PUMaC 2022 · 团队赛 · 第 9 题
PUMaC 2022 — Team Round — Problem 9
题目详情
- In the complex plane, let z , z , z be the roots of the polynomial p ( x ) = x − ax + bx − ab . 1 2 3 4 4 4 Find the number of integers n between 1 and 500 inclusive that are expressible as z + z + z 1 2 3 for some choice of positive integers a, b . 3 2
解析
- In the complex plane, let z , z , z be the roots of the polynomial p ( x ) = x − ax + bx − ab . 1 2 3 4 4 4 Find the number of integers n between 1 and 500 inclusive that are expressible as z + z + z 1 2 3 for some choice of positive integers a, b . Proposed by Sunay Joshi Answer: 51 3 2 4 2 2 2 For all j ∈ { 1 , 2 , 3 } , we have z = az − bz + ab . Multiplying by z , we find z = ( a − b ) z + a b . j j j j j j P P 2 2 4 4 2 Summing over j and using the fact that z = a − 2 b , we find z = a + 2 b . In other j j 4 2 words, it suffices to find the number of n ∈ [1 , 500] of the form a + 2 b for a, b ≥ 1. We first count the total number of pairs ( a, b ) satisfying the condition. 2 4 a = 1: this implies 2 b ≤ 500 − 1 , hence b ≤ 15. This yields 15 solutions. 2 4 a = 2: this implies 2 b ≤ 500 − 2 , hence b ≤ 15. This yields 15 solutions. 2 4 a = 3: this implies 2 b ≤ 500 − 3 , hence b ≤ 14. This yields 14 solutions. 2 4 a = 4: this implies 2 b ≤ 500 − 4 , hence b ≤ 11. This yields 11 solutions. 4 2 4 2 Next, we eliminate duplicates. Note that if a + 2 b = c + 2 d , then a ≡ b (mod 2). Hence it suffices to check the cases ( a, c ) = (1 , 3) and ( a, c ) = (2 , 4). 4 2 4 2 2 2 If ( a, c ) = (1 , 3), then 1 + 2 b = 3 + 2 d , implying b − d = 40. Thus the pair ( b − d, b + d ) can either be (2 , 20) or (4 , 10). These yield b = 11 and b = 7 respectively, which correspond to the duplicate solutions n = 243 and n = 99. 4 2 4 2 2 2 If ( a, c ) = (2 , 4), then 2 + 2 b = 4 + 2 d , implying b − d = 24. Thus the pair ( b − d, b + d ) can either be (2 , 12) or (4 , 6). These yield b = 7 and b = 5 respectively, which correspond to the duplicate solutions n = 114 and n = 66. Subtracting the 4 duplicates from our original count of 55 = 15 + 15 + 14 + 11, we find our answer of 51. 3 2