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 $(x-19)(x-99)$?
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$
According to the problem statement, there are polynomials $Q(x)$ and $R(x)$ such that $P(x) = Q(x)(x-19) + 99 = R(x)(x-99) + 19$.
From the last equality we get $Q(x)(x-19) + 80 = R(x)(x-99)$.
The value $x=99$ is a root of the polynomial on the right hand side, therefore it must be a root of the one on the left hand side as well. Substituting, we get $Q(99)(99-19) + 80 = 0$, from which $Q(99)=-1$. This means that $99$ is a root of the polynomial $Q(x)+1$. In other words, there is a polynomial $S(x)$ such that $Q(x)+1 = S(x)(x-99)$.
Substituting this into the original formula for $P(x)$ we get 
\[P(x) = Q(x)(x-19) + 99 = (S(x)(x-99) - 1)(x-19) + 99 =\]
\[= S(x)(x-99)(x-19) - (x-19) + 99\]
Therefore when $P(x)$ is divided by $(x-19)(x-99)$, the remainder is $\boxed{-x + 118}$.