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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 $\cfrac{x(x+1)(x+2)}{6}$; 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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marked as duplicate by Martin Sleziak, Daniel Fischer Dec 5 '15 at 14:07

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    $\begingroup$ Convince yourself that one of those three numbers is divisible by $3$, and at least one is divisible by $2$. $\endgroup$ – user61527 Feb 11 '14 at 22:28
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    $\begingroup$ See also: math.stackexchange.com/questions/527300/… $\endgroup$ – Martin Sleziak Dec 4 '15 at 12:52
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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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  • $\begingroup$ I see what you are saying, Is there any way to write it as a formal proof or no? $\endgroup$ – Lil Feb 11 '14 at 22:30
  • $\begingroup$ @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 $3$. $\endgroup$ – Thomas Feb 11 '14 at 22:31
  • $\begingroup$ 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? $\endgroup$ – Lil Feb 11 '14 at 22:32
  • $\begingroup$ @Lil: Note that if you are given $x$, then either $x$ is even or $x+1$ is even, so ... $\endgroup$ – Thomas Feb 11 '14 at 22:33
  • $\begingroup$ Thomas, how would one prove rigorously that in a list of three consecutive integers, one (and only one) of the three is divisible by $3$? Or in general, in a list of $n$ integers, one (and only one) of them is divisible by $n$? $\endgroup$ – Mathemanic Apr 15 '15 at 23:34
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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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  • $\begingroup$ Peter, how would one prove rigorously that in a list of three consecutive integers, one (and only one) of the three is divisible by $3$? Or in general, in a list of $n$ integers, one (and only one) of them is divisible by $n$? $\endgroup$ – Mathemanic Apr 15 '15 at 23:04
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Hint $\displaystyle\ \ n(n\!+\!1)(n\!+\!2)\, =\, 6 { n+2 \choose 3}$

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  • $\begingroup$ It's the same as asking why ${ n+2 \choose 3}$ is an integer. $\endgroup$ – Paracosmiste Dec 4 '15 at 17:51

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