Proving $k$ is a common divisor of $a$ and $b$ iff $k\,|\gcd(a,b)$ -- converse

Question

I'm currently trying to solve the converse of this statement is true after proving the normal version is true. If $k$ is a common divisor of $a$ and $b$, then $k \,| \gcd(a, b)$

So far I know the converse states that if $k \,| \gcd(a, b)$ then $k$ is a common divisor of $a$ and $b$.

Can anyone help me out and thanks!

Answer 1

I'll start you off. Let $d = \text{gcd}(a,b)$. Then by definition $d|a$ and $d|b$. Now suppose $k|d$. Can you use what you know about division to prove the result?