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? 

One number among $n$, $n+1$, $n+2$ is a multiple of 3.
Since $5 \pmod 3 \equiv 2$, among $n$, $n+1$ and $n+5$, one is a multiple of 3.
Either $n$ or $n+1$ must be an even number (since they are consecutive numbers).
Therefore, n(n+1)(n+5) is a multiple of 2.
As $n(n+1)(n+5)$ is a has both 3 and 2 as factors, it is a multiple of 6. 
