Consider 6 variables $a,b,c,x,y,z\in\mathbb Z$. We have two ratios, $a:b:c=1:2:3$ and $x:y:z=1:2:3$. We also have that $\gcd(a,x)=2$. What is $\gcd(a+b+c,x+y+z)$? I know that $\gcd(a,x)=2$, which means $a$ and $x$ are even, but I'm not too sure how to use this information to solve the problem. I also tried to use the ratios in some way, but I couldn't get far. I noted that $y=2x$, $z=3x$. $b=2a$, and $c=3a$, so $\gcd(a+b+c,x+y+z)=\gcd(a+2a+3a,x+2x+3x)=\gcd(6a,6x)$, but I'm not sure how to continue from here.
Could someone point me in the right direction to solve this problem? Thanks.
Let's roll out the tanks.
$b = 2a; c=3a; y=2x;z=3x$.
So $a+b + c =6a$ and $x+y+z=6x$.
So you are asked to find the $\gcd(6a,6x)$ give that $\gcd(a,x)=2$.
Can you take it from there?