The Canterbury Puzzles, and Other Curious Problems

THE MERRY MONKS OF RIDDLEWELL

41 (#pgepubid00052).—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 (#pgepubid00053).—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 (#pgepubid00054).—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 (#pgepubid00055).—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 (#pgepubid00056).—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 (#pgepubid00057).—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 × 73

– 1 = 776

. This is a particular case of the so-called "Pellian Equation," respecting which see A. in M., p. 164.

47 (#pgepubid00058).—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 (#x_1_i117).

Treated generally, this problem consists in finding the factors, if any, of numbers of the form (10

– 1)/9.

Lucas, in his L'Arithmétique 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 Alkalaçadi, 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.[2 - 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.] 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).

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) × 111 × 91

n = 10 = (11) × 11,111 × (9,091)

n = 14 = (11) × 1,111,111 × (909,091)

n = 18 = (11) × 111,111,111 × 90,909,091

Or we may put the factors this way:—

n = 2 = (11)

n = 6 = 111 × 1,001

n = 10 = 11,111 × 100,001

n = 14 = 1,111,111 × 10,000,001

n = 18 = 111,111,111 × 1,000,000,001

In the above two tables n is of the form 4m + 2. When n is of the form 4m the factors may be written down as follows:—

n= 4 = (11) × (101)

n = 8 = (11) × (101) × 10,001

n = 12 = (11) × (101) × 100,010,001

n = 16 = (11) × (101) × 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

. 37 . 333,667. Therefore we know that factors of n = 18 are 11. 3

. 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 (#pgepubid00059).—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 3n