persons on each side of the building, and twice as many above as below.

43.--_The Riddle of the Tiled Hearth._

The correct answer is shown in the illustration on page 196. No tile is in line (either horizontally, vertically, or diagonally) with another tile of the same design, and only three plain tiles are used. If after placing the four lions you fall into the error of placing four other tiles of another pattern, instead of only three, you will be left with four places that must be occupied by plain tiles. The secret consists in placing four of one kind and only three of each of the others.

44.--_The Riddle of the Sack of Wine._

The question was: Did Brother Benjamin take more wine from the bottle than water from the jug? Or did he take more water from the jug than wine from the bottle? He did neither. The same quantity of wine was transferred from the bottle as water was taken from the jug. Let us assume that the glass would hold a quarter of a pint. There was a pint of wine in the bottle and a pint of water in the jug. After the first manipulation the bottle contains three-quarters of a pint of wine, and the jug one pint of water mixed with a quarter of a pint of wine. Now, the second transaction consists in taking away a fifth of the contents of the jug--that is, one-fifth of a pint of water mixed with one-fifth of a quarter of a pint of wine. We thus leave behind in the jug four-fifths of a quarter of a pint of wine--that is, one-fifth of a pint--while we transfer from the jug to the bottle an equal quantity (one-fifth of a pint) of water.

45.--_The Riddle of the Cellarer._

There were 100 pints of wine in the cask, and on thirty occasions John the Cellarer had stolen a pint and replaced it with a pint of water. After the first theft the wine left in the cask would be 99 pints; after the second theft the wine in the cask would be 9801/100 pints (the square of 99 divided by 100); after the third theft there would remain 970299/10000 (the cube of 99 divided by the square of 100); after the fourth theft there would remain the fourth power of 99 divided by the cube of 100; and after the thirtieth theft there would remain in the cask the thirtieth power of 99 divided by the twenty-ninth power of 100. This by the ordinary method of calculation gives us a number composed of 59 figures to be divided by a number composed of 58 figures! But by the use of logarithms it may be quickly ascertained that the required quantity is very nearly 73-97/100 pints of wine left in the cask. Consequently the cellarer stole nearly 26.03 pints. The monks doubtless omitted the answer for the reason that they had no tables of logarithms, and did not care to face the task of making that long and tedious calculation in order to get the quantity "to a nicety," as the wily cellarer had stipulated.

By a simplified process of calculation, I have ascertained that the exact quantity of wine stolen would be

26.0299626611719577269984907683285057747323737647323555652999

pints. A man who would involve the monastery in a fraction of fifty-eight decimals deserved severe punishment.

46.--_The Riddle of the Crusaders._

The correct answer is that there would have been 602,176 Crusaders, who could form themselves into a square 776 by 776; and after the stranger joined their ranks, they could form 113 squares of 5,329 men--that is, 73 by 73. Or 113 x 73^2 - 1 = 776^2. This is a particular case of the so-called "Pellian Equation," respecting which see _A. in M._, p. 164.

47.--_The Riddle of St. Edmondsbury._

The reader is aware that there are prime numbers and composite whole numbers. Now, 1,111,111 cannot be a prime number, because if it were the only possible answers would be those proposed by Brother Benjamin and rejected by Father Peter. Also it cannot have more than two factors, or the answer would be indeterminate. As a matter of fact, 1,111,111 equals 239 x 4649 (both primes), and since each cat killed more mice than there were cats, the answer must be 239 cats. See also the Introduction, p. 18.

Treated generally, this problem consists in finding the factors, if any, of numbers of the form (10^_n_ - 1)/9.

Lucas, in his _L'Arithmetique Amusante_, gives a number of curious tables which he obtained from an arithmetical treatise, called the _Talkhys_, by Ibn Albanna, an Arabian mathematician and astronomer of the first half of the thirteenth century. In the Paris National Library are several manuscripts dealing with the _Talkhys_, and a commentary by Alkalacadi, who died in 1486. Among the tables given by Lucas is one giving all the factors of numbers of the above form up to _n_ = 18. It seems almost inconceivable that Arabians of that date could find the factors where _n_ = 17, as given in my Introduction. But I read Lucas as stating that they are given in _Talkhys_, though an eminent mathematician reads him differently, and suggests to me that they were discovered by Lucas himself. This can, of course, be settled by an examination of _Talkhys_, but this has not been possible during the war.

