返回题库

HMMT 二月 2021 · ALGNT 赛 · 第 3 题

HMMT February 2021 — ALGNT Round — Problem 3

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Among all polynomials P ( x ) with integer coefficients for which P ( − 10) = 145 and P (9) = 164, compute the smallest possible value of | P (0) | .
解析
  1. Among all polynomials P ( x ) with integer coefficients for which P ( − 10) = 145 and P (9) = 164, compute the smallest possible value of | P (0) | . Proposed by: Carl Schildkraut Answer: 25 Solution: Since a − b | P ( a ) − P ( b ) for any integer polynomial P and integers a and b , we require that 10 | P (0) − P ( − 10) and 9 | P (0) − P (9). So, we are looking for an integer a near 0 for which a ≡ 5 mod 10 , a ≡ 2 mod 9 . The smallest such positive integer is 65, and the smallest such negative integer is − 25. This is achiev- 2 able, for example, if P ( x ) = 2 x + 3 x − 25, so our answer is 25.