Why is a number that is divisible by $2$ and $3$, also divisible by $6$? Why is it that a number such as $108$, that is divisible by $2$ and $3$, is also divisible by $6$? 
Is this true that all numbers divisible by two integers are divisible by their product?

Let $a$ be divisible by $2$ and $3$. So $$a=2^{r_{0}}\cdot p^{r_{1}}_{1}\cdot p^{r_{2}}_{2}\cdot\ldots=3^{s_{0}}\cdot q^{s_{1}}_{1}\cdot q^{s_{2}}_{2}\cdot\ldots$$ where $p_{i}$ and $q_{j}$ are distinct prime numbers, $r_{i}$ and $s_{j}$ are possibly $0$ for $1\leqslant i,j<\infty$ and $r_{0},s_{0}\geqslant 1$. We know from the fundamental theorem that these two must be unique factorisations (up to re-ordering). Hence there must be some $p_{k}^{r_{k}}=3^{s_{0}}$ and some $q_{\ell}^{s_{\ell}}=2^{r_{0}}$. Therefore $a$ has a factor of $6$: \begin{align*}
a&=2^{r_{0}}\cdot p_{1}^{r_{1}}\ldots\cdot p_{k}^{r_{k}}\cdot\ldots\\
 &=2^{r_{0}}\cdot p_{1}^{r_{1}}\ldots\cdot3^{s_{0}}\cdot\ldots\\
 &=6\cdot(2^{r_{0}-1}\cdot3^{s_{0}-1}\cdot p_{1}^{r_{1}}\cdot\ldots).
\end{align*} (The argument works on the other side as well).
This might be a little explicit but I hope it helps.
