Problem:
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)+⋯+E(100).
Answer Choices:
A. 200
B. 360
C. 400
D. 900
E. 2250
Solution:
Since E(100)=E(00), the result is the same as E(00)+E(01)+E(02)+ E(03)+⋯+E(99), which is the same as
E(00010203…99).
There are 200 digits, and each digit occurs 20 times, so the sum of the even digits is 20(0+2+4+6+8)=20(20)=400.