PUMaC 2023 · 代数（A 组） · 第 6 题
PUMaC 2023 — Algebra (Division A) — Problem 6
题目详情
- A polynomial p ( x ) = a x with real coefficients is called mountainous if n ≥ 2 and there j j =1 exists a real number k such that the polynomial’s coefficients satisfy a = 1, a − a = k for 1 j +1 j 1 ≤ j ≤ n − 1, and a − a = − k for n ≤ j ≤ 2 n − 2; we call k the step size of p ( x ). A real j +1 j number k is called good if there exists a mountainous polynomial p ( x ) with step size k such b that p ( − 3) = 0. Let S be the sum of all good numbers k satisfying k ≥ 5 or k ≤ 3. If S = c for relatively prime positive integers b, c , find b + c .
解析
- A polynomial p ( x ) = a x with real coefficients is called mountainous if n ≥ 2 and there j j =1 exists a real number k such that the polynomial’s coefficients satisfy a = 1, a − a = k for 1 j +1 j 2 1 ≤ j ≤ n − 1, and a − a = − k for n ≤ j ≤ 2 n − 2; we call k the step size of p ( x ). A real j +1 j number k is called good if there exists a mountainous polynomial p ( x ) with step size k such b that p ( − 3) = 0. Let S be the sum of all good numbers k satisfying k ≥ 5 or k ≤ 3. If S = c for relatively prime positive integers b, c , find b + c . Proposed by Sunay Joshi Answer: 101 7 61 We claim that the only good values of k are k = and , corresponding to n = 2 and n = 3 3 12 89 respectively. This yields S = and an answer of 101. 12 To see this, note that a generic mountainous polynomial p ( x ) can be written as 2 n n 2 x − x ( x − 1) p ( x ) = (1 − k ) + kx 2 x − 1 ( x − 1) n 2 2 n ( x − 1) x − x 2 2 n − 1 if x ̸ = 1. This follows from the observation that = x + x + . . . + x and = 2 x − 1 ( x − 1) n − 1 n − 2 2 2 n n +1 2 n − 2 ( x + x + . . . + 1) = x + 2 x + . . . + nx + ( n − 1) x + . . . + x . Hence p ( x ) = 0 n 2 2 n ( x − 1) x − x implies that (1 − k ) + kx = 0. Rearranging and solving for k , we find 2 x − 1 ( x − 1) 1 n x + − 2 n x k = 1 − 1 n − 1 x + − 2 n − 1 x As n → ∞ , k = k ( n ) tends to 1 − x . In our case x = − 3, so the limit equals 4. It follows that there are only finitely many n such that | k − 4 | ≥ 1. Calculating k ( n ) for n = 2 , 3 , 4, we find k (2) = 7 / 3, k (3) = 61 / 12. We claim that for n ≥ 4, | k ( n ) − 4 | < 1, so that n = 2 , 3 are the only valid cases. Note that 8 8 + n ( − 3) | k ( n ) − 4 | = 1 n − 1 ( − 3) + − 2 n − 1 ( − 3) We split into the cases when n is even ( n ≥ 4) and n is odd ( n ≥ 5). If n is even, then 8 8 + n 3 | k ( n ) − 4 | = 1 n − 1 3 + + 2 n − 1 3 1 1 2 n n 2 The inequality | k ( n ) − 4 | < 1 is equivalent to 3 − 6 · 3 − 5 > 0, i.e. x − 6 x − 5 > 0 for 3 3 x ≥ 81, which is true. If n is odd, then 8 8 − n 3 | k ( n ) − 4 | = 1 n − 1 3 + − 2 n − 1 3 1 2 n n 1 2 The inequality | k ( n ) − 4 | < 1 is equivalent to 3 − 10 · 3 + 11 > 0, i.e. x − 10 x + 11 > 0 3 3 for x ≥ 243, which is true. The result follows.