HMMT 二月 2026 · 冲刺赛 · 第 4 题

HMMT February 2026 — Guts Round — Problem 4

[5] Over all nonnegative integers a , b , c , and d such that ab + cd = 31 and ac + bd = 29 , compute the minimum possible value of a + b + c + d . ©2026 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2026, February 14, 2026 — GUTS ROUND Organization Team Team ID# ◦

解析

[5] Over all nonnegative integers a , b , c , and d such that ab + cd = 31 and ac + bd = 29 , compute the minimum possible value of a + b + c + d . Proposed by: Derek Liu Answer: 17 Solution: Let x = a + d and y = b + c . Adding the two given equations gives us xy = 60 , and subtracting them gives us ( a − d )( b − c ) = 2 . The latter tells us one of a − d and b − c is ± 1 and the other is ± 2 ; in particular, they have opposite parity. Since x and a − d differ by 2 d , they have the same parity. Likewise, y and b − c have the same parity. Thus, x and y have opposite parity. Since xy = 60 , the smallest possible sum of a + b + c + d = x + y is 5 + 12 = 17 . To show this is achievable, we note that a + d = 5 , a − d = 1 , b + c = 12 , and b − c = 2 should give us a solution, and indeed ( a, b, c, d ) = (3 , 7 , 5 , 2) works. ◦