Divisibility by 45 implies divisibility by 15

Statement

Prove that if $45$ divides an integer $n$, then $15$ also divides $n$.

Solution

Since $45 = 15\cdot 3$, from $45\mid n$ we get $n = 45\,k = 15\,(3k)$ for some integer $k$. Hence $n$ is a multiple of $15$, i.e. $15\mid n$.