HMMT February

First note that all the integers must be at least 2, because the greatest common divisor of any two positive integers is at least 1. Let $n$ be the largest integer in the circle. The greatest common divisor of its two neighbors is $n-1$. Therefore, each of the two neighbors is at least $n-1$ but at most $n$, so since $n-1 \nmid n$ for $n-1 \geq 2$, they must both be equal to $n-1$. Let $m$ be one of the numbers on the other side of $n-1$ from $n$. Then $\operatorname{gcd}(n, m)=n-2$. Since $n-2 \geq 0, n-2 \mid n$ only for $n=3$ or 4. If $n=3$, each number must be 2 or 3, and it is easy to check that there is no solution. If $n=4$, then it is again not hard to find that there is a unique solution up to rotation, namely 4322343223. The only possible sum is therefore 28.