So I'm unsure as to how to prove this:
If $a \mid b$ and $b \mid c$, then $a \mid (b+c)$.
I'm aware of the divisibility properties such as: if $a \mid b$, then $b=ak$ for some integer $k$.
I also know the Transitivity of Divisibility: Let $a$, $b$, and $c$ be integers. If $a \mid b$ and $b \mid c$, then $a \mid c$.
Any help as to how to approach this implication, or even some hints would be really appreciated.
Thanks.
As you said, $a|c$, that means $c = ap$ for some integer $p$. Then:
$$b+c = ak+ap = a(k+p)$$
As both $k,p$ are integers, $k+p$ is an integer. By definition of divisibility, $a|b+c$