Prove: if a rational number has a repeating decimal and the denominator is not divisible by 3 then the block of digits that repeats, taken as a decimal, is divisible by 9.

Examples:

7/13 = .538461 538461 538461 ...

538461 = 9*59829

15/22 = .6 81 81 81 ...

81 = 9*9

Let .cdddddd... = a/b

where a and b are integers of any length,

c is an integer of length n (where n might be 0)

d is an integer of length m (where m > 1)

For instance, if a/b = 15/22, then c = 6 and d = 81, n = 1 and m = 2.

Multiply by b*(10^n)

Then a*(10^n) = bc + b*(.ddddddd...) = bc + bd/((10^m)-1).

(That last step is done by treating each d as a term in a geometric series which is 1/(10^m) times the previous term.)

a*(10^n) is an integer, and so is bc, so bd/((10^m)-1) must be an integer also. Note that 10^m - 1 is divisible by 9 for all values of m > 1. If b is not divisible by 3, then in order to make the whole thing an integer it follows necessarily that d is divisible by 9.

*q.e.d.*

*p.s. -- It follows, contrapositively, that if d is not a multiple of 9, then b is a multiple of 3.*

*Edited on ***March 24, 2013, 10:10 am**