Currently I'm working on a proof to prove that the product of any four consecutive integers is divisible by 8. In trying to figure it out I found a similar question online and in its proof states "If 3 divides $n$, then 3 divides the product $n$($n$+1)($n$+2)." I cant seem to understand exactly why this is true, can someone explain the concept or theorem behind this?
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2Welcome to Math Stack Exchange. If $a$ divides $b$ then $a$ divides $bc$ (Use the definition of divides) – J. W. Tanner Nov 11 '19 at 16:53
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4It is worth pointing out that for $n$ natural, regardless of whether or not $3$ divides $n$ you will have that $3$ will divide $n(n+1)(n+2)$. This can be seen as a direct consequence of the quotient-remainder theorem noting that exactly one of $n, n+1, n+2$ will have to be a multiple of three making the product of these a multiple of three as well. This is analogous to how two consecutive integers will be such that one is even and the other odd, here just the result when talking about how close to being a multiple of three or not something is rather than $2$. – JMoravitz Nov 11 '19 at 16:57
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And see this for a generalization. – Jyrki Lahtonen Nov 12 '19 at 04:37
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For a bit more you can use binomial coefficients: $$n(n+1)(n+2)=6\cdot\binom{n+2}3.$$ That binomial coefficient is an integer, so this number is divisible by six, hence also by three. – Jyrki Lahtonen Nov 12 '19 at 04:38
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Maybe that could be helpful for you: If $n=3k$ then you can write $$n(n+1)(n+2) = 3 \cdot \underline{k(n+1)(n+2)}$$ Look that underlined numbers are also equal to some $p$ so $$ 3 \cdot \underline{k(n+1)(n+2)} = 3 \cdot \underline{p}$$ so $$n(n+1)(n+2) = 3p $$
mvxxx
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If $3$ divides $n$ then $n=3k$ and our product is given by $$3k(3k+1)(3k+2)$$ is clearly divisible by $3$
Dr. Sonnhard Graubner
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$3$ divides $n$ means $n=3k$ for some integer $k$.
Therefore $n(n+1)(n+2)=3k(n+1)(n+2)=3j, $ where $j=k(n+1)(n+2)$;
that means $3$ divides $n(n+1)(n+2)$.
J. W. Tanner
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