Question

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?

Answer 1

Answer:

$R=900$

Step-by-step explanation:

You can add all the digits one by one, or count how many of each digit you have in total, for example: how many digits one are in the digits of the integers from 0 to 99:

First, the ones that are in the units: 1,11,21,31,41,51,61,72,81,91, there are 10 in total (not counting the digit one in the decimals  in 11).

Then the ones that are in the decimals:

10,11,12,13,14,15,16,17,18,19 they are 10 in total (not counting the digit one in the units in 11).

This means they are 20 digit ones in total.  And is the same for the twos, the threes, etc. The sum of all will be:

$R=20*1+20*2+20*3+20*4+20*5+20*6+20*7+20*8+20*9$

Factorizing:

$R=20(1+2+3+4+5+6+7+8+9)=20(45)=900$