The difficulty lies wholly with those cases where _n_ is a prime number. If _n_ = 2, we get the prime 11. The factors when _n_ = 3, 5, 11, and 13 are respectively (3 . 37), (41 . 271), (21,649 . 513,239), and (53 . 79 . 265371653). I have given in these pages the factors where _n_ = 7 and 17. The factors when _n_= 19, 23, and 37 are unknown, if there are any.[B] When _n_ = 29, the factors are (3,191 . 16,763 . 43,037. 62,003 . 77,843,839,397); when _n_ = 31, one factor is 2,791; and when _n_ = 41, two factors are (83 . 1,231).

[B] Mr. Oscar Hoppe, of New York, informs me that, after reading my statement in the Introduction, he was led to investigate the case of _n_ = 19, and after long and tedious work he succeeded in proving the number to be a prime. He submitted his proof to the London Mathematical Society, and a specially appointed committee of that body accepted the proof as final and conclusive. He refers me to the _Proceedings_ of the Society for 14th February 1918.

As for the even values of _n_, the following curious series of factors will doubtless interest the reader. The numbers in brackets are primes.

_n_ = 2 = (11)

_n_ = 6 = (11) x 111 x 91

_n_ = 10 = (11) x 11,111 x (9,091)

_n_ = 14 = (11) x 1,111,111 x (909,091)

_n_ = 18 = (11) x 111,111,111 x 90,909,091

Or we may put the factors this way:--

_n_ = 2 = (11)

_n_ = 6 = 111 x 1,001

_n_ = 10 = 11,111 x 100,001

_n_ = 14 = 1,111,111 x 10,000,001

_n_ = 18 = 111,111,111 x 1,000,000,001

In the above two tables _n_ is of the form 4_m_ + 2. When _n_ is of the form 4_m_ the factors may be written down as follows:--

_n_= 4 = (11) x (101)

_n_ = 8 = (11) x (101) x 10,001

_n_ = 12 = (11) x (101) x 100,010,001

_n_ = 16 = (11) x (101) x 1,000,100,010,001.

When _n_ = 2, we have the prime number 11; when _n_ = 3, the factors are 3 . 37; when _n_ = 6, they are 11 . 3 . 37 . 7. 13; when _n_ = 9, they are 3^2 . 37 . 333,667. Therefore we know that factors of _n_ = 18 are 11. 3^2 . 37 . 7 . 13 . 333,667, while the remaining factor is composite and can be split into 19 . 52579. This will show how the working may be simplified when _n_ is not prime.

48.--_The Riddle of the Frogs' Ring._

The fewest possible moves in which this puzzle can be solved are 118. I will give the complete solution. The black figures on white discs move in the directions of the hands of a clock, and the white figures on black discs the other way. The following are the numbers in the order in which they move. Whether you have to make a simple move or a leaping move will be clear from the position, as you never can have an alternative. The moves enclosed in brackets are to be played five times over: 6, 7, 8, 6, 5, 4, 7, 8, 9, 10, 6, 5, 4, 3, 2, 7, 8, 9, 10, 11 (6, 5, 4, 3, 2, 1), 6, 5, 4, 3, 2, 12, (7, 8, 9, 10, 11, 12), 7, 8, 9, 10, 11, 1, 6, 5, 4, 3, 2, 12, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 8, 9, 10, 11, 4, 3, 2, 10, 11, 2. We thus have made 118 moves within the conditions, the black frogs have changed places with the white ones, and 1 and 12 are side by side in the positions stipulated.

The general solution in the case of this puzzle is 3_n_^{2} + 2_n_ - 2 moves, where the number of frogs of each colour is _n_. The law governing the sequence of moves is easily discovered by an examination of the simpler cases, where _n_ = 2, 3, and 4.

If, instead of 11 and 12 changing places, the 6 and 7 must interchange, the expression is _n_^{2} + 4_n_ + 2 moves. If we give _n_ the value 6, as in the example of the Frogs' Ring, the number of moves would be 62.

For a general solution of the case where frogs of one colour reverse their order, leaving the blank space in the same position, and each frog is allowed to be moved in either direction (leaping, of course, over his own colour), see "The Grasshopper Puzzle" in _A. in M._, p. 193.

THE STRANGE ESCAPE OF THE KING'S JESTER

Although the king's jester promised that he would "thereafter make the manner thereof plain to all," there is no record of his having ever done so. I will therefore submit to the reader my own views as to the probable solutions to the mysteries involved.

