# Pillow-Problems: Problem #44

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## Problem:

44.

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.

[18/3/81

## Solution:

44.

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 10m, 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.

 1 7
= .·14285·7 =
 142857 106 − 1
.

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.