8) Little Rita plays this game: for every integer between 10 and 99 inclusive, she subtracts the units digit from the tens digit and writes the result on a piece of paper (for example, for 21 she writes 1, i.e., $2-1$, while for 37 she writes -4, i.e., $3-7$). At the end, she sums all the numbers she has written on the paper; what result does she find?
(A) 0,
(B) -30
(C) $45, \quad$ (D) -50
(E) 100.
8. The answer is $\mathbf{( C )}$.

Consider a two-digit number between 10 and 99 (inclusive) that does not have 0 as its second digit. Let $D$ be the tens digit and $U$ be the units digit of this number. For this number, Rita writes down $D-U$ on the paper. If the two digits are the same, then $D-U=0$. If the two digits are different, the number with the digits swapped, i.e., $U$ as the tens digit and $D$ as the units digit, is also between 10 and 99, and for this number, Rita writes down $U-D$, which is the opposite of the first number. The two numbers $D-U$ and $U-D$ written by Rita on the paper and added together will give a zero contribution to the final sum. Therefore, to find the required sum, we can consider only the multiples of ten between 10 and 90. It is easy to calculate that the sum of the contributions from these numbers is 45.

[Problem proposed by X.Y. Lu.]