49.--_The Mysterious Rope._

When the jester "divided his rope in half," it does not follow that he cut it into two parts, each half the original length of the rope. No doubt he simply untwisted the strands, and so divided it into two ropes, each of the original length, but one-half the thickness. He would thus be able to tie the two together and make a rope nearly twice the original length, with which it is quite conceivable that he made good his escape from the dungeon.

50.--_The Underground Maze._

How did the jester find his way out of the maze in the dark? He had simply to grope his way to a wall and then keep on walking without once removing his left hand (or right hand) from the wall. Starting from A, the dotted line will make the route clear when he goes to the left. If the reader tries the route to the right in the same way he will be equally successful; in fact, the two routes unite and cover every part of the walls of the maze except those two detached parts on the left-hand side--one piece like a U, and the other like a distorted E. This rule will apply to the majority of mazes and puzzle gardens; but if the centre were enclosed by an isolated wall in the form of a split ring, the jester would simply have gone round and round this ring.

See the article, "Mazes, and How to Thread Them," _in A. in M._

51.--_The Secret Lock._

This puzzle entailed the finding of an English word of three letters, each letter being found on a different dial. Now, there is no English word composed of consonants alone, and the only vowel appearing anywhere on the dials is Y. No English word begins with Y and has the two other letters consonants, and all the words of three letters ending in Y (with two consonants) either begin with an S or have H, L, or R as their second letter. But these four consonants do not appear. Therefore Y must occur in the middle, and the only word that I can find is "PYX," and there can be little doubt that this was the word. At any rate, it solves our puzzle.

52.--_Crossing the Moat._

No doubt some of my readers will smile at the statement that a man in a boat on smooth water can pull himself across with the tiller rope! But it is a fact. If the jester had fastened the end of his rope to the stern of the boat and then, while standing in the bows, had given a series of violent jerks, the boat would have been propelled forward. This has often been put to a practical test, and it is said that a speed of two or three miles an hour may be attained. See W. W. Rouse Ball's _Mathematical Recreations_.

53.--_The Royal Gardens._

This puzzle must have struck many readers as being absolutely impossible. The jester said: "I had, of a truth, entered every one of the sixteen gardens once, and never more than once." If we follow the route shown in the accompanying diagram, we find that there is no difficulty in once entering all the gardens but one before reaching the last garden containing the exit B. The difficulty is to get into the garden with a star, because if we leave the B garden we are compelled to enter it a second time before escaping, and no garden may be entered twice. The trick consists in the fact that you may enter that starred garden without necessarily leaving the other. If, when the jester got to the gateway where the dotted line makes a sharp bend, his intention had been to hide in the starred garden, but after he had put one foot through the doorway, upon the star, he discovered it was a false alarm and withdrew, he could truly say: "I entered the starred garden, because I put my foot and part of my body in it; and I did not enter the other garden twice, because, after once going in I never left it until I made my exit at B." This is the only answer possible, and it was doubtless that which the jester intended.

See "The Languishing Maiden," in _A. in M._

54.--_Bridging the Ditch._

The solution to this puzzle is best explained by the illustration. If he had placed his eight planks, in the manner shown, across the angle of the ditch, he would have been able to cross without much trouble. The king's jester might thus have well overcome all his difficulties and got safely away, as he has told us that he succeeded in doing.

THE SQUIRE'S CHRISTMAS PUZZLE PARTY

_HOW THE VARIOUS TRICKS WERE DONE_

The record of one of Squire Davidge's annual "Puzzle Parties," made by the old gentleman's young lady relative, who had often spent a merry Christmas at Stoke Courcy Hall, does not contain the solutions of the mysteries. So I will give my own answers to the puzzles and try to make them as clear as possible to those who may be more or less novices in such matters.

55.--_The Three Teacups._

Miss Charity Lockyer clearly must have had a trick up her sleeve, and I think it highly probable that it was conceived on the following lines. She proposed that ten lumps of sugar should be placed in three teacups, so that there should be an odd number of lumps in every cup. The illustration perhaps shows Miss Charity's answer, and the figures on the cups indicate the number of lumps that have been separately placed in them. By placing the cup that holds one lump inside the one that holds two lumps, it can be correctly stated that every cup contains an odd number of lumps. One cup holds seven lumps, another holds one lump, while the third cup holds three lumps. It is evident that if a cup contains another cup it also contains the contents of that second cup.

