Consider the set of all integers from
0 to 99 written as a 3-digit numbers, i.e. adding leading zeroes to numbers
below 100. We get 10^3 triplets with all
decimal digits equally appearing, averaging 4.5 .
Thus a sum of 3000*4.5 = 13500 is reached.
In the next thousand each triplet is
prefixed by a digit 1, rending the accrued sumof 14500.
Add 90 (accounting for 2000-2014) to
reach the final result:13,500+14,500+90 = 28090.
solution
