HMMT 二月 2019 · 冲刺赛 · 第 6 题

HMMT February 2019 — Guts Round — Problem 6

[ 4 ] The pairwise products ab, bc, cd, and da of positive integers a, b, c, and d are 64 , 88 , 120 , and 165 in some order. Find a + b + c + d.

解析

[ 4 ] The pairwise products ab, bc, cd, and da of positive integers a, b, c, and d are 64 , 88 , 120 , and 165 in some order. Find a + b + c + d. Proposed by: Anders Olsen Answer: 42 The sum ab + bc + cd + da = ( a + c )( b + d ) = 437 = 19 · 23 , so { a + c, b + d } = { 19 , 23 } as having either pair sum to 1 is impossible. Then the sum of all 4 is 19 + 23 = 42 . (In fact, it is not difficult to see that the only possible solutions are ( a, b, c, d ) = (8 , 8 , 11 , 15) or its cyclic permutations and reflections.)