CodalSearch this book — or all of Codal…⌘K
nydus/The Canterbury Puzzles, and Other Curious ProblemsPublic
Page 221 of 306
Table of Contents

47 .— The Riddle of St. Edmondsbury.

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.

221