The product of three consecutive natural numbers is divisible by $6$ Please give me feedback for my answer to this question.

Prove or ﬁnd a counterexample: The product of any three consecutive
  natural numbers is divisible by $6$

My answer: True. Suppose $n$ is a natural number, such that the $3$ consecutive natural numbers is $n, n+1, n+2$. Then, 
$$\begin{align}\frac{n(n+1)(n+2)}{6} 
&= \frac{1(1+1)(1+2)}{6} \\
&= \frac 66 \\
&= 1\end{align}$$
Thus, $n(n+1)(n+2)$ is divisble by 6.

You have proved that $1\times2\times3$ is divisible by six, not that the product of any 3 consecutive natural numbers is divisible by $6$.
If a number is divisible by $6$, then it must be divisible by both $2$ and $3$. Your product is $$n(n+1)(n+2)$$ so you could try showing that at least one of $n$, $n+1$ or $n+2$ is a multiple of 3, and at least one is even.
