HMMT 二月 2015 · 代数 · 第 1 题

1. Let Q be a polynomial n Q ( x ) = a + a x + · · · + a x , 0 1 n where 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 4 a ≤ 4 for each i . Thus, a = 2. Also, since 5 > 152 = Q (5), a , a , . . . , a = 0. 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 1 2 3 a + a + a = 2 1 2 3 to get a + 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

1