M7MBA wrote: ↑

Thu Oct 29, 2020 12:59 pm

The sum of the digits used to write the sum 10 + 11 + 12 + 13 is 10. What is the sum of the digits used to write the sum of the integers from 1 to 110, inclusive?

A. 900
B. 911
C. 955
D. 957
E. 1001

Answer: D

Solution:

We can break the integers from 1 to 110 into groups of 10 (except that the first group has 9 numbers and the last group includes only the number 110)

The sum of the digits of the integers from 1 to 9 is 1 + 2 + 3 + … + 9 = 45.

The sum of the digits of the integers from 10 to 19 is:

(1 + 0) + (1 + 1) + (1 + 2) + … + (1 + 8) + (1 + 9) = 1 + 2 + 3 + … + 9 + 10 = 55

As we can see 55 is 10 more than 45 (the previous sum) because the tens digit 1 appears 10 times (notice the units digit 0 appears once, but it won’t contribute more to the sum).

Therefore, the sum of the digits of the integers from 20 to 29 is 65, from 30 to 39 is 75, and so on. The last group that is less than 100 (i.e., 90 to 99) will have a sum of 135. Therefore, the sum of the digits of all the integers from 1 to 99 is:

45 + 55 + 65 + 75 + … + 135 = (45 + 135)/2 x 10 = 180/2 x 10 = 900

The first group that is greater than 100 (i.e., 100 to 109) has a sum of:

(1 + 0 + 0) + (1 + 0 + 1) + (1 + 0 + 2) + … + (1 + 0 + 9) = 1 + 2 + 3 + … + 10 = 55

The last group is just the number 110, which has a sum of 1 + 1 + 0 = 2. Therefore, the sum of the digits of all the integers from 1 to 110 is:

900 + 55 + 2 = 957

Answer: D