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!

If $k = 6a+b$,
where $0 \le b \le 5$,
then
$\begin{array}\\
k(k+1)(k+2)
&= k(k^2+3k+2)\\
&= k^3+3k^2+2k\\
&= (6a+b)^3+3(6a+b)^2+2(6a+b)\\
&= (6a)^3+3((6a)^2b+6ab^2)+b^3+3(36a^2+12ab+b^2)+12a+2b\\
&= 6(6^2a^3+18a^2b+ab^218a^2+6ab+2a)+b^3+3b^2+2b\\
&= 6(6^2a^3+18a^2b+ab^218a^2+6ab+2a)+b(b+1)(b+2)\\
\end{array}
$
By directly computing
$b(b+1)(b+2)$
for $0 \le b \le 5$,
the values are all divisible by $6$
(they are
$0, 6, 24, 60, 120, 210$).
Therefore,
the product is
always divisible by $6$,
being the sum
of two terms
each of which
is divisible by $6$.
