Two numbers n, m satisfy $0 < |2n−3m| ≤ 20$. What is maximum possible $gcd (m, n)$? Explain your answer.
I got 20, with my explanation being that if it was greater than 20, such as 21, then 2n-3m will have to be 21 or a multiple of 21, which would violate the less than or equal to 20 rule.
Hint: One way to construct the gcd $m$ and $n$ is that it is the minimum positive element in the set $$ G:=\{am+bn:a,b\in\mathbb{Z}\}. $$ In your case, observe that $2n-3m$ and $-2n+3m$ are both in $G$. Can you use the given inequality to bound the minimum positive element of $G$?
P.S. Your answer is correct and the reasoning is OK, but it only shows that it must be less than or equal to $20$, not that $20$ is actually possible. For that, you need an example.