HMMT 二月 2026 · ALGNT 赛 · 第 8 题

HMMT February 2026 — ALGNT Round — Problem 8

Let a , a , a , . . . be the unique sequence of nonnegative integers less than 397 with a = 1 and 0 1 2 0 2 a ( a + 1) ≡ a (mod 397) n +1 n n for all nonnegative integers n . Given that a = 9 , compute the remainder when a + a + · · · + a 2026 0 1 2026 is divided by 397 .

解析

Let a , a , a , . . . be the unique sequence of nonnegative integers less than 397 with a = 1 and 0 1 2 0 2 a ( a + 1) ≡ a (mod 397) n +1 n n for all nonnegative integers n . Given that a = 9 , compute the remainder when a + a + · · · + a 2026 0 1 2026 is divided by 397 . Proposed by: Tiger Zhang Answer: 279 Solution: We work in modulo 397 , and we will freely use division. (Formally, we will use the notation a − 1 − 1 to refer to ab , where b is the multiplicative inverse of b modulo 397 . Also, all equalities are in b modulo 397 .) First, we rewrite the given equation to 2 a ( a + 1) = a n +1 n n 2 a a + 2 a a + a = a n +1 n +1 n n +1 n n 1 1 a + 2 + = n a a n n +1 1 1 a = − − 2 . n a a n +1 n In particular, we have that 1 1 a = − − 2 0 a a 1 0 1 1 a = − − 2 1 a a 2 1 . . . 1 1 a = − − 2 , 2026 a a 2027 2026 so adding them all gives 1 1 a + a + · · · + a = − − 2 · 2027 . 0 1 2026 a a 2027 0 a 9 2026 Hence, since a = = , we have 2027 2 ( a +1) 100 2026 100 a + a + · · · + a = − 1 − 2 · 2027 0 1 2026 9 = 364 − 1 − 84 = 279 .