HMMT 十一月 2021 · 团队赛 · 第 2 题

HMMT November 2021 — Team Round — Problem 2

[25] Joey wrote a system of equations on a blackboard, where each of the equations was of the form a + b = c or a · 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: x z = 15 x y = 12 x x = 36 If x, y, z are integer solutions to the original system, find the sum of all possible values of 100 x +10 y + z .

解析

[25] Joey wrote a system of equations on a blackboard, where each of the equations was of the form a + b = c or a · 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: x z = 15 x y = 12 x x = 36 If x, y, z are integer solutions to the original system, find the sum of all possible values of 100 x +10 y + z . Proposed by: David Vulakh Answer: 2037 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.