How to show that $a$ can be divided by $6$ if and only if it can be divided by both $2$ and $3$? Prove that for: $a \in\mathbb Z$, $a$ is divisible by 2 and $a$ is divisible by 3 if and only if $a$ is divisible by 6.
EDIT: Sorry, I wasn't aware of how exactly this site worked. This is pretty much a question from my Discrete Math course that I was having a little trouble understanding. Apologies for not being clear about the background and only posting a question with no context.
 A: Hint: We write $a\mid b$ for "$a$ divides $b$".


*

*If $p$ is a prime number and $p\mid a\cdot b$ then either $p\mid a$ or $p\mid b$; 

*if $x\mid y$ and $y\mid z$ then $x\mid z$ as well.

*$2$ and $3$ are prime numbers; $6=2\cdot 3$.

A: Hint: $\ \dfrac{a}6\, =\, \dfrac{a}2\, -\, \dfrac{a} 3$
A: $\Rightarrow$
Assume a is divisible by $2$ and $3$. By Euclid's algorithm
$$
a=6q+r
$$
where $0\le |r|\le 5$
But $r=a-6q$ and since each of $6$ and $a$ are divisible by both $2$ and $3$, $r$ is divisible by both $2$ and $3$ and so must be $0$.
Therefore, $a$ is divisible by $6$.
$\Leftarrow$
If $a$ is divisible by $6$ then $a=6n$ for some $n$ so $$a=6n=2(3n) = 3(2n)$$
, therefore $a$ is divisible by both $2$ and $3$.
A: Hint:  $(2)(3) = 6$
LCM of 2 and 3 is 6.
This doesn't act as a proof, but it should help you make sense of it.
A: If $(p,q)=1$, then there exist $u,v$ such that $up+vq=1$. If we now suppose that $p$ and $q$ divide $a$, then we may write $ps=a=qt$. Multiplying the equation $up+vq=1$ by $a$ we obtain $upa+vqa=a$ or $upqt+vqps=a$. Hence $pq$ divides $a$.
Clearly if we can $a$ by $pq$ then we can divide $a$ by $p$ and $q$.
