The sum of the digits used to write the sum 10 + 11 + 12 + 1

We need to find the sum of all digits used in the following expression: 1 + 2 + 3 + 4 + ... + 109 + 110.

Look at the sum from 1 to 99, inclusive:
01
02
03
...
98
99

Each digit above is used 10 times for units digits and 10 times for tens digits, so 20 times in total. Therefore the sum of the digits used to write the sum of the integers from 1 to 99, inclusive is 20(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = (something with 0 as units digit).

Now, the sum of the digits used to write the sum of the integers from 100 to 110, inclusive (100 + 101 + 102 + 103 + 104 + 105 + 106 + 107 + 108 + 109 + 110) is 11 + 1 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 57.

Total: (something with 0 as units digit) + 57 = (something with 7 as units digit).

Answer: D.

Hope it's clear.
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Break the Numbers into multiple pieces

From 0 - 9 Sum of the digits = 45
From 10 - 19 Sum of the UNIT digits = 45
From 20 - 29 Sum of the UNIT digits = 45
and so on...
i.e. There are 11 such series of 10 numbers from 1 through 110 and
Sum of the Unit digits in all 11 series = 11*45 = 495


From 0 - 9 Sum of the TENS digits = 0
From 10 - 19 Sum of the TENS digits = 1*10 = 10
From 20 - 29 Sum of the TENS digits = 2*10 = 20
From 30 - 39 Sum of the TENS digits = 3*10 = 30
and so on...
i.e. There are 11 such series of 10 numbers from 1 through 110 and
Sum of the TENS digits in all 11 series = 10+20+30+40+50+60+70+80+90+1(100 to 110)=451

Sum of HUNDREDS digit from Numbers from 1 through 110 = 11 (only 100 to 110 have Hundreds Digit)

TOTAL DIGIT SUM = 495+451+11 = 957

Answer: Option D

I did not understand the question....
Sum of any arithmatic series is \(\frac{n}{2}(2a+(n-1)d)\)
So sum of 10, 11, 12, 13 is 64, so sum of digits of 64 is 10.... makes sense until now....
Likewise, sum of 1,2,... 110 is \(\frac{110}{2}((2*1)+(110-1)1)\) = 55*(2+109) = 55*111 = 6105.... So sum of digits of 6105 is 12.... What is the question even asking then?

sum from 1 to 9 = 45
sum from 10 to 19 = 45+10=55
sum from 20 to 29 = 45+20=65
Therefore sum from 1 to 99 = 45+55+65+75+.....+135=5*(45+135)=5*180=900
sum from 100 to 110=45+1+11=57
Hence, sum of all digits = 957
The correct option is D

alternative method

sum of all unit digits= 11(1+2+3+....+9)=11*45=495
sum of all ten's digits=10(1+2+3+...+9)+1=451 (+1 is in 110)
sum of all hundred's digits=11
Hence, sum of all digits = 495+451+11=957

Okay..here's a quick approach that worked for me..

I solved this problem using the digit sum principle..Let me first elucidate what we will be doing with a simple example..

Consider numbers chosen at random(and you can check this by picking any number of random numbers)

15,45,67,89

The actual total sum of these numbers is 216..is this number divisible by 3? Yes..9? Yes.

We could check the same by individually adding all the digits of all these numbers

1+5+4+5+6+7+8+9 = 45..is this number divisible by 3? Yes..9? Yes.

Even if these two sums left a remainder with 3 and 9, it would be the same in both the cases.

Now lets come to the problem at hand..

Sum of the numbers from 1 to 110 inclusive is calculated by the formula

\(\frac{n(n+1)}{2}\) , here..n=110

= \(\frac{110*111}{2}\)

= \(55*111\)

The remainder that this number will leave when divided by 3 or 9, will be the same if we added the individual digits of every number in this range.

Is this number divisible by 3?..Yes..9? No.

Now check options..only (D) satisfies these conditions.

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

Hi Brent,

I have a question. I have used the formula for sum from 1 to 110.

n(n+1)/2 =110*111/2=55 * 111=6105. Then sum of digits 6+1+0+5=12.

I did the above base don example provided in the beginning of the stem.

10+11+12+13=46. Sumf of digits =6+4= 10.

Where did I go wrong?

I'm totally confused

I think you're answering a different question.
You're first finding the SUM of the integers from 1 to 110, inclusive, and then you're finding the sum of the digits in that SUM
This is not what the question is asking.

Unfortunately, the person who wrote the question didn't realize that the sum of ten applies to the sum of all integers AND the sum of the digits in the sum of the integers.
The sum of the digits used to write the sum 10 + 11 + 12 + 13 is 10
In other words, 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 = 10

Here's what the question asking: Let's say you write the list of all integers from 1 to 110, inclusive. Then, you start adding every individual digit. What is the sum of all of those individual digits?

Cheers,
Brent

Bunuel
Won't there be 9 0's from 1 to 99? I am not sure what I am missing! Although that won't affect the answer, I am asking just for clarity!

Think of the first 100 numbers as:

00, 01, 02, 03, 04, ... 10, 11, 12, ... 97, 98, 99

This brings in complete uniformity. Each digit is used 10 times in each place (tens as well as units place).
So the sum of all digits will be
2*10*(0 + 1 + 2 + 3 + ...9) = 900

Now we have to worry about numbers from 100 to 110 only.
Again, I would like to consider 10 numbers at a time only. From 100 to 109.
10 1s in hundreds place give us a sum of 10. All tens places are 0. The sum of all units digits is 0 + 1 + 2 + 3 + ...9 = 45

Now only 1 number left to consider 110.

