HMMT 十一月 2017 · 冲刺赛 · 第 32 题

HMMT November 2017 — Guts Round — Problem 32

[ 24 ] Let P be a polynomial with integer coefficients such that P (0) + P (90) = 2018. Find the least possible value for | P (20) + P (70) | .

解析

[ 24 ] Let P be a polynomial with integer coefficients such that P (0) + P (90) = 2018. Find the least possible value for | P (20) + P (70) | . Proposed by: Michael Tang Answer: 782 2 First, note that P ( x ) = x − 3041 satisfy the condition and gives | P (70)+ P (20) | = | 4900+400 − 6082 | =
To show that 782 is the minimum, we show 2800 | P (90) − P (70) − P (20) + P (0) for every P , since − 782 is the only number in the range [ − 782 , 782] that is congruent to 2018 modulo 2800. n n n n 0 Proof: It suffices to show that 2800 | 90 − 70 − 20 + 0 for every n ≥ 0 (having 0 = 1). n n n n 3 Let Q ( n ) = 90 − 70 − 20 + 0 , then we note that Q (0) = Q (1) = 0 , Q (2) = 2800, and Q (3) = (9 − 3 3 3 4 n n n n 7 − 2 ) · 10 = 378000 = 135 · 2800. For n ≥ 4, we note that 400 divides 10 , and 90 + 0 ≡ 70 + 20 (mod 7). Therefore 2800 | Q ( n ) for all n .