AMC8 2007 8, Problem 18

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 $B$?

Solution: 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 $\fbox{8}$ This is a direct multiplication way.

Answer: 8