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

My first proof shows that if d is an integer equal to the repeating string produced by the fraction a/b, then b*d is divisible by 9.

We have already concluded, as required, that

1) if b is not divisible by 3, then d is divisible by 9.

Using the same logic, we conclude that

2) if d is not divisible by 3, then b is divisible by 9

And the contrapositives are of course also true:

3) if d is not divisible by 9, then b is divisible by 3

4) if b is not divisible by 9, then d is divisible by 3

It is also worth noting that the fraction a/b does not uniquely determine d.

For example,

7/13 = .538461 538461 538461 ...

= .5 384615 384615 384615 ...

= .53 846153 846153 846153 ...

=

.538 461538 461538 461538 ... = .5384 615384 615384 615384 ...

=

.53846 153846 153846 153846 ...

giving rise to 6 different values of d.

Also, d can be 12 or 18 or 24, etc. digits long. For instance,

7/13 = .5 384615384615 384615384615 384615384615 ...

In fact, there is an infinite (but countable) number of values of d generated by any value of b.

The conclusions above apply to all of the infinite values of d.

*Edited on ***March 24, 2013, 10:37 am**