Prove the following
For any integer a, 14|a if and only if 2|a and 7|a.
I am stuck on how to prove this for the backwards implication case, that is if 2|a and 7|a then 14|a.
I get a=2k1 and a=7k2 for integers k1,k2,and a. Not sure where to go from there (multiply?)
I'll assume you can prove the first direction of the implication. For the second, note that if $2 \mid a$ and $7\mid a$, then $\exists x, y \in \mathbb{Z}$ s.t. $a = 2x = 7y$, which means that $x = 7y/2$, or $x = 7 \cdot (y/2)$. For this to be an integer, $y/2$ must be an integer, so we must have that for some $t \in \mathbb{Z}$, $y = 2t$.
In this case $a = 7y = 14t$, and we can conclude.