sum of consecutive integers

[PGMAT]

Any formula for solving these kind of problems?

The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8 + 1+9 + 2+0 + 2+1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

450
810
900
1000
1100


OA is C

[Anurag@Gurome]

The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8 + 1+9 + 2+0 + 2+1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

450
810
900
1000
1100

Consider all one-digit integers as two-digit integer with zero as the first digit.

Now, sum of all the unit's digit of all integers from 0 to 99 = 10*(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 10*45 = 450

Also, sum of all the tens digit of all integers from 0 to 99 = 10*(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 10*45 = 450

Hence, sum of all the digits of the integers from 0 to 99 inclusive = (sum of all the unit's digits of the integers from 0 to 99 inclusive) + (sum of all the tens digits of the integers from 0 to 99 inclusive) = (450 + 450) = 900

The correct answer is C.