Prove that integer not divisible by $2$ or $3$ is not divisible by $6$

Question

How to prove that any integer $n$ which is not divisible by $2$ or $3$ is not divisible by $6$?

The point was to prove separately inverse, converse and contrapositive statements of the given statement: "for all integers $n$, if $n$ is divisible by $6$, then $n$ is divisible by $3$ and $n$ is divisible by $2$". I have the proof for converse and inverse similar to that given in comments. I have trouble only with the proof that integer not divisible by $2$ or $3$ is not divisible by $6$.

As I review my proof for inverse statement, I'm not sure of it as well. "For all integers $n$, if $n$ is not divisible by $6$, $n$ is not divisible by $3$ or $n$ is not divisible by $2$."

$n = 6x$ where $x$ is not an integer
$n = 2\cdot 3\cdot x$
$n/2 = 3x$ and $n/3 = 2x$ where $2x$ or $3x$ is not an integer,
so $n$ is not divisible by $2$ or $3$

Answer 1

If $6\mid n$, then $n=6\cdot k$ for some $k$. So $n=2\cdot 3\cdot k$. Thus $n$ is divisible by $2$ and $3$.

Answer 2

The contra positive statement is that every multiple of $6$ is a multiple of $2$ and a multiple of $3$

Can you prove the contrapositive?