HMMT 二月 2025 · 冲刺赛 · 第 24 题

HMMT February 2025 — Guts Round — Problem 24

[12] For any integer x , let 2 3 100 x x x f ( x ) = 100! 1 + x + + + · · · + . 2! 3! 100! 2 A positive integer a is chosen such that f ( a ) − 20 is divisible by 101 . Compute the remainder when 2 f ( a + 101) is divided by 101 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2025, February 15, 2025 — GUTS ROUND Organization Team Team ID#

解析

[12] For any integer x , let 2 3 100 x x x f ( x ) = 100! 1 + x + + + · · · + . 2! 3! 100! 2 A positive integer a is chosen such that f ( a ) − 20 is divisible by 101 . Compute the remainder when 2 f ( a + 101) is divided by 101 . Proposed by: Pitchayut Saengrungkongka Answer: 1939 Solution 1: By the binomial theorem, n n n n − 1 n n − 1 2 ( a + 101) ≡ a + a 101 = a + 101 na (mod 101 ) . 1 2 Using this gives (all congruences are modulo 101 ) 100 n X ( a + 101) f ( a + 101) = 100! n ! n =0 100 n n − 1 X a 101 na ≡ 100! + n ! n ! n =0 100 n − 1 X a ≡ f ( a ) + 100! · 101 ( n − 1)! n =1 100 a ≡ f ( a ) + 101 f ( a ) − 100! · 101 100! ≡ f ( a ) + 101( f ( a ) − 1) 2 ≡ 20 + 101(20 − 1) = 1939 (mod 101 ) . Solution 2: The above solution can be viewed as a consequence of Hensel’s lemma as follows. Because 101 is prime, for any integer x not divisible by 101, we have that 2 99 x x ′ 100 f ( x ) = 100! 1 + x + + · · · + = f ( x ) − x ≡ f ( x ) − 1 (mod 101) . 2! 99! Clearly 101 ∤ a . Hence, by Hensel’s lemma, we get that ′ 2 f ( a + 101) ≡ f ( a ) + 101 f ( a ) ≡ 20 + 101 · 19 ≡ 1939 (mod 101 ) . 2 Remark. All possible a ’s are a ≡ 1012, 6670, and 9885 (mod 101 ). They all lead to the same answer.