Problem Solving

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

GmatKiss wrote: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

Hello!

On this type of question you always want to look for a pattern. In this case, let's separate the units and tens digits.

Overall, we have 10 sets of numbers here: 0-9, 10-19, 20-29, ..., 90-99.

In each set of 10, we have 1 of each units digit, so that's 10 of each.
For each set of 10, we have 1 of the 9 digits 10 times (e.g. from 10-19 we have "1" in the tens digit 10 times). So, that's another 10 of each.

Accordingly, we have 20 of each digit from 1-9.

If we add up 1-9, we get 45. 20*45=900... choose C!