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I'm not sure if my proof is correct:

$7|(100a+10b+b)$

$7|(100a+11b)-7(14a+b)$

$7|(2a+4b)$ => $7|2(a+2b)$

I'm not sure if my notation is universal but $abb$ symbolizes three digit numbers who's two last digits are the same.

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  • $\begingroup$ To finish you need to justify why $\ 7\mid 2n\,\Rightarrow\, 7\mid n\ \ $ $\endgroup$ – Bill Dubuque Apr 6 '17 at 16:08
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Using your notation:

$$7|abb \iff 100a+10b+b=7k,\quad k\in\mathbb{N}, \quad 1\leq a,b\leq9$$

We need to prove that $$2a+b=7k,\quad k\in\mathbb{N}$$

We have that:

$$100a+11b=7k\implies 98a+2a+4b+7b = 7k$$

$$7\cdot(14a+b)+2a+4b=7k\implies 2\cdot(a+2b)=7k', \quad k'=k-(14a+b)$$

Since $7\not | 2$, we have that $7|a+2b$ as required.

Edit: Upon re-reading your question, yes your proof does seem correct. Treat the above as an alternate method.

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