There are in all fifteen different solutions to this puzzle. Here they are:--

1 0 9 1 4 5 9 0 1
3 0 7 7 0 3 7 2 1
1 2 7 5 2 3 5 4 1
5 0 5 3 4 3 3 6 1
3 2 5 1 6 3 1 8 1

The first two numbers in a triplet represent respectively the number of lumps to be placed in the inner and outer of the two cups that are placed one inside the other. It will be noted that the outer cup of the pair may itself be empty.

56.--_The Eleven Pennies._

It is rather evident that the trick in this puzzle was as follows:--From the eleven coins take five; then add four (to those already taken away) and you leave nine--in the second heap of those removed!

57.--_The Christmas Geese._

Farmer Rouse sent exactly 101 geese to market. Jabez first sold Mr. Jasper Tyler half of the flock and half a goose over (that is, 50-1/2 + 1/2, or 51 geese, leaving 50); he then sold Farmer Avent a third of what remained and a third of a goose over (that is, 16-2/3 + 1/3, or 17 geese, leaving 33); he then sold Widow Foster a quarter of what remained and three-quarters of a goose over (that is, 8-1/4 + 3/4 or 9 geese, leaving 24); he next sold Ned Collier a fifth of what he had left and gave him a fifth of a goose "for the missus" (that is, 4-4/5 + 1/5 or 5 geese, leaving 19). He then took these 19 back to his master.

58.--_The Chalked Numbers._

This little jest on the part of Major Trenchard is another trick puzzle, and the face of the roguish boy on the extreme right, with the figure 9 on his back, showed clearly that he was in the secret, whatever that secret might be. I have no doubt (bearing in mind the Major's hint as to the numbers being "properly regarded") that his answer was that depicted in the illustration, where boy No. 9 stands on his head and so converts his number into 6. This makes the total 36--an even number--and by making boys 3 and 4 change places with 7 and 8, we get 1278 and 5346, the figures of which, in each case, add up to 18. There are just three other ways in which the boys may be grouped: 1368--2457, 1467--2358, and 2367--1458.

59.--_Tasting the Plum Puddings._

The diagram will show how this puzzle is to be solved. It is the only way within the conditions laid down. Starting at the pudding with holly at the top left-hand corner, we strike out all the puddings in twenty-one straight strokes, taste the steaming hot pudding at the end of the tenth stroke, and end at the second sprig of holly.

Here we have an example of a chess rook's path that is not re-entrant, but between two squares that are at the greatest possible distance from one another. For if it were desired to move, under the condition of visiting every square once and once only, from one corner square to the other corner square on the same diagonal, the feat is impossible.

There are a good many different routes for passing from one sprig of holly to the other in the smallest possible number of moves--twenty-one--but I have not counted them. I have recorded fourteen of these, and possibly there are more. Any one of these would serve our purpose, except for the condition that the tenth stroke shall end at the steaming hot pudding. This was introduced to stop a plurality of solutions--called by the maker of chess problems "cooks." I am not aware of more than one solution to this puzzle; but as I may not have recorded all the tours, I cannot make a positive statement on the point at the time of writing.

60.--_Under the Mistletoe Bough._

Everybody was found to have kissed everybody else once under the mistletoe, with the following additions and exceptions: No male kissed a male; no man kissed a married woman except his own wife; all the bachelors and boys kissed all the maidens and girls twice; the widower did not kiss anybody, and the widows did not kiss each other. Every kiss was returned, and the double performance was to count as one kiss. In making a list of the company, we can leave out the widower altogether, because he took no part in the osculatory exercise.

7 Married couples 14
3 Widows 3
12 Bachelors and Boys 12
10 Maidens and Girls 10
Total 39 Persons

Now, if every one of these 39 persons kissed everybody else once, the number of kisses would be 741; and if the 12 bachelors and boys each kissed the 10 maidens and girls once again, we must add 120, making a total of 861 kisses. But as no married man kissed a married woman other than his own wife, we must deduct 42 kisses; as no male kissed another male, we must deduct 171 kisses; and as no widow kissed another widow, we must deduct 3 kisses. We have, therefore, to deduct 42+171+3=216 kisses from the above total of 861, and the result, 645, represents exactly the number of kisses that were actually given under the mistletoe bough.

