HMMT 二月 2025 · ALGNT 赛 · 第 4 题

HMMT February 2025 — ALGNT Round — Problem 4

Let ⌊ z ⌋ denote the greatest integer less than or equal to z . Compute 1000 X 2025 . j + 0 . 5 j = − 1000 √ √

解析

Let ⌊ z ⌋ denote the greatest integer less than or equal to z . Compute 1000 X 2025 . j + 0 . 5 j = − 1000 Proposed by: Linus Yifeng Tang Answer: − 984 j k 2025 2025 Solution: The key idea is to pair up the terms and . There are 1000 such pairs and one − x x 2025 lone term, = 2. Thus, 1000 . 5 1000 X X 2025 2025 2025 = 2 + + . j + 0 . 5 x − x j = − 1000 x ∈{ 0 . 5 , 1 . 5 ,..., 999 . 5 } We note that ( 0 if a is an integer. ⌊ a ⌋ + ⌊− a ⌋ = − 1 otherwise. Therefore, ( 2025 2025 0 if 2 x divides 4050

= x − x − 1 otherwise . As x ranges in the set { 0 . 5 , 1 . 5 , 2 . 5 , . . . , 999 . 5 } , 2 x ranges in the set { 1 , 3 , 5 , . . . , 1999 } . This set includes j k 2025 2025 all 15 odd divisors of 4050 except for 2025. Thus, there are 14 values of x for which + x − x evaluates to 0, and the remaining 1000 − 14 = 986 values of x make it evaluate to − 1. Therefore, 1000 X X 2025 2025 2025 = 2 + + = 2 + 986 · ( − 1) = − 984 . j + 0 . 5 x − x j = − 1000 x ∈{ 0 . 5 , 1 . 5 ,..., 999 . 5 } √ √