elementary number theory - Show via direct proof that $k(k+1)(k+2)$ is divisible by $6$.

How do I show via direct proof that $k(k+1)(k+2)$ is divisible by $6$. I showed it was divisible by $2$ because at least one of the multiples is even but could not figure out how to show it is divisible by $3$. I tried making $k$ even or odd and substituting $2q$ or $2q+1$ but have not made much progress. Does anyone have any tips as to what direction I should take? Thanks!

Answer

By the division algorithm, $k$ divided by $3$ yields a remainder of $0$, $1$, or $2$. In other words, there are some integers $q,r$ such that $k=3q+r$ where $r=0,1,$ or $2$.

If $r=0$, then $k=3q$ is divisible by $3$. If $r=1$, then $k+2=(3q+1)+2=3(q+1)$ is divisible by $e$. If $r=2$, then $k+1=(3q+2)+1=3(q+1)$ is divisible by $3$. Therefore, in all cases, at least one of $k$, $k+1$, and $k+2$ is divisible by $3$.