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I'm new to number theory and was wondering if someone could help me with this proof. Prove: The product of any three consecutive integers is divisible by 6.

So far I have x(x+1)(x+2)/6 how would I go about proving this? should I replace x with k and then k with k+1 and see if the statement is true?

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Convince yourself that one of those three numbers is divisible by $3$, and at least one is divisible by $2$. –  user61527 Feb 11 '14 at 22:28

3 Answers 3

Of $n$, $n +1$, $n +2$, one must be even, so divisible by 2 (why?). One must be divisible by 3 (why?). So their product must be divisible by $2 \times 3$ (why?) ...

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Hint $\displaystyle\ \ n(n\!+\!1)(n\!+\!2)\, =\, 6 { n+2 \choose 3}$

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Hint: Note that the product of two consecutive integers is divisible by $2$ because one of them is even. Note then that the product of three consecutive integers is divisible by $3$ (this about it). Now $2$ and $3$ are prime, so the prodcut is divisible by $2\cdot 3 = 6$.

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I see what you are saying, Is there any way to write it as a formal proof or no? –  Lil Feb 11 '14 at 22:30
@Lil: I almost wrote down the formal proof. You just have to provide the justification for the claim that the product of three consecutive integers is divisible by $3$. –  Thomas Feb 11 '14 at 22:31
is the any way to justify that the product of two consecutive integers, x(x+1) is divisible by 2 using mathematical induction and replacing x with k+1? –  Lil Feb 11 '14 at 22:32
@Lil: Note that if you are given $x$, then either $x$ is even or $x+1$ is even, so ... –  Thomas Feb 11 '14 at 22:33

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