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.
7/13 = .538461 538461 538461 ...
538461 = 9*59829
15/22 = .6 81 81 81 ...
81 = 9*9
We can divide all numbers into those that exactly divide some power of 10, say 10^n, and those that don't. If a number, x, exactly divides 10^n, then its decimal representation will by definition terminate: 2 divides 10; 1/2=5/10=0.50000...., with the string of 0's just emphasising that there is no further remainder to deal with.
Lots of numbers don't exactly divide 10^n. But they do divide (10^n-1) for some n, so we can use the equality 0.9999...=1 to derive as close an approximation as we wish for the decimal value of the fraction 1/x. Since 3 divides 9 exactly, if 3 divides x, we can't make any exact prediction about the sum of the repeating digits. But in all other cases, the block of digits that repeats in the simple fraction 1/x must be divisible by 9, since x of those blocks add up to 0.999999... (to as close an approximation as we wish). And if the simple fraction 1/x is so divisible, then so is 2/x,3/x...etc.
Edited on March 24, 2013, 12:25 am
Posted by broll
on 2013-03-24 00:17:18