Find the greatest common divisor of $ \frac {12\gcd(a,b)}{d}$ and $\frac cd $

Question

Given that the greatest common divisor of $a$, $b$ and $c$ is $d$. And 3 doesn't divide $\frac cd$, $\frac cd$ is an even integer, $\frac {c}{2d}$ is an odd integer. Find the greatest common divisor of $\frac {12\gcd(a,b)}{d}$ and $\frac cd$

I've been struggling with this question for a while that appeared on my elementary number theory course in college and I initially thought that I should follow and use the definition of the greatest common divisor ($d=ax+by$) but now it seems to me like i should try and find some relations about $ \frac {12\gcd(a,b)}{d}$ and $\frac cd $ but i'm not quite sure how to connect the dots.

My initial guess was that, it is trivial to show $ \frac {12\gcd(a,b)}{d}$ is a multiple of $12$ but from that I'm lost. I did initially thought that $\frac cd $ is in the forms $12k+2$, $12k+6$, $12k+10$,(by using the given properties of the fraction) which meant I'd have to find a general insight of the greatest common divisor of those forms and the multiple of $12$ or $12k$ form which I'm yet to be wise of.

Answer 1

Simplify the problem by writing $$\gcd(a,b)=de,\quad c=2df.$$ The hypothesis become $$\gcd(e,2f)=1\quad\text{and}\quad\gcd(f,6)=1.$$

Now, $\gcd(e,2f)=1\implies\gcd(e,f)=1,$ and $$\gcd(f,6)=1\implies\gcd(f,6e)=\gcd(f,e)$$

so that $$\gcd(12e,2f)=2\gcd(6e,f)=2.$$

Answer 2

Hint: if $\text{gcd}(a, b, c) = 1$, the only prime numbers that can divide simultaneously $12 a, 12b, c$ are 2 and 3.