18. Let $P(x)$ be a polynomial with integer coefficients such that $P(-4)=5$ and $P(5)=-4$. What is the maximum possible remainder when $P(0)$ is divided by 60 ?
18. 41
Solution.
$$
\begin{array}{c}
0-(-4) \mid P(0)-P(-4) \text { or } 4 \mid P(0)-5, \text { so } P(0) \equiv 5(\bmod 4) \\
5-0 \mid P(5)-P(0) \text { or } 5 \mid-4-P(0), \text { so } P(0) \equiv-4(\bmod 5)
\end{array}
$$

By the Chinese Remainder Theorem, there is a solution $r$ that satisfies both of the previous equations, and this solution is unique modulo $4 \cdot 5=20$. It is easy to verify that this solution is 1 . Thus, $P(0) \equiv 1(\bmod 20)$. This implies that $P(0)$ can be 1,21 , or $41(\bmod 60)$. The largest remainder 41 is indeed achievable, for example by the polynomial $-2(x+4)(x-5)+1-x$.