$29 \cdot 59$ Let $P(x)$ be a polynomial. When $P(x)$ is divided by $x-19$, the remainder is 99, and when $P(x)$ is divided by $x-99$, the remainder is 19. When $P(x)$ is divided by $(x-19)(x-99)$, the remainder is
(A) $-x+80$.
(B) $x+80$.
(C) $-x+118$.
(D) $x+118$.
(E) 0.
(50th American High School Mathematics Examination, 1999)
[Solution] Let $p(x)=(x-19)(x-99) Q(x)+a x+b$, where $Q(x)$ is a polynomial of degree 2 less than that of $p(x)$, and $a, b$ are constants to be determined.
Given $p(19)=99, p(99)=19$, we have
$$
\left\{\begin{array}{l}
19 a+b=99 \\
99 a+b=19 .
\end{array}\right.
$$

Solving these equations, we get $a=-1, b=118$.
Thus, $-x+118$ is the remainder when $p(x)$ is divided by $(x-19)(x-99)$. Therefore, the answer is $(C)$.