HMMT 二月 2022 · ALGNT 赛 · 第 3 题

HMMT February 2022 — ALGNT Round — Problem 3

Let x , x , . . . , x be nonzero real numbers. Suppose that x + < 0 for each 1 ≤ k ≤ 2022, 1 2 2022 k x k +1 where x = x . Compute the maximum possible number of integers 1 ≤ n ≤ 2022 such that x > 0. 2023 1 n

解析

Let x , x , . . . , x be nonzero real numbers. Suppose that x + < 0 for each 1 ≤ k ≤ 2022, 1 2 2022 k x k +1 where x = x . Compute the maximum possible number of integers 1 ≤ n ≤ 2022 such that x > 0. 2023 1 n Proposed by: Akash Das Answer: 1010 Solution: Let the answer be M . If M > 1011 , there would exist two consecutive positive terms 1 x , x which contradicts the assumption that x + < 0. Thus, M ≤ 1011. If M = 1011 , then k k +1 k x k +1 the 2022 x s must alternate between positive and negative. WLOG, assume x > 0 and x < 0 for i 2 k − 1 2 k each k . Then, we have 1 x + < 0 = ⇒ | x x | < 1 , 2 k − 1 2 k − 1 2 k x 2 k 1 x + < 0 = ⇒ | x x | > 1 . 2 k 2 k 2 k +1 x 2 k +1 Q 2022 Multiplying the first equation over all k gives us | x | < 1 , while multiplying the second equation i i =1 Q 2022 over all k gives us | x | > 1 . Thus, we must have M < 1011. i i =1 M = 1010 is possible by the following construction: 1 1 1 1 , − , 3 , − , . . . , 2019 , − , − 10000 , − 10000 . 2 4 2020