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.
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+1 and thank you so much for providing context! This site has been around for a while and many of the common questions have already been asked, so unfortunately there is a duplicate of your question. – Toby Mak May 13 '19 at 13:55
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Well I would prefer to know if there is a way to answer this question without the use of induction, i.e., only with the use of congruences. – user208480 May 13 '19 at 13:59
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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 $2\cdot 3=6$.
Dave
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3@Arthur The OP pointed out that $$(7n+30)(13n+7)(n+2)\equiv n(n+1)(n+2)\pmod 6$$ – Dave May 13 '19 at 14:01