Prove $3|n(n+1)(n+2)$ by induction I tried proving inductively but I didn't really go anywhere. So I tried:

Let $3|n(n+1)(n+2)$.
Then $3|n^3 + 3n^2 + 2n \Longrightarrow  3|(n(n(n+3)) + 2)$

But then?

Here's a somewhat tedious (but fairly elementary) way. By the division algorithm, we can write $n=3k+r$ for some integers $k,r$ with $0\le r\le 2,$ whence $$\begin{align}n(n+1)(n+2) &= (3k+r)(3k+r+1)(3k+r+2)\\ &= (3k+r)(9k^2+6kr+9k+r^2+3r+2)\\ &= 27k^3+27k^2r+27k^2+9kr^2+18kr+6k+r^3+3r^2+2r\\ &= 3(9k^3+9k^2r+9k^2+3kr^2+6kr+2k+r^2)+r^3+2r.\end{align}$$ All that's left is to show that $r^3+2r$ is a multiple of $3$ for $r=0,1,2$.
