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 Repeated part divisible by 9 (Posted on 2013-03-22)
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

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 Proof by geometric series (spoiler) | Comment 1 of 3
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
 Posted by Steve Herman on 2013-03-24 00:11:03

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