BTGModeratorVI wrote: ↑

Sat Aug 08, 2020 7:03 am

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
Source: Veritas Prep

IMPORTANT: Always check the answer choices before beginning any solution. The answer choices may hint at an approach and/or suggest that you can skip tedious calculations.

Here, the units digits in the answer choices are all different, which suggests that I may be able to avoid some "grunt" work.

First, let's examine the numbers from 00 to 99 inclusive (i.e., 00, 01, 02, .... 97, 98, 99) [note: adding 00 to the mix doesn't change the final answer]
Notice that there are 100 digits from 00 to 99 inclusive.
Also notice that the digits in the tens and units position are equally distributed

So, in the UNITS position, there will be ten 0's, ten 1's, ten 2's . . . ten 8's and ten 9's
In the TENS position, there will be ten 0's, ten 1's, ten 2's . . . ten 8's and ten 9's
So, the sum of ALL DIGITS from 00 to 99 will equal 20(1+2+3+...7+8+9)

IMPORTANT: We don't need to calculate 20(1+2+3+...7+8+9). We need only recognize that the units digit will equal 0. That is 20(1+2+3+...7+8+9) = ??0

From here, we need to add the digits in 100 to 110 inclusive.
To do so, we can use Rich's approach, or we might even list the values and add them in our head, since there aren't many to add here.
When we add the digits in 100 to 110 inclusive, we get 57

So, the sum of the digits from 00 to 110 inclusive = ??0 + 57 = ??7

Since only D has a 7 in the units position, this is the correct answer.

Cheers,
Brent

BTGModeratorVI wrote: ↑

Sat Aug 08, 2020 7:03 am

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 the first group has 9 numbers and the last group is just 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