Calculate $\gcd(2a+4b, 2a +8b)$, if $a\equiv b \pmod{\! 5},\ 6a+11b = 5$ Given:

*

*$a$ is even


*$6a+11b=5$


*$a-b=0\pmod 5$
Q: Calculate $\gcd(2a+4b,2a+8b)$

My try:
We know there is some $i$ such that $a=2i$, plus from 3 we know there is some $j$ such that: $a-b=5j$ which means $b=a-5j=2i-5j$. From 2, we get: $34i-55j=5$
So, $$\gcd(2a+4b,2a+8b)=\gcd(12i-20j,20i-40j).$$
I'm stuck here, how to continue?

This helps with the last step after @Buraian's answer.
Now $\gcd(a,b) \in \{1,5\}$ because there is an integral linear combination of $a$ and $b$ that sums to $5$; in particular, $6a+11b=5$.
Meanwhile, as $a$ is even, it follows that $\frac{a}{2}$ is an integer, and as $\gcd(2,5)=1$, if $5|a$ then $5|\frac{a}{2}$. [In fact, let $k$ be any integer satisfying both $\gcd(k,5)=1$ and $k|a$, such as $k=2$. Then as $\gcd(a,b) \in \{1,5\}$, it follows that $\gcd\left(\frac{a}{k},b\right) = \gcd(a,b)$.] So from this it follows that $\gcd\left(\frac{a}{2},b\right)=\gcd(a,b)$.
So let us now calculate $\gcd(a,b)=\gcd\left(\frac{a}{2},b\right)$. Then from the first answer by @Buraian, $4\times \gcd(a,b) = \gcd(2a+4b,2a+8b)$.
Now, we claim that both $a$ and $b$ divide $5$. [Indeed, let $a \pmod 5 = r$. Then $b \pmod 5 =r$ as well, by the condition that $5|(a-b)$. However, on the one hand, (a) $6a+11b \pmod 5$ is $a \pmod 5 + b\pmod 5$ which is $2r$. As, on the other hand, (b) $6a+11b = 5 \equiv_5 0$ by hypothesis, it follows from putting (a) and (b) together that $2r \equiv_5 0$, so $r$ must be $0$, and thus indeed, both $a$ and $b$ divide $5$.] So $\gcd(a,b)$ must be a multiple of $5$. However, as noted already, $\gcd(a,b)$ is either $1$ or $5$, and so $\gcd(a,b)$ must indeed be exactly $5$.
As noted above, $\gcd(2a+4b,2a+8b)=4\gcd(a,b)$, so $\gcd(2a+4b,2a+8b)$ must be $4 \times 5 = 20$.