Total sum = 900 + 10 + 45 + 2 = 957

Answer (D)
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I really dislike Karishma's solution.

This solution doesn't even make sense. The first 100 numbers sum to 910 not 900. Buy maybe you just made a typo.

Anyways, here is how I did it. I don't think we can think of how you did it on the spot since that isn't a trick at all. That would be something thought of in the spur of the moment and clearly you wouldn't have enough time to notice your pattern on the spot. And even if you did, you still have a lot of math work ahead of you.

1) You may notice the first Bar adds up to 46
2) The second bar adds up to 56.

Therefore, you can assume that each bar is 10 higher and there are 10 bars.

We can use the counting formula
An= N1+d(n-1) D=10 as it goes up by 10 on each rung
AN= 46+10(9) = 136

So we can now use the summing formula of 10(46+136)/2. which is 182/2 or 91 X 10 which is 910.

3) Now all we need to do is sum the final row which is 46 if you remember + 1 extra one.


I agree with the above, this problem is excessively mean and it doesn't test ability.

Hi Bunuel,

100 = 1
101 = 2
102 = 3
103 = 4
104 = 5
105 = 6
106 = 7
107 = 8
108 = 9
109 = 10
110 = 11

If I sum 1+2+3+4+5+6+7+8+9+10+11 = 66. How to get 57?

If 10 = 1, 11= 2, 45 + 1 + 2 = 48?
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110 = 1 + 1 + 0 = 2.
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Hi All,

These types of questions will almost always involve some type of pattern in the numbers (or in this case, the digits of the numbers). Here's one way to solve this problem (based on the patterns in the digits):

We're asked to deal with the numbers 1 to 110, inclusive. I'm going to break this into two groups: 1 - 99 and 100 - 110

The numbers from 1 to 99 follow an interesting pattern: each digit appears 10 times in the "tens spot" and 10 times in the "units spot"

For example, the digit 1:

10, 11, 12, …..19 ---> 10 times in the "tens spot"
1, 11, 21, 31….91 ---> 10 times in the "units spot"

So, we have to multiply each digit by 20 to get it's total sum. There's a faster way to do it though:

1+2+3+4+5+6+7+8+9 = 45

45 x 20 = 900

So the sum of the digits of the numbers from 1 to 99 is 900

Now, for the numbers from 100 to 110

We have 11 "ones" in the "hundreds spot" = 11
We have the digits 1 to 9 in the "ones spot" = 45
We have one more 1 in the "tens spot" = 1

11+ 45 + 1 + 900 = 957

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Now, the sum of the digits used to write the sum of the integers from 100 to 110, inclusive (100 + 101 + 102 + 103 + 104 + 105 + 106 + 107 + 108 + 109 + 110) is 11 + 1 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 57.

Bunuel can you please explain this calculation

11 there is the sum of the hundreds digits (100, 101, 102, 103, 104, 105, 106, 107, 108, 109,110);

1 is the sum of the tens digits (only one number there with non-zero tens digit: 110)

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 is the sum of the units' digits: (101, 102, 103, 104, 105, 106, 107, 108, 109).

Hope it's clear.
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Number of time digit 1 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) + 13 (100-1,101-2,102-109-1 (8 cases) & 110-2) = 33
Number of time digit 2 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) +1 (102) = 21
Number of time digit 3 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) + 1 (103) = 21
Number of time digit 4 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) + 1 (104) = 21
Number of time digit 5 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) + 1 (105) = 21
Number of time digit 6 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) + 1 (106) = 21
Number of time digit 7 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) + 1 (107) = 21
Number of time digit 8 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) + 1 (108) = 21
Number of time digit 9 appears in integers 1 to 110, inclusive = 1 (single digit) + 19 (2 digits) + 1 (109) = 21

Sum of digits = 12*1 + 21 *(1+2+3+4+5+6+7+8+9) = 11 + 21*45 = 12+945 = 957


IMO D

Given: The sum of the digits used to write the sum 10 + 11 + 12 + 13 is 10.

Asked: What is the sum of the digits used to write the sum of the integers from 1 to 110, inclusive?

Sum of digits for 1 + 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Sum of digits for 10 + 11+.... + 19 = 45 + 10 = 55
Sum of digits for 20 + 21+.... + 29 = 45 + 20 = 65
Sum of digits for 30 + 31+.... + 39 = 45 + 30 = 75
Sum of digits for 40 + 41+.... + 49 = 45 + 40 = 85
Sum of digits for 50 + 51+.... + 59 = 45 + 50 = 95
Sum of digits for 60 + 61+.... + 69 = 45 + 60 = 105
Sum of digits for 70 + 71+.... + 79 = 45 + 70 = 115
Sum of digits for 80 + 81+.... + 89 = 45 + 80 = 125
Sum of digits for 90 + 91+.... + 99 = 45 + 90 = 135
Sum of digits for 100 + 101+.... + 109 = 45 + 10 = 55
Sum of digits for 110 = 2

Sum of digits for 1 + ... + 110 = 45 + 55+ 65 + 75 + 85 + 95 + 105 + 115 + 125+ 135 + 55 + 2 = 5 (180) + 57 = 900 + 57 = 957

IMO 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
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