HMMT 十一月 2013 · 冲刺赛 · 第 14 题

HMMT November 2013 — Guts Round — Problem 14

[ 9 ] Find all triples of positive integers ( x, y, z ) such that x + y − z = 100 and x + y − z = 124. √ ◦ ◦ ◦

解析

[ 9 ] Find all triples of positive integers ( x, y, z ) such that x + y − z = 100 and x + y − z = 124. Answer: (12 , 13 , 57) Cancel z to get 24 = ( y − x )( y + x − 1). Since x, y are positive, we have y + x − 1 ≥ 1 + 1 − 1 > 0, so 0 < y − x < y + x − 1. But y − x and y + x − 1 have opposite parity, so ( y − x, y + x − 1) ∈ { (1 , 24) , (3 , 8) } yields ( y, x ) ∈ { (13 , 12) , (6 , 3) } . 2 Finally, 0 < z = x + y − 100 forces ( x, y, z ) = (12 , 13 , 57). √ ◦ ◦ ◦