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
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