<s>[INST] Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$. 
The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed? 
 [/INST] You just needed one more step.
$$\begin{align}n(n+1)(n+5) = & (n+1)(n+2)(n+6) \\ = & n^3+9 n^2+20 n+12 \\ = & (n^3+6n^2+5n)+3n^2+15n+12 \\ = & n(n+1)(n+5)+\underbrace{3n(n+5)}_{\star}+12\end{align}$$
$\bigstar$ If $n$ is even, then $3n$ is divisible by 6, otherwise $n$ is odd and $3(n+5)$ is divisible by 6.

Of course you could have done this immediately.


*

*$n(n+1)$ is divisible by $2$.

*If neither $n$ nor $n+1$ are divisible by $3$, then $n+2$ and $n+5$ will be.


*

*exactly one of the three factors, $n$, $n+1$, $n+5$ is divisible by $3$.


*Therefore $n(n+1)(n+5)$ is divisible by $2$ and $3$, and thus by $6$.

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