I'm trying to prove that for every integer $n$, $15\mid n$ iff $3\mid n$ and $5\mid n$. The first part of this bi-conditional was easy for me to prove, but I'm having problems with the second. Here is what I have so far:
Suppose $15\mid n$. Then we can let $k$ be some integer such that $15k=n$. Since $3(5k)=15k=n, 3\mid n$. Also, since $5(3k)=15k=n$, it follows that $5\mid n$.
Now I'm stuck trying to prove it the other way. I've assumed that for some integers $k$ and $j$, $3k=n$ and $5j=n$, and am trying to come up with something of the form $15p=n$ for some integer $p$, but I can only seem to come up with fractions, which seems wrong. Any hints?