Question

The sum of all the digits used to write the whole numbers 10 through 13 is $1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 = 10$. what is the sum of all the digits used to write the whole numbers 1 through 110, inclusive?

Answer 1

957

Explanation:

We first calculate the sum of all digits used to write the whole numbers 0 through 99. If we consider all such numbers as two-digit (e.g. write 04 instead of 4), the sum of digits will be unchanged. Then we see that each digit appears an equal number of times in the ones place, and similarly for the tens place, meaning it appears a total of $2\cdot \frac{100}{10} = 20$ times. Thus the sum of all digits used to write the whole numbers 00 through 99 is $20\cdot (0 + 1 +\cdots + 8 + 9) = 900$. The sum of the digits from 100 to 110 is just $(1 + 0) + (1 + 1) + (1 + 2) + \cdots + (1+9) + 2 = 57$. Thus our final answer is $900 + 57 = \boxed{957}$.

Answer 2

First get the sum of digits 0 through 9: 0+1+2+...+9=45. Call this sum S for short.

Then notice that the sum of the digits for 10 through 19 is the sum S + 10.

Then notice that the sum of digits 20 through 29 is S + 20, and do so until you hit 100. 100 through 109 is the same as 10 though 19 which you know is S + 10, and finally add the sum of digits of 110, i.e., 2.

Total = S + (S + 10) + (S + 20) + ... + (S + 90) + (S + 10) +2 =

= 11 * S + 10 * (S + 1) + 2 = 11 * 45 + 10 * 46 + 2 = 957