HMMT 二月 2015 · 代数 · 第 1 题

英文解析

1. Let Q be a polynomialn
   Q ( x ) = a + a x + · · · + a x ,
   0 1 nwhere a , . . . , a are nonnegative integers. Given that Q (1) = 4 and Q (5) = 152, find Q (6).
   0 n
   Answer: 254 Since each a is a nonnegative integer, 152 = Q (5) ≡ a (mod 5) and Q (1) = 4 = ⇒
   i 0
   a ≤ 4 for each i . Thus, a = 2. Also, since 5 > 152 = Q (5), a , a , . . . , a = 0.4
   i 0 4 5 n
   Now we simply need to solve the system of equations
   2 2 3 3
   5 a + 5 a + 5 a = 150
   2 31
   a + a + a = 2
   1 2 3
   to geta + 6 a = 7 .
   2 3
   3 2
   Since a and a are nonnegative integers, a = 1, a = 1, and a = 0. Therefore, Q (6) = 6 + 6 + 2 =
   2 3 2 3 1
2. 1