HMMT 十一月 2024 · 冲刺赛 · 第 15 题

HMMT November 2024 — Guts Round — Problem 15

[9] Compute the sum of the three smallest positive integers n for which 1 + 2 + 3 + · · · + (2024 n − 1) + 2024 n 1 + 2 + 3 + · · · + (4 n − 1) + 4 n is an integer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2024, November 09, 2024 — GUTS ROUND Organization Team Team ID#

解析

[9] Compute the sum of the three smallest positive integers n for which 1 + 2 + 3 + · · · + (2024 n − 1) + 2024 n 1 + 2 + 3 + · · · + (4 n − 1) + 4 n is an integer. Proposed by: David Wei Answer: 89 Solution: We simplify the expression as follows: (2024 n )(2024 n + 1) / 2 506 · (2024 n + 1) = (4 n )(4 n + 1) / 2 4 n + 1 506 · (506 · (4 n + 1) − 505) = 4 n + 1 506 · 505 2 = 506 − 4 n + 1 2 · 5 · 11 · 23 · 101 2 = 506 − . 4 n + 1 Thus, the expression is an integer if and only if 4 n + 1 divides 5 · 11 · 23 · 101. The smallest divisors of 5 · 11 · 23 · 101 are 1 , 5 , 11 , 23 , 55 , 101 , 253 . Since 4 n + 1 > 1 and is 1 modulo 4, the three smallest values it can take are 5, 101, and 253. Hence, the three smallest values of n are 1, 25, and 63, giving the answer of 1 + 25 + 63 = 89 .