$n \in \mathbb{N}$
If n is divided into sets of three digits, prove that n is divisible by 7 if and only if an alternating series of sets of three digits is divisible by 7.
e.g. 123456789 = 123,456,789
For it to be divisible by 7,
123 - 456 + 789 has to be divisible by 7.

I think the fact that $7 * 11 * 13 = 1001 ? -1 (mod 7)$ is the key to solving this, since you can create an alternating series with -1.

$7\:\vert \:(a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + ... + a_1 \cdot 10 + a_0)<br /> <br /> 7 \vert \: (a_0 + 10a_1 + 100a_2)<br /> <br /> 7 \vert \: (a_3 \cdot 10^3 + a_4 \cdot 10^4 + a_5 \cdot 10^5)$
..
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Then I don't know how to continue..
Can someone point me in the right direction?