The Canterbury Puzzles Part 24

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.

[Ill.u.s.tration]

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

The correct answer is shown in the ill.u.s.tration 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.

[Ill.u.s.tration]

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 quant.i.ty of wine was transferred from the bottle as water was taken from the jug. Let us a.s.sume that the gla.s.s 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 quant.i.ty (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 quant.i.ty 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 quant.i.ty "to a nicety," as the wily cellarer had stipulated.

By a simplified process of calculation, I have ascertained that the exact quant.i.ty 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 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 ma.n.u.scripts 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) 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 4_m_ + 2. When _n_ is of the form 4_m_ 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 3^2 . 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.