TITLE: How can we prove that a binomial coefficient $n\choose m$is divisible by the ratio of $n$ and $\gcd(n,m)$?
QUESTION [17 upvotes]: 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.

REPLY [6 votes]: Here is a conceptual way to derive this. Below we show that it is a special case of the well-known $\rm\color{#0a0}{Lemma}$ below that if a fraction $q\,$ can be written with denominators $\,n\,$ and $\,m,\,$ then it can also be written with denominator being their gcd $ = (n,m).\,$ This makes the proof obvious, viz.
$$\begin{align}  nq,\ mq\in\Bbb Z\,\ &\Rightarrow\, (n,m)q \in\Bbb Z,\  \text{ so for  }\ \color{#c00}q = \frac{1}{n}{n\choose m}\\
n\color{#c00}q = {n\choose m}^{\vphantom{|^{|^|}}}\in\Bbb Z,\,\ m\color{#c00}q= {n\!-\!1\choose m\!-\!1}\in\Bbb Z\,\ &\Rightarrow\ (n,m)q = \dfrac{(n,m)}n{{{n\choose m}}}\in \Bbb Z\end{align}\qquad$$

$\rm\color{#0a0}{Lemma}\ $ If a fraction $\,q = c/d\,$ can be written with denominators $\,n\,$ and $\,m,\,$ then it can also be written with denominator being their gcd $ =(n,m)$.
Proof $\ $  We give a few  proofs of this basic result since that proves instructive.
$(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 $\rm\color{#c00}{Bezout}$
$(4)\ \ \,q = \dfrac{C}{D} = \dfrac{c}d\Rightarrow d(C,c) = (dC,dc) = (cD,dc) = c(D,d),\,$ so $\ \dfrac{c}d = \dfrac{(C,c)}{(D,d)}$

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.