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? 

Suppose $n$ is a multiple of 6. Then we are done. Suppose $n$ is neither a multiple of 2 or 3. Then $n+1$ must be even, and if $n+1$ is still not a multiple of 3 then $n+5$ is. Note $n+5\equiv n+2$ (mod 3) and since neither $n$ nor $n+1$ is a multiple of 3, then $n+2$ is. If $n$ is even but not a multiple of 3, we proceed the same way to show either $n+1$ or $n+2$ is. The only case left is when $n$ is an odd multiple of 3 then $n+1$ is even and we are done. In any case, we get a factor of 2 and a factor of 3.