61.--_The Silver Cubes._

There is no limit to the number of different dimensions that will give two cubes whose sum shall be exactly seventeen cubic inches. Here is the answer in the smallest possible numbers. One of the silver cubes must measure 2-23278/40831 inches along each edge, and the other must measure 11663/40831 inch. If the reader likes to undertake the task of cubing each number (that is, multiply each number twice by itself), he will find that when added together the contents exactly equal seventeen cubic inches. See also No. 20, "The Puzzle of the Doctor of Physic."

THE ADVENTURES OF THE PUZZLE CLUB

62.--_The Ambiguous Photograph._

One by one the members of the Club succeeded in discovering the key to the mystery of the Ambiguous Photograph, except Churton, who was at length persuaded to "give it up." Herbert Baynes then pointed out to him that the coat that Lord Marksford was carrying over his arm was a lady's coat, because the buttons are on the left side, whereas a man's coat always has the buttons on the right-hand side. Lord Marksford would not be likely to walk about the streets of Paris with a lady's coat over his arm unless he was accompanying the owner. He was therefore walking with the lady.

As they were talking a waiter brought a telegram to Baynes.

"Here you are," he said, after reading the message. "A wire from Dovey: 'Don't bother about photo. Find lady was the gentleman's sister, passing through Paris.' That settles it. You might notice that the lady was lightly clad, and therefore the coat might well be hers. But it is clear that the rain was only a sudden shower, and no doubt they were close to their destination, and she did not think it worth while to put the coat on."

63.--_The Cornish Cliff Mystery._

Melville's explanation of the Cornish Cliff Mystery was very simple when he gave it. Yet it was an ingenious trick that the two criminals adopted, and it would have completely succeeded had not our friends from the Puzzle Club accidentally appeared on the scene. This is what happened: When Lamson and Marsh reached the stile, Marsh alone walked to the top of the cliff, with Lamson's larger boots in his hands. Arrived at the edge of the cliff, he changed the boots and walked backwards to the stile, carrying his own boots.

This little manoeuvre accounts for the smaller footprints showing a deeper impression at the heel, and the larger prints a deeper impression at the toe; for a man will walk more heavily on his heels when going forward, but will make a deeper impression with the toes in walking backwards. It will also account for the fact that the large footprints were sometimes impressed over the smaller ones, but never the reverse; also for the circumstance that the larger footprints showed a shorter stride, for a man will necessarily take a smaller stride when walking backwards. The pocket-book was intentionally dropped, to lead the police to discover the footprints, and so be put on the wrong scent.

64.--_The Runaway Motor-Car._

Russell found that there are just twelve five-figure numbers that have the peculiarity that the first two figures multiplied by the last three--all the figures being different, and there being no 0--will produce a number with exactly the same five figures, in a different order. But only one of these twelve begins with a 1--namely, 14926. Now, if we multiply 14 by 926, the result is 12964, which contains the same five figures. The number of the motor-car was therefore 14926.

Here are the other eleven numbers:--24651, 42678, 51246, 57834, 75231, 78624, 87435, 72936, 65281, 65983, and 86251.

Compare with the problems in "Digital Puzzles," section of _A. in M._, and with Nos. 93 and 101 in these pages.

65.--_The Mystery of Ravensdene Park._

The diagrams show that there are two different ways in which the routes of the various persons involved in the Ravensdene Mystery may be traced, without any path ever crossing another. It depends whether the butler, E, went to the north or the south of the gamekeeper's cottage, and the gamekeeper, A, went to the south or the north of the hall. But it will be found that the only persons who could have approached Mr. Cyril Hastings without crossing a path were the butler, E, and the man, C. It was, however, a fact that the butler retired to bed five minutes before midnight, whereas Mr. Hastings did not leave his friend's house until midnight. Therefore the criminal must have been the man who entered the park at C.

66.--_The Buried Treasure._

The field must have contained between 179 and 180 acres--to be more exact, 179.37254 acres. Had the measurements been 3, 2, and 4 furlongs respectively from successive corners, then the field would have been 209.70537 acres in area.

