If $7|(100a+b)$ prove $7|(2a+b)$

Question

It looks pretty easy but I can't see a way out of this problem.

If $7|100a+b$, prove $7|2a+b$, $a, b \in \mathbb{Z}$

Answer 1

We have that

$$100a+b\equiv 0 \mod 7$$

$$\implies 100a-7*14a+b\equiv 0 \mod 7$$

$$\implies 2a+b\equiv 0 \mod 7$$

Answer 2

Note that $7|(100a+b)$ and also $7|7a$.

Hence $7|(14\times 7a) \implies 7|98a$.

Thus, we can say that $\boxed {7|(100a+b)-98a \implies 7|(2a+b)}$.