HMMT 二月 2026 · ALGNT 赛 · 第 10 题
HMMT February 2026 — ALGNT Round — Problem 10
题目详情
- Let ( ) 2026 ∑ 2 k k S = k 2 . k k =0 Compute the remainder when S is divided by 2027 . (Note that 2027 is prime.) ©2026 HMMT
解析
- Let ( ) 2026 ∑ 2 k k S = k 2 . k k =0 Compute the remainder when S is divided by 2027 . (Note that 2027 is prime.) ©2026 HMMT Proposed by: Jacopo Rizzo, Pitchayut Saengrungkongka Answer: 289 Solution: We show, more generally, that for any prime p > 2 , ( ) p − 1 ∑ p − 3 2 k k 2 k 2 ≡ 4 · ( − 7) mod p. k k =0 This implies with p = 2027 , 2027 − 3 2 S ≡ 4 · ( − 7) ≡ 289 mod 2027 . ( ) p − 1 2 k It suffices to only consider the sum from k = 0 to as all larger terms have divisible by p . 2 k Claim 1. Over F [ x ] , p p − 1 ( ) 2 ∑ p − 1 2 k k 2 (1 − 4 x ) = x . k k =0 (Here, we take F [ x ] as the set of polynomials with coefficients taken modulo p .) p Proof. All polynomial equalities below are taken over F [ x ] . First, by the Binomial Theorem, p p − 1 ( ) 2 p − 1 ∑ p − 1 k 2 2 (1 − 4 x ) = ( − 4 x ) . k k =0 p − 1 ( ) 2 Each can then be reduced mod p as follows: k ( ) ( ) ( ) k p − 1 p − 1 p − 1 1 1 − 2 k k ( ) · . . . · ( − k + 1) ( − ) · . . . · ( ) ( − 1) (2 k − 1)!! 1 2 k 2 2 2 2 2 = ≡ ≡ ≡ − mod p. k k k ! k ! 2 · k ! 4 k The claim immediately follows. Now, take the formal derivative of both sides in the above identity to obtain p − 1 ( ) 2 ∑ p − 3 p − 1 2 k k − 1 2 · ( − 4) · (1 − 4 x ) = kx . 2 k k =1 In particular, plugging x = 2 , we get ( ) p − 3 p − 1 2 S = 2 · · ( − 4) · (1 − 4 · 2) , 2 p − 3 2 thus S ≡ 4 · ( − 7) mod p . Remark. One can avoid using formal derivatives by noting that ( ) ( ) ( ) 1 3 2 k − 1 − 2 2 k ≡ k ≡ − · mod p k k 2 k − 1 p − 3 2 and proceeding with the polynomial (1 − 4 x ) . ©2026 HMMT