One method of solving this problem is as follows. Find the area of triangle APB in terms of _x_, the side of the square. Double the result=_xy_. Divide by _x_ and then square, and we have the value of _y_^{2} in terms of _x_. Similarly find value of _z_^{2} in terms of _x_; then solve the equation _y_^{2}+_z_^{2}=3^{2}, which will come out in the form _x_^{4}-20_x_^{2}=-37. Therefore _x_^{2}=10+(sqrt{63})=17.937254 square furlongs, very nearly, and as there are ten acres in one square furlong, this equals 179.37254 acres. If we take the negative root of the equation, we get the area of the field as 20.62746 acres, in which case the treasure would have been buried outside the field, as in Diagram 2. But this solution is excluded by the condition that the treasure was buried in the field. The words were, "The document ... states clearly that the field is square, and that the treasure is buried in it."

THE PROFESSOR'S PUZZLES

67.--_The Coinage Puzzle._

The point of this puzzle turns on the fact that if the magic square were to be composed of whole numbers adding up 15 in all ways, the two must be placed in one of the corners. Otherwise fractions must be used, and these are supplied in the puzzle by the employment of sixpences and half-crowns. I give the arrangement requiring the fewest possible current English coins--fifteen. It will be seen that the amount in each corner is a fractional one, the sum required in the total being a whole number of shillings.

68.--_The Postage Stamps Puzzles._

The first of these puzzles is based on a similar principle, though it is really much easier, because the condition that nine of the stamps must be of different values makes their selection a simple matter, though how they are to be placed requires a little thought or trial until one knows the rule respecting putting the fractions in the corners. I give the solution.

[1/2 d]
[4-1/2 d] [1 d] [3 d]

[2 d] [3 d] [4 d]

[2-1/2 d] [5 d] [1-1/2 d] ]

[4 d] [1/2 d]
[3 d] [1-1/2 d]
[9 d]

[10 d] [6 d] [2 d]

[1 d] [1 s.] [5 d] ]

I also show the solution to the second stamp puzzle. All the columns, rows, and diagonals add up 1_s._ 6_d._ There is no stamp on one square, and the conditions did not forbid this omission. The stamps at present in circulation are these:--1/2_d._, 1_d._, 1-1/2_d._, 2_d._, 2-1/2_d._, 3_d._, 4_d._, 5_d._, 6_d._, 9_d._, 10_d._, 1_s._, 2_s._ 6_d._, 5_s._, 10_s._, L1, and L5. In the first solution the numbers are in arithmetical progression--1, 1-1/2, 2, 2-1/2, 3, 3-1/2, 4, 4-1/2, 5. But any nine numbers will form a magic square if we can write them thus:--

1 2 3
7 8 9
13 14 15

where the horizontal differences are all alike and the vertical differences all alike, but not necessarily the same as the horizontal. This happens in the case of the second solution, the numbers of which may be written:--

0 1 2
5 6 7
10 11 12

Also in the case of the solution to No. 67, the Coinage Puzzle, the numbers are, in shillings:--

2 2-1/2 3
4-1/2 5 5-1/2
7 7-1/2 8

If there are to be nine _different_ numbers, 0 may occur once (as in the solution to No. 22). Yet one might construct squares with negative numbers, as follows:--

-2 -1 0
5 6 7
12 13 14

69.--_The Frogs and Tumblers._

It is perfectly true, as the Professor said, that there is only one solution (not counting a reversal) to this puzzle. The frogs that jump are George in the third horizontal row; Chang, the artful-looking batrachian at the end of the fourth row; and Wilhelmina, the fair creature in the seventh row. George jumps downwards to the second tumbler in the seventh row; Chang, who can only leap short distances in consequence of chronic rheumatism, removes somewhat unwillingly to the glass just above him--the eighth in the third row; while Wilhelmina, with all the sprightliness of her youth and sex, performs the very creditable saltatory feat of leaping to the fourth tumbler in the fourth row. In their new positions, as shown in the accompanying diagram, it will be found that of the eight frogs no two are in line vertically, horizontally, or diagonally.

70.--_Romeo and Juliet._

This is rather a difficult puzzle, though, as the Professor remarked when Hawkhurst hit on the solution, it is "just one of those puzzles that a person might solve at a glance" by pure luck. Yet when the solution, with its pretty, symmetrical arrangement, is seen, it looks ridiculously simple.

It will be found that Romeo reaches Juliet's balcony after visiting every house once and only once, and making fourteen turnings, not counting the turn he makes at starting. These are the fewest turnings possible, and the problem can only be solved by the route shown or its reversal.

