Pillow-Problems: Problem #44

Problem #41 | Problem #45
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

Problem:

If a, b be two numbers prime to each other, a value may be found for n which will make (aⁿ − 1) a multiple of b.

Solution:

Let k be a No. not containing 2 or 5 as a factor, i.e. let it be prime to 10. Then, if

(1)/

(k)

be reduced to a circulating decimal, and that to a vulgar fraction, the digits of the denominator will be a certain number of 9’s; i.e. it will be of the form (10ⁿ − 1). And since this fraction =

(1)/

(k)

, and that k is prime to 10, and so prime to 10^m, the factor (10ⁿ − 1) must be a multiple of k.

This evidently holds good in any other scale of notation. Hence, if a be the radi× of the scale of notation, and b be a No. prime to a, a value may be found for n, which will make (aⁿ − 1) a multiple of b.

Q.E.D.

Examples (not thought out).

(1) With radix 10, find a value for n, which will make (10ⁿ − 1) a multiple of 7.

Ans. n = 6.

(2) Let the two given Nos. be 8, 9.

Taking 8 as radix, we get

1

9

= .·0·7 =

7

8² − 1

.

Ans. n = 2.

Let the two given Nos. be 7, 13.

Taking 7 as radix, we get

1

13

= .·03524563142·1 =

35245631421

712 − 1

.

Ans. n = 12.