HMMT 二月 2017 · ALGNT 赛 · 第 1 题
HMMT February 2017 — ALGNT Round — Problem 1
题目详情
英文原题
- 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
英文解析
- 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 .