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Prove that the expression $$\frac{\gcd(m,n)}{n} \binom{n}{m} \tag{1}$$ is an integer for all pairs of integers $n \geqslant m \geqslant 1.$

This problem appeared in the famous William Lowell Putnam Mathematical Competition on December $2, 2000.$ The solution I have seen uses Bézout's identity and is quite straightforward - for example, see this question and the solution here. I wish to know if it is possible to realize the expression in $(1)$ as the size of some finite group $G$. Since there are no restrictions besides $n \geqslant m\geqslant 1$, it is hard to believe that quotient groups are the way to go (if that were the case, I would expect a divisibility relation between $m$ and $n$.)

I apologize if this question is not concrete enough, but I trust you are equally interested in knowing if the expression in $(1)$ is the size of some finite combinatorial structure, such as a finite group.

Bill Dubuque
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stoic-santiago
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    I wish to know if it is possible to realize the expression in $(1)$ as the size of some finite group $G$ - Yes, this is trivially possible. Take the cyclic group of this order, so $C_{f(n,m)}$. I suppose, it is more helpful to use that the dimension of the exterior algebra $\bigwedge^m \mathbb{R}^n$ is the binomial coefficient $\binom{n}{m}$. – Dietrich Burde Jun 01 '24 at 10:10
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    @DietrichBurde That is not the point of the question! The point is to show that $\frac{\gcd(m,n)}{n} \binom{n}{m} \in \Bbb Z$ using the existence of a finite group of size $|G| = \frac{\gcd(m,n)}{n} \binom{n}{m}$. Does that make sense? – stoic-santiago Jun 01 '24 at 10:14
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    Yes, this makes sense. So you definitely should correct this sentence then, which I have cited, and clarify this in your post, not only in the comments. My comment is more about the idea, not to use groups, but algebras. But of course, the solution in B-2 of your link is so easy, that one hardly can do better. This is what elementary number theory is made for. – Dietrich Burde Jun 01 '24 at 10:15
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    Actually, your idea has been tried already, see for example this post, with the comments by anon. But it didn't work. It looks like that the elementary number theory proofs are the best ones. One could also look for a combinatorial proof, without using groups. – Dietrich Burde Jun 01 '24 at 10:28

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