Talk:Factorial Divides Product of Successive Numbers

I'm afraid I don't see what the problem is, LF. At which stage in the proof is there a problem? I'm invoking a result I'm planning on putting up today (the red link), that a set of $n$ successive numbers contains a multiple of $n$. Surely that's true, and then this follows? --GFauxPas (talk) 16:58, 12 November 2014 (UTC)


 * The problem lies therein that if it contains a multiple of $2$ and one of $6$, then it does not necessarily contain one of $2 \cdot 6$... I hope that's clear enough. &mdash; Lord_Farin (talk) 17:09, 12 November 2014 (UTC)


 * But the product of the numbers surely is a multiple of $12$? --GFauxPas (talk) 17:17, 12 November 2014 (UTC)


 * Unless both factors would come from the same number (and this doesn't happen, which is what we have to prove). &mdash; Lord_Farin (talk) 17:57, 12 November 2014 (UTC)


 * Is the proof salvageable? --GFauxPas (talk) 18:09, 12 November 2014 (UTC)

I don't know yet. I'm thinking about it. &mdash; Lord_Farin (talk) 18:16, 12 November 2014 (UTC)


 * I have a proof using primes, but it's not very elegant. I tried to continue with your idea but it is hard to define how to choose one of the possible elements of $S$ in a suitable way. &mdash; Lord_Farin (talk) 19:45, 12 November 2014 (UTC)


 * Perhaps comment the current proof out, put up yours, and I'll see if I think of a way to fix it. --GFauxPas (talk) 22:34, 12 November 2014 (UTC)


 * As has been pointed out, Divisibility of Product of Consecutive Integers is a proof of the same thing:
 * $m^{\overline n} = \dfrac {\left({m + n - 1}\right)!} {\left({m - 1}\right)!} = n! \dfrac {\left({m + n - 1}\right)!} {\left({m - 1}\right)! n!} = n! \displaystyle \binom {m + n - 1} {m - 1}$
 * As $\binom {m + n - 1} {m - 1}$ is an integer (there's a proof somewhere), the result follows. --prime mover (talk) 22:55, 12 November 2014 (UTC)