1. Compute the sum of all possible distinct values of $m+n$ if $m$ and $n$ are positive integers such that
$$
\operatorname{lcm}(m, n)+\operatorname{gcd}(m, n)=2(m+n)+11
$$
Answer: 32
Solution: Let $d=\operatorname{gcd}(m, n)$. We consider the following cases:
(a) Suppose $d=1$. Then the equation becomes $m n+1=2 m+2 n+11$ or $(m-2)(n-2)=14$. As $m$ and $n$ are coprime in this case, exactly one of them is even. This implies that exactly one of the factors $m-2$ and $n-2$ of 14 is even. Thus, we have $(m-2, n-2) \in\{(14,1),(7,2),(2,7),(1,14)\}$ so that $(m, n) \in\{(16,3),(9,4),(4,9),(3,16)\}$. The sum of all possible values of $m+n$ is $19+13=32$.
(b) Suppose $d \geq 2$. As $d \mid \operatorname{lcm}(m, n)$, we see that $d$ divides $2(m+n)+11$. But $d$ divides both $m$ and $n$, so $d \mid 11$. This forces $d=11$. Plugging $m=11 x$ and $n=11 y$ into the given equation, where $\operatorname{gcd}(x, y)=1$, we have $\operatorname{lcm}(11 x, 11 y)+\operatorname{gcd}(11 x, 11 y)=2(11 x+11 y)+11$, which is equivalent to
$$
11 \operatorname{cm}(x, y)+11=2(11 x+11 y)+11 \Longrightarrow x y=2(x+y) \Longrightarrow(x-2)(y-2)=4 \text {. }
$$

Exactly one of the factors $x-2$ and $y-2$ of 4 is even. We then get $(x-2, y-2) \in\{(4,1),(1,4)\}$ and $(x, y) \in\{(6,3),(3,6)\}$, contradicting $\operatorname{gcd}(x, y)=1$.

Hence, the sum of all possible values of $m+n$ satisfying $\operatorname{lcm}(m, n)+\operatorname{gcd}(m, n)=2(m+n)+11$ is 32.