HMMT 二月 2023 · 冲刺赛 · 第 27 题
HMMT February 2023 — Guts Round — Problem 27
题目详情
- [20] Suppose m > n > 1 are positive integers such that there exist n complex numbers x , x , . . . , x for 1 2 n which k k k • x + x + · · · + x = 1 for k = 1, 2, . . . , n − 1; 1 2 n n n n • x + x + · · · + x = 2; and 1 2 n m m m • x + x + · · · + x = 4. 1 2 n Compute the smallest possible value of m + n . ◦ ◦
解析
- [20] Suppose m > n > 1 are positive integers such that there exist n complex numbers x , x , . . . , x 1 2 n for which k k k • x + x + · · · + x = 1 for k = 1, 2, . . . , n − 1; 1 2 n n n n • x + x + · · · + x = 2; and 1 2 n m m m • x + x + · · · + x = 4. 1 2 n Compute the smallest possible value of m + n . Proposed by: Rishabh Das Answer: 34 n ∑ k Solution: Let S = x , so S = S = · · · = S = 1, S = 2, and S = 4. The first of these k 1 2 n − 1 n m j j =1 n n − 1 conditions gives that x , . . . , x are the roots of P ( x ) = x − x − c for some constant c . Then 1 n n − 1 n x = x + c , and thus i i 2 = S = S + cn = 1 + cn, n n − 1 1 so c = . n S j k − n Thus, we have the recurrence S = S + . This gives S = 2 + for 0 ≤ j ≤ n − 1, and then k k − 1 n + j n n 2 2 j +1 j + j 1 S = 3 + . Then S = 3 + + for 0 ≤ j ≤ n − 1. In particular, S > 4, so we have 2 n 2 n + j 2 3 n − 1 n n 2 n m ∈ [2 n, 3 n − 1]. Let m = 2 n + j . Then 2 2 j + 1 j + j 2 2 3 + + = 4 = ⇒ 2 n − 2 n (2 j + 1) − ( j + j ) = 0 . 2 n 2 n 2 2 2 2 Viewing this as a quadratic in n , the discriminant 4(2 j +1) +8( j + j ) = 24 j +24 j +4 = 4(6 j +6 j +1) 2 must be a perfect square, so 6 j + 6 j + 1 is a square. Then 2 2 2 2 2 2 6 j + 6 j + 1 = y = ⇒ 12 j + 12 j + 2 = 2 y = ⇒ 3(2 j + 1) − 2 y = 1 . The case j = 0 gives n = 1, a contradiction. After this, the smallest j that works is j = 4 (and y = 11). Plugging this back into our quadratic, 2 2 2 n − 18 n − 20 = 0 = ⇒ n − 9 n − 10 = 0 , so n = 10. Then m = 2 n + j = 24, so m + n = 34. ◦ ◦