<s>[INST] Prove that if n is divisible by $2$ and $3$, then n is divisible by $6$ 
For every integer n,  if $2 | n$ and $3 | n$ then  $6 | n$

! Note: x | y means y is divisible by x.
!! Note: I know that there are way better ways to prove it. However, I am just curious whether the proof below, admittedly peculiar, is correct.
Since 2 | n and 3 | n, we can write $\frac{n}{2} = x $ and $\frac{n}{3} = y$ where $x,y \in \mathbb Z$. Therefore
$$\tag1 \frac{n}{2} + \frac{n}{3} = x + y$$
$$\tag2 \frac{5n}{6} = x + y $$
$$\tag3 5\cdot\frac{n}{6} = x + y$$
Since $x, y \in \mathbb Z$, it follows that $x + y$ in integer and $5\cdot\frac{n}{6}$ is integer is as well. Need to prove that $\frac{n}{6} \in \mathbb Z$.
Suppose $\frac{n}{6} \notin \mathbb Z$. Since $5 \cdot \frac{n}{6}$ is an integer, $\frac{n}{6}$ can be rewritten as $\frac{n}{6} = a + 0.2$, where $a \in \mathbb Z$. But then it will imply that $n = 6a + 1.2$, meaning that $n \notin \mathbb Z$, hence a contradiction. Therefore, $\frac{n}{6} \in \mathbb Z$
Is it correct? 
 [/INST] Simpler: $\ \dfrac{n}2\in\Bbb Z,\, \dfrac{n}3\in \Bbb Z\,\Rightarrow\, \dfrac{n}6 = \dfrac{n}2-\dfrac{n}3\in\Bbb Z.\ $  Turning to your argument:

Suppose $\frac{n}{6} \notin \mathbb Z$. Since $5 \cdot \frac{n}{6}$ is an integer, $\frac{n}{6}$ can be rewritten as $\frac{n}{6} = a + 0.2$, where $a \in \mathbb Z$.

This claim is unfounded.
Remark $ $ More generally $\,a,b\mid n\iff {\rm lcm}(a,b)\mid n\ $ and this can be proved as above.
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