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How can I prove that prove that the product of $4$ consecutive positive integers is divisible by $24$, ie for any positive integer $n$ : $n(n+1)(n+2)(n+3)$ is divisible by $24$. I've noticed that: $24$ = $2^3 * 3$

$n(n+1)(n+2)(n+3)$ is divisible by $2*2$ so by $4$ (as there are at least 2 even numbers, obvious)

$n(n+1)(n+2)(n+3)$ is divisible by $3$ (as there is the product of 3 consecutive integers, easily provable using congruences)

so I've proved that you can divide it by $12$ until now. What am I missing here?

sammy210
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    One of your terms is divisible by $4$ and another is divisible by $2$. – lulu Dec 31 '20 at 20:21
  • like lulu said if you look at this $\pmod 4$, one of them is $4k$ and another is $4k+2$, meaning $8$ divides the expression. Since $(3,8)=1$ and $3$ divides the expression as you've already shown, the result follows . – Derek Luna Dec 31 '20 at 20:23
  • Welcome to Mathematics Stack Exchange. I think you mean the product of four consecutive integers – J. W. Tanner Dec 31 '20 at 20:27
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    @J.W.Tanner edited, forgot to mention it in the title – sammy210 Dec 31 '20 at 20:28
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    Hint: $\binom{n+3}{4}$ is an integer. – Daniel Schepler Dec 31 '20 at 20:51
  • @DanielSchepler I was going to say the same - your comment has the nice property that it invites appropriate generalisation for the product of $r$ consecutive integers being divisible by $r!$ – Mark Bennet Dec 31 '20 at 21:26

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Notice that among $4$ consecutive numbers exactly one is divisible by 4 and one more divisible by 2 (say $n$ and $n+2$ or $n+1$ and $n+3$).

Notice also that among $3$ consecutive numbers exactly one is divisible by 3.

So the number is divisible by $3\cdot 8$.

nonuser
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  • Answer is very simple: Since the product of $n$ consecutive integers viz. $n(n-1)(n-2) ... 3.2.1 = n!$ is divisible by $n!$, so the product of any four consecutive integers must be divisible by $4! = 24$. – Khalid Naeem Oct 17 '21 at 15:06