HMMT 二月 2026 · ALGNT 赛 · 第 9 题
HMMT February 2026 — ALGNT Round — Problem 9
题目详情
- Compute ∞ ( ) ∑ −⌊ 101 k/ 1 ⌋ −⌊ 101 k/ 2 ⌋ −⌊ 101 k/ 100 ⌋ 2 + 2 + · · · + 2 . k =1
解析
- Compute ∞ ( ) ∑ −⌊ 101 k/ 1 ⌋ −⌊ 101 k/ 2 ⌋ −⌊ 101 k/ 100 ⌋ 2 + 2 + · · · + 2 . k =1 ©2026 HMMT Proposed by: Albert Wang, Isaac Zhu, Karthik Venkata Vedula ( ) 1 Answer: 50 1 − 101 2 − 1 Solution: For all positive integers n , let f ( n ) be the number of pairs ( k, j ) of positive integers with ⌊ ⌋ 101 k k ≥ 1 and j ∈ { 1 , 2 , . . . , 100 } such that = n . We show that: j • f ( n ) = 100 when 101 | n , • f ( n ) = 0 when 101 | n + 1 , • f ( n ) = 50 otherwise. The answer is then ( ) ∑ ∑ ∑ ∑ f ( n ) 50 50 50 1 = − + = 50 1 − . n n 101 n − 1 101 n 101 2 2 2 2 2 − 1 n ≥ 1 n ≥ 1 n ≥ 1 n ≥ 1 ⌊ ⌋ 101 k First, = n is equivalent to j 101 k ( n + 1) j nj n + 1 > ≥ n ⇐⇒ > k ≥ . j 101 101 For fixed j and n , note that there is at most one integer k satisfying the above inequality. We now take cases on n modulo 101 . ( n +1) j nj nj • 101 | n . Here, is an integer and < + 1 , thus for each j ∈ { 1 , 2 , . . . , 100 } there is 101 101 101 exactly one integer k satisfying the above inequality. In this case, f ( n ) = 100 . ( n +1) j nj • 101 | n + 1 . Since < − 1 , it is clear no such k exists, so f ( n ) = 0 . 101 101 • 101 ∤ n ( n + 1) . We prove the following claim. Claim 1. Suppose 101 ∤ n ( n + 1) , and let j ∈ { 1 , 2 , . . . , 100 } . Then, there exists an integer [ ) ( n +1) j nj in the interval I = , if and only if there does not exist an integer in the interval 1 101 101 [ ) n (101 − j ) ( n +1)(101 − j ) I = , . 2 101 101 [ ) ( n +1) j nj Proof. Rewrite the interval I = n − , n + 1 − . Let a, b ∈ { 0 , 1 , 2 , . . . , 100 } be the 2 101 101 reductions of nj and ( n + 1) j modulo 101 , respectively. We have a ̸ = b since 0 < j < 101 , and nj +101 − a both a and b are nonzero since 101 ∤ n ( n + 1) . By definition, is the smallest integer 101 nj nj − a nj greater than , and n − is the smallest integer greater than n − . 101 101 101 ( n +1) j − b nj +101 − a If a > b , then = is the only integer in I , and there are no integers in I . 1 2 101 101 ( nj − a ) ( n +1) j − b Analogously, if a < b , then n − = n − is the only integer in I , and there are no 2 101 101 integers in I . The claim follows. 1 By the above claim, exactly one integer j from each pair { i, 101 − i } for i ∈ { 1 , 2 , 3 , . . . , 50 } has ⌊ ⌋ 101 k the property property that some k satisfies = n . Thus, f ( n ) = 50 . j