HMMT 二月 2017 · ALGNT 赛 · 第 1 题

HMMT February 2017 — ALGNT Round — Problem 1

Let Q ( x ) = a + a x + · · · + a x be a polynomial with integer coefficients, and 0 ≤ a < 3 for all 0 1 n i 0 ≤ i ≤ n. √ √ Given that Q ( 3) = 20 + 17 3, compute Q (2) .

解析

Let Q ( x ) = a + a x + · · · + a x be a polynomial with integer coefficients, and 0 ≤ a < 3 for all 0 1 n i 0 ≤ i ≤ n. √ √ Given that Q ( 3) = 20 + 17 3, compute Q (2) . Proposed by: Yang Liu Answer: 86 One can evaluate √ √ 2 2 Q ( 3) = ( a + 3 a + 3 a + . . . ) + ( a + 3 a + 3 a + . . . ) 3 . 0 2 4 1 3 5 Therefore, we have that 2 2 ( a + 3 a + 3 a + . . . ) = 20 and ( a + 3 a + 3 a + . . . ) = 17 . 0 2 4 1 3 5 3 4 5 This corresponds to the base-3 expansions of 20 and 17. This gives us that Q ( x ) = 2+2 x +2 x +2 x + x , so Q (2) = 86 .