How can we prove that a binomial coefficient $n\choose m$is divisible by the ratio of $n$ and $\gcd(n,m)$?

Prove that for every $n\geq m \geq1$ natural numbers, the following number is an integer:

$${\gcd(n,m)\over n}\cdot{n\choose m}$$

Where $\gcd$ is the greatest common divisor.

I tried to make it simpler by cancelling the $n$ from the left side, and making it $(n-1)!$ on the right: $\gcd(n, m) \cdot \frac{(n-1)!}{m!(n-m)!}$, but can't really go further.

This was problem B-2 on the 2000 Putnam exam.

Write $nx+my=\gcd(n,m)$, with $x,y\in\mathbb{Z}$

Then:

$$\frac{\gcd(n,m)}{n}\binom{n}{m}=\frac{nx+my}{n}\binom{n}{m}=x\binom{n}{m}+y\,\frac{m}{n}\binom{n}{m}=x\binom{n}{m}+y\binom{n-1}{m-1}\in\mathbb{Z}$$

Here is a conceptual way to derive this. We will show that it is a special case of the well-known $\rm\color{#0a0}{Lemma}$ that if a fraction $q\,$ can be written with denominators $\,n\,$ and $\,m,\,$ then it can also be written with denominator $\,\gcd(n,m),\,$ i.e.

$$\begin{align} nq,\ mq\in\Bbb Z\,\ &\Rightarrow\, (n,m)q \in\Bbb Z,\ \text{ so for }\ \color{#c00}q = \frac{1}{n}{n\choose m}\\ n\color{#c00}q = {n\choose m}^{\vphantom{|^{|^|}}}\in\Bbb Z,\,\ m\color{#c00}q= {n\!-\!1\choose m\!-\!1}\in\Bbb Z\,\ &\Rightarrow\ (n,m)q = \dfrac{(n,m)}n{{{n\choose m}}}\in \Bbb Z\end{align}\qquad$$

Remark $ $ The $\rm\color{#0a0}{Lemma}$ is a denominator form of this ubiquitous group theory theorem:

$\qquad$ If $\,q^m\! = 1 = q^n\,$ then $\,q^{(m,n)}=1,\ $ by $\ {\rm ord}(q)\mid m,n\Rightarrow {\rm ord}(q)\mid (m,n)$

The least denominator of a fraction is its order in $\,\Bbb Q/\Bbb Z,\,$ so the Lemma is a special case of this result. For more on this viewpoint (denominator and order ideals) see here

$\rm\color{#0a0}{Proof}$: for completeness here are a few proofs of the Lemma. Recall $\,(x,y):=\gcd(x,y)$

$(1)\ $ Recall that a fraction can be written with denominator $\,n\,$ iff its least denominator $\,d\mid n.\,$ Therefore $\,m,n\,$ are denoms $\iff d\mid m,n\iff d\mid (m,n)\iff (m,n)\:$ is a denom.

$(2)\ \ \dfrac{mc}d,\dfrac{nc}d\in\Bbb Z\iff d\mid mc,nc\iff d\mid (mc,nc)=(m,n)c\iff\! \dfrac{(m,n)c}d\in\Bbb Z$

$(3)\ \ \dfrac{mc}d, \dfrac{nc}d\in\Bbb Z\,\Rightarrow \dfrac{jmc}d,\, \dfrac{knc}d\in\Bbb Z\,\Rightarrow\,\dfrac{(jm\!+\!kn)c}d\,\overset{\large \color{#c00}{\exists\, j,k}_{\phantom{1^{1^{1}}\!\!\!\!\!}}} = \dfrac{(m,n)c}d\in\Bbb Z\ $ by $\rm\color{#c00}{Bezout}$