HMMT 二月 2024 · 冲刺赛 · 第 12 题

HMMT February 2024 — Guts Round — Problem 12

[7] Compute the number of quadruples ( a, b, c, d ) of positive integers satisfying 12 a + 21 b + 28 c + 84 d = 2024 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2024, February 17, 2024 — GUTS ROUND Organization Team Team ID#

解析

[7] Compute the number of quadruples ( a, b, c, d ) of positive integers satisfying 12 a + 21 b + 28 c + 84 d = 2024 . Proposed by: Rishabh Das Answer: 2024 ′ Solution: Looking at the equation mod 7 gives a ≡ 3 (mod 7) , so let a = 7 a + 3 . Then mod 4 gives ′ ′ b ≡ 0 (mod 4) , so let b = 4 b . Finally, mod 3 gives c ≡ 2 (mod 3) , so let c = 3 c + 2 . Now our equation yields ′ ′ ′ ′ ′ ′ 84 a + 84 b + 84 c + 84 d = 2024 − 3 · 12 − 2 · 28 = 1932 = ⇒ a + b + c + d = 23 . ′ ′ ′ Since a, b, c, d are positive integers, we have a and c are nonnegative and b and d are positive. Thus, ′′ ′ ′ ′ ′′ ′ ′ let b = b + 1 and d = d + 1 , so a , b , c , d are nonnegative integers summing to 21 . By stars and ( ) 24 bars, there are = 2024 such solutions. 3