Suppose $a \in \mathbb{Z}.$ Prove that $5 \mid 2^na$ implies $5 \mid a$ for any $n \in \mathbb{N}$ This question is supposed to be solved by induction, however I'm unsure of where to get my base case from exactly, because the question is asking about both $a$ and $n$. I started with my base case being $n = 1$, but then I get $5\mid 2a$, and I'm unsure what to do from there. Am I just barking up the wrong tree here? Or how should I go about getting a start to this problem?

$1)$ assume $5\nmid a$
$2)$ by the Division Algorithm $a=5q+r$, $\space\space r\lt 5$,$r<q\space\space\space\space\space\space\space 1)$
$3)$ $2^na=2^n(5q+r)=2^n(5q)+2^nr=5(2^nq)+2^nr$,  $\space\space r<5\space\space\space\space\space\space\space 2)$
$4)$ $5\nmid 2^na\space\space\space\space\space\space\space1),2),3)$
$5)$ $F_0$(contradiction)$\space\space\space\space\space\space\space 4)$
$6)$ $5\nmid a\rightarrow F_0\space\space\space\space\space\space\space 5)$
$7)$ $5|a\space\space\space\space\space\space\space 6)$