71.--_Romeo's Second Journey._

In order to take his trip through all the white squares only with the fewest possible turnings, Romeo would do well to adopt the route I have shown, by means of which only sixteen turnings are required to perform the feat. The Professor informs me that the Helix Aspersa, or common or garden snail, has a peculiar aversion to making turnings--so much so that one specimen with which he made experiments went off in a straight line one night and has never come back since.

72.--_The Frogs who would a-wooing go._

This is one of those puzzles in which a plurality of solutions is practically unavoidable. There are two or three positions into which four frogs may jump so as to form five rows with four in each row, but the case I have given is the most satisfactory arrangement.

The frogs that have jumped have left their astral bodies behind, in order to show the reader the positions which they originally occupied. Chang, the frog in the middle of the upper row, suffering from rheumatism, as explained above in the Frogs and Tumblers solution, makes the shortest jump of all--a little distance between the two rows; George and Wilhelmina leap from the ends of the lower row to some distance N. by N.W. and N. by N.E. respectively; while the frog in the middle of the lower row, whose name the Professor forgot to state, goes direct S.

73.--_The Game of Kayles._

To win at this game you must, sooner or later, leave your opponent an even number of similar groups. Then whatever he does in one group you repeat in a similar group. Suppose, for example, that you leave him these groups: o.o.ooo.ooo. Now, if he knocks down a single, you knock down a single; if he knocks down two in one triplet, you knock down two in the other triplet; if he knocks down the central kayle in a triplet, you knock down the central one in the other triplet. In this way you must eventually win. As the game is started with the arrangement o.ooooooooooo, the first player can always win, but only by knocking down the sixth or tenth kayle (counting the one already fallen as the second), and this leaves in either case o.ooo.ooooooo, as the order of the groups is of no importance. Whatever the second player now does, this can always be resolved into an even number of equal groups. Let us suppose that he knocks down the single one; then we play to leave him oo.ooooooo. Now, whatever he does we can afterwards leave him either ooo.ooo or o.oo.ooo. We know why the former wins, and the latter wins also; because, however he may play, we can always leave him either o.o, or o.o.o.o, or oo.oo, as the case may be. The complete analysis I can now leave for the amusement of the reader.

74.--_The Broken Chessboard._

The illustration will show how the thirteen pieces can be put together so as to construct the perfect board, and the reverse problem of cutting these particular pieces out will be found equally entertaining.

Compare with Nos. 293 and 294 in _A. in M._

75.--_The Spider and the Fly._

Though this problem was much discussed in the _Daily Mail_ from 18th January to 7th February 1905, when it appeared to create great public interest, it was actually first propounded by me in the _Weekly Dispatch_ of 14th June 1903.

Imagine the room to be a cardboard box. Then the box may be cut in various different ways, so that the cardboard may be laid flat on the table. I show four of these ways, and indicate in every case the relative positions of the spider and the fly, and the straightened course which the spider must take without going off the cardboard. These are the four most favourable cases, and it will be found that the shortest route is in No. 4, for it is only 40 feet in length (add the square of 32 to the square of 24 and extract the square root). It will be seen that the spider actually passes along five of the six sides of the room! Having marked the route, fold the box up (removing the side the spider does not use), and the appearance of the shortest course is rather surprising. If the spider had taken what most persons will consider obviously the shortest route (that shown in No. 1), he would have gone 42 feet! Route No. 2 is 43.174 feet in length, and Route No. 3 is 40.718 feet.

I will leave the reader to discover which are the shortest routes when the spider and the fly are 2, 3, 4, 5, and 6 feet from the ceiling and the floor respectively.

76.--_The Perplexed Cellarman._

Brother John gave the first man three large bottles and one small bottleful of wine, and one large and three small empty bottles. To each of the other two men he gave two large and three small bottles of wine, and two large and one small empty bottle. Each of the three then receives the same quantity of wine, and the same number of each size of bottle.

77.--_Making a Flag._

The diagram shows how the piece of bunting is to be cut into two pieces. Lower the piece on the right one "tooth," and they will form a perfect square, with the roses symmetrically placed.

It will be found interesting to compare this with No. 154 in _A. in M._

78.--_Catching the Hogs._

A very short examination of this puzzle game should convince the reader that Hendrick can never catch the black hog, and that the white hog can never be caught by Katruen.

Each hog merely runs in and out of one of the nearest corners and can never be captured. The fact is, curious as it must at first sight appear, a Dutchman cannot catch a black hog, and a Dutchwoman can never capture a white one! But each can, without difficulty, catch one of the other colour.