1. 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 $Q(2)$.
Answer:
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 $Q(x)=2+2 x+2 x^{3}+2 x^{4}+x^{5}$, so $Q(2)=86$.