The product of the two $99$-digit numbers
$303,030,303,...,030,303$ and $505,050,505,...,050,505$
has thousands digit $A$ and units digit $B$. What is the sum of $A$ and $B$?
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$
We can first make a small example to find out $A$ and $B$. So, 
$303\times505=153015$
The ones digit plus thousands digit is $5+3=8$.
Note that the ones and thousands digits are, added together, $8$. (and so on...) So the answer is $\boxed{\textbf{(D)}\ 8}$
This is a direct multiplication way.