Let $m$ and integer and $d$ divisor of $m$. How to prove that $\gcd$ of certain numbers is $m/d$?

I'm trying to prove something about divisilibity and got stuck for long hours in the following:

All the integers mentioned below are $\geq 0$.

Let $q$ and $m$ be integers and let $d$ be a divisor of $m$.

Show that if $q<d$ and $\operatorname{gcd}(q,d)=1$, then $\operatorname{gcd}(q\cdot\frac{m}{d},m)=\frac{m}{d}$.

My attempt:

Well, clearly $\frac{m}{d}$ divides both. Suppose that there is a $k>\frac{m}{d}$ such that $k$ divides both.

I'm trying to conclude then that $k$ must divide $\frac{m}{d}$, but the fact that $\frac{m}{d}$ and $q$ are not necessarily coprime is making things really difficult. By doing some examples and seeing what was going on, I came up with the formula $$m-q\cdot\frac{m}{d}=\frac{m}{d}(d-q),$$ which gives the question for $q=d-1$.

How can I finish this proof? Thanks.

• pls. correct the title – G Cab Apr 7 '16 at 17:11
• What is wrong? Can you be specific? – Shoutre Apr 7 '16 at 17:13

Since $\gcd(q,d)=1$, there are $x,y \in \mathbb Z$ such that $qx+dy=1$.
Multiplying both sides by $\dfrac md$,we get $q\dfrac md x +my = \dfrac md$.
This proves that $\dfrac md$ is a multiple of $\gcd(q\dfrac md,m)$.
Since $\dfrac md$ is a common divisor of $q\dfrac md$ and $m$, it must divide $\gcd(q\dfrac md,m)$.
Therefore, $\dfrac md=\gcd(q\dfrac md,m)$.