Use congruence to show that the output of a natural number function is always divisible by 6

Question

I have been asked to show that for any natural number $n$, $(7n+30)(13n+7)(n+2)$ is divisible by 6. I can show that this function is congruent to $n(n+1)(n+2)$. I noticed that this function is the sequence of tetrahedral numbers multiplied by 6 which already proves it but I have been told by a tutor that introducing a sequence that is not discussed in the course is insufficient. I am not sure if I am able to use induction as the question only asks for me to use congruences to show that it is divisible by 6. I would like to know if this is possible without induction in a way that specifically uses modulo congruence techniques.

Answer 1

You know that in any triple of consecutive integers, say $n$, $n+1$, and $n+2$, at least one is divisible by $2$ and at least one is divisible by $3$. Since $2$ and $3$ are coprime, the product $n(n+1)(n+2)$ is divisible by $2\cdot 3=6$.