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, 2017

Algebra Difficulty 4.9 AIME Prove it United States

Problem:

Let $Q(x) = a_{0} + a_{1} x + \cdots + a_{n} x^{n}$ be a polynomial with integer coefficients, and $0 \leq a_{i} < 3$ for all $0 \leq i \leq n$ .
Given that $Q(\sqrt{3}) = 20 + 17 \sqrt{3}$ , compute $Q(2)$ .

Solution

Solution:

One can evaluate
$Q(\sqrt{3}) = \left(a_{0} + 3 a_{2} + 3^{2} a_{4} + \ldots\right) + \left(a_{1} + 3 a_{3} + 3^{2} a_{5} + \ldots\right) \sqrt{3}$
Therefore, we have that
$\left(a_{0} + 3 a_{2} + 3^{2} a_{4} + \ldots\right) = 20 \text{ and } \left(a_{1} + 3 a_{3} + 3^{2} a_{5} + \ldots\right) = 17$
This corresponds to the base-3 expansions of $20$ and $17$ . This gives us that $Q(x) = 2 + 2x + 2x^{3} + 2x^{4} + x^{5}$ , so $Q(2) = 86$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.