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I've tried this for very long yet can't get much progress. I thought of letting $n=ab$ for $1<a<b<n$ then ended up with $(n-1)!=(n-1)...(n-a)(n-b)...2*1=2(n-1)...(ab-a)(ab-b)...1=2(n-1)...a(b-1)b(a-1)$ which is the same point as other solutions on aops, (credits to huashiliao2020) from which I copied and pasted the latex. Everything was going great till I realized that when splitting our original question into factors such as $n$ and $2$ how can we be sure $(n-1)!$ contains enough factors of $2$ for the claim to hold and such that overlaps in factors of $2$ in $n$ don't matter? It seems obvious but I'm truly not sure how to do it.

Also other solutions would be welcome with strict number theory, thank you.

This is 2006 JBMO problem #1.

Bill Dubuque
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Mathisgreat
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    the answer to "how can we be sure (−1)! contains enough factors of 2 for the claim to hold and such that overlaps in factors of 2 in don't matter?" is that $n > 4$ – Aniruda Suswaram Jun 11 '25 at 09:40
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    Note: Writing $0<a<b<ab$ excludes squares. For instance, if $p$ is prime, then $n=p^2$ can't be written that way. – lulu Jun 11 '25 at 10:10
  • Oh yeah I see thanks, but n>4 isn't necessarily strong enough for every case, some grumpy graders might dock for it. Yes it is trivial, but proof by Legendre's or something similar might be better. – Mathisgreat Jun 11 '25 at 12:20
  • Wait @BillDubuque you said wlog a>2, yet in the factorization we have $b$ as greater than $2$ as well right. What if either does equal $2$ in this instance though. Or does this still work either way. – Mathisgreat Jun 11 '25 at 14:46
  • Since if $b=2$, then it clear that the $(a+1)(a+2)$ factor still works and if b>2, then clearly it trivially works similarly? – Mathisgreat Jun 11 '25 at 14:52
  • Tweak the proofs in the dupe, e.g. for this one, if $,\color{#c00}a>\color{darkorange}2,$ then we get an extra factor of $,\color{darkorange}2,$ as here $,\color{darkorange}2:!\color{#c00}a:!\color{#0a0}b\mid 1!\cdot! \color{darkorange}2\cdots\color{#C00} a:\color{#0A0}{(a!+!1), (a!+!2) \cdots (a!+!b)}\cdots (\color{#90f}{ab!-!1})=(n!-!1)!,,$ else $,a=2,,$ so if $,b\ge 4,$ the green product has at least $4$ terms so at least $2$ even terms so we get the extra factor of $,2,$ from one of them. $\ \ $ – Bill Dubuque Jun 11 '25 at 15:25
  • "how can we be sure (n−1)! contains enough factors of 2 for the claim to hold and such that overlaps in factors of 2 in n don't matter?" You only need one. If $2 < a$ you have it. If $2=a$ then we just need there to exist one even number that is not equal to $b$. If $b=3$ then we can get it from $4$. If $b\ge 3$ then there are at least $4$ even number up to $n=ab=2b$. $2$ is one of them. $b$ is another and $n=2b$ is a third. But there has to be a fourth. – fleablood Jun 11 '25 at 15:52
  • You can easily exhaust all cases. case 1: $n=ab=p^2; a=b=p> 2;p\ prime$. case 2: $n=ab; 2< a<b$. case 3 $n=ab=2b;a=2 2<b; b\ne 4$. and case 4: $n=ab=8; a=2,b=4$. That's all four possibilities and they all in their own way deal with the potential doubling of the twos (and the first deals with the doubling of the $a$ which you hadn't considered but @lulu pointed out) – fleablood Jun 11 '25 at 16:06
  • @fleablood Why repeat my prior comment? (I presumed obvious the case $,a,b = 2,3)\ \ $ – Bill Dubuque Jun 11 '25 at 16:07

1 Answers1

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One perhaps cumbersome way to argue is by dividing up in cases. First let us assume that $n=ab$ with$2<a<b<n$ then as in your computation $$(n-1)!=1 \cdots 2 \cdots a \cdots b\cdots (n-1)$$ hence $2n=2ab$ divides $(n-1)!$.

Now if $2=a<b<n$ instead, then $2n=4b$ and we can check the cases $b\leq 4$ which is when $n=6,8$ by hand.

If $2=a$ and $4<b<n$ then $$(n-1)!=1 \cdot 2 \cdot 3 \cdot 4 \cdots b \cdots (n-1).$$ Hence $2n=4b$ divides $(n-1)!$.

If $2<a=b<n$ so $n=a^2$ then $2a<n=a^2$ hence $$(n-1)!=1 \cdot 2 \cdots a \cdots 2a \cdots (n-1)$$ therefore $2n=2a^2$ divides $(n-1)!$.

Ito Yukis
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