Let $E(n)$ denote the sum of the even digits of $n$. For example, $E(5681) = 6+8 = 14$.  Find $E(1)+E(2)+E(3)+\cdots+E(100)$
$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$
The problem is asking for the sum of all the even digits in the numbers $1$ to $100$.  We can remove $100$ from the list, add $00$ to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits.  This gives the list:
$00, 01, 02, 03, ..., 10, 11, ..., 98, 99$.
There are $2\cdot 100 = 200$ digits on that list, and each digit appears $\frac{200}{10} = 20$ times.
Thus, each even digit appears $20$ times, and the sum of all the even digits is $0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400$, and the correct answer is $\boxed{C}$.