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Theorem. For every integer n, $6 \vert n$ $\ \text iff$ $2 \vert n$ and $3 \vert n$. 

Proof. 
Let $n$ be an arbitrary integer.

$(\rightarrow)$ Suppose $6 \vert n$. $\exists k \in \mathbb{Z}(6k = n)$, so $3(2k) = n$ and $n \vert3$. Likewise, $6k = 2(3k) = n$, so $2 \vert n$. Thus, $2 \vert n$ and $3 \vert n$.

$(\leftarrow)$ Suppose $n \vert 3$ and $n \vert 2$. $\exists j \in \mathbb{Z} (3j = n)$ and $\exists k \in \mathbb{Z} (2k = n)$. Thus $6(k - j) = 6k - 6j = 3(2k) - 2(3j) = 3n - 2n = n$. Becaus $k - j$ is an integer, it follows that $n \vert 6$.

$\therefore$ $\forall n \in \mathbb{Z}$, $6 \vert n$ $\ \text iff$ $2 \vert n$ and $3 \vert n$.

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