HMMT-Nov 2021 guts, Problem 18

Let $x, y, z$ be real numbers satisfying \[ \frac{1}{x}+y+z=x+\frac{1}{y}+z=x+y+\frac{1}{z}=3 \] The sum of all possible values of $x+y+z$ can be written as $\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$. Find $100 m+n$.

Solution: Solution: The equality $\frac{1}{x}+y+z=x+\frac{1}{y}+z$ implies $\frac{1}{x}+y=x+\frac{1}{y}$, so $x y=-1$ or $x=y$. Similarly, $y z=-1$ or $y=z$, and $z x=-1$ or $z=x$. If no two elements multiply to -1 , then $x=y=z$. which implies $2 x+\frac{1}{x}=3$ and so $(x, y, z) \in$ $\left\{(1,1,1),\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\right\}$. Otherwise, we may assume $x y=-1$, which implies $z=3$ and $x+y=\frac{8}{3}$, whence $\{x, y, z\}=\left\{-\frac{1}{3}, 3,3\right\}$. The final answer is $(1+1+1)+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+3+3\right)=\frac{61}{6}$. $\fbox{6106}$.

Answer: 6106