HMMT-Nov 2021 team, Problem 2

Joey wrote a system of equations on a blackboard, where each of the equations was of the form $a+b=c$ or $a \cdot b=c$ for some variables or integers $a, b, c$. Then Sean came to the board and erased all of the plus signs and multiplication signs, so that the board reads: \[ \begin{array}{ll} x & z=15 \\ x & y=12 \\ x & x=36 \end{array} \] If $x, y, z$ are integer solutions to the original system, find the sum of all possible values of $100 x+10 y+z$.

Solution: Solution: The bottom line gives $x=-6, x=6$ or $x=18$. If $x=-6, y$ can be -2 or 18 and $z$ must be 21 , so the possible values for $100 x+10 y+z$ are -599 and -399 . If $x=6, y$ can be 2 or 6 and $z$ must be 9 , so the possible values are 629 and 669 . If $x=18, y$ must be -6 and $z$ must be -3 , so the only possible value is 1737 . The total sum is 2037 . $\fbox{2037}$.

Answer: 2037