HMMT 十一月 2010 · GEN1 赛 · 第 5 题

HMMT November 2010 — GEN1 Round — Problem 5

[ 5 ] A polynomial P is of the form ± x ± x ± x ± x ± x ± x ± 1. Given that P (2) = 27, what is P (3)? 2

解析

[ 5 ] A polynomial P is of the form ± x ± x ± x ± x ± x ± x ± 1. Given that P (2) = 27, what is P (3)? Answer: 439 We use the following lemma: n n − 1 n Lemma. The sign of ± 2 ± 2 ± · · · ± 2 ± 1 is the same as the sign of the 2 term. n n n − 1 Proof. Without loss of generality, let 2 be positive. (We can flip all signs.) Notice that 2 ± 2 ± n − 2 n n − 1 n − 2 2 ± · · · 2 ± 1 ≥ 2 − 2 − 2 − · · · − 2 − 1 = 1, which is positive. We can use this lemma to uniquely determine the signs of P . Since our desired sum, 27, is positive, the 6 5 4 coefficient of x must be positive. Subtracting 64, we now have that ± 2 ± 2 ± . . . ± 2 ± 1 = − 37, so the 5 6 5 4 3 2 sign of 2 must be negative. Continuing in this manner, we find that P ( x ) = x − x − x + x + x − x +1, 6 5 4 3 2 so P (3) = 3 − 3 − 3 + 3 + 3 − 3 + 1 = 439. 2