The Riddle Of St Edmondsbury
(
THE MERRY MONKS OF RIDDLEWELL)
"It used to be told at St. Edmondsbury," said Father Peter on one occasion, "that many years ago they were so overrun with mice that the good abbot gave orders that all the cats from the country round should be obtained to exterminate the vermin. A record was kept, and at the end of the year it was found that every cat had killed an equal number of mice, and the total was exactly 1,111,111 mice. How many cats do you suppose there were?"
"Methinks one cat killed the lot," said Brother Benjamin.
"Out upon thee, brother! I said 'cats.'"
"Well, then," persisted Benjamin, "perchance 1,111,111 cats each killed one mouse."
"No," replied Father Peter, after the monks' jovial laughter had ended, "I said 'mice;' and all I need add is this—that each cat killed more mice than there were cats. They told me it was merely a question of the division of numbers, but I know not the answer to the riddle."
The correct answer is recorded, but it is not shown how they arrived at it.
Answer:
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. .
Treated generally, this problem consists in finding the factors, if any, of numbers of the form (10n - 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. 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).
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 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 32 . 37 . 333,667. Therefore we know that factors of n = 18 are 11. 32 . 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.
Informational
