Since the divisor $(x-19)(x-99)$ is a quadratic, the degree of the remainder is at most linear.  We can write $P(x)$ in the form $$ P(x)=Q(x)(x-19)(x-99)+c x+d $$ where $c x+d$ is the remainder.  By the Remainder Theorem, plugging in 19 and 99 gives us a system of equations:  $$ \begin{aligned} \& 99 c+d=19 \textbackslash\textbackslash{} \& 19 c+d=99\end{aligned} $$  Solving gives us $c=-1$ and $d=118$, thus, our answer is $(\text{C})-x+118$. 
Let $P(x)$ be a polynomial such that 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. What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$? (1999 AHSME Problems, Question \#17) 

(1) $-x+80$ 
(2) $x+80$ 
(3) $-x+118$ 
(4) $x+118$ 
(5) $$0$$ 