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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?

BLAZE
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Lil
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    Convince yourself that one of those three numbers is divisible by $3$, and at least one is divisible by $2$. –  Feb 11 '14 at 22:28
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    See also: http://math.stackexchange.com/questions/527300/prove-that-6-divides-nn-1n-2 – Martin Sleziak Dec 04 '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 $2\cdot 3 = 6$.

Thomas
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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
  • 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 $n$? – EthanAlvaree Apr 15 '15 at 23:34
  • @EthanAlvaree: Consider a list of three consecutive integers: $m$, $m+1$, $m+2$. Now dividing $n$ by $3$ will give you a unique remainder $0\leq r < 3$. That is, $r= 0, 1$, or $2$. Now you can just do the three cases. If $r= 0$, then $n$ is divisible by $3$. If $r=1$, then $n+2$ is divisible by $3$. If $r=2$, then $n+1$ is divisible by $3$. The general case you can do likewise with some type of induction argument on the remainder. – Thomas Apr 16 '15 at 17:11
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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?) ...

Peter Smith
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  • 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 $n$? – EthanAlvaree Apr 15 '15 at 23:04
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    @Mathemanic let the product be P=n(n+1)(n+2). Now n=3m or 3m+1 or 3m+2. In these cases P= 3m(3m+1)(3m+2) or (3m+1)(3m+2)(3)(m+1) or (3m+2)(3)(m+1)(3(m+1)+1) respectively. Henceforth the proof is self evident. – aditya gupta Apr 04 '20 at 06:26
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Hint $\displaystyle\ \ n(n\!+\!1)(n\!+\!2)\, =\, 6 { n+2 \choose 3}$

Bill Dubuque
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