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

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The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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Updated on: 01 Jul 2014, 02:50
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Question Stats:

39% (02:43) correct 61% (02:50) wrong based on 451 sessions

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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

Originally posted by honchos on 26 Dec 2013, 18:06.
Last edited by Bunuel on 01 Jul 2014, 02:50, edited 1 time in total.
Edited the OA.
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The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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27 Dec 2013, 01:59
9
20
honchos wrote:
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

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).

Hope it's clear.
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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26 Dec 2013, 19:09
1
1
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?
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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26 Dec 2013, 20:01
Amateur wrote:
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?

what is the sum of digit of following-
10, 11

1+0+1+1 = 3
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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27 Dec 2013, 11:13
honchos wrote:
what is the sum of digit of following-
10, 11

1+0+1+1 = 3

Then shouldn't the question structure be sum of digits of first 110 positive integers..... or sum of digits of integers between 1 and 110.....

By stating "The sum of the digits used to write the sum of " what is used to write the sum of 10,11,12,13? Number 64....
Gmat is trying to kill my time by using ambiguous statements.... So I boycott the Verbal section, please ask GMAT to take that section away from my exam
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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02 Aug 2015, 11:20
6
6
honchos wrote:
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

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

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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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15 Aug 2015, 08:56
Can anyone tell why cant we use (n-1)! X Sum of digits X 111...N in this question.
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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16 Aug 2015, 02:47
1 digit numbers
Sum for 1 digit numbers is 5 in average
9 one digit numbrs
9x5 = 45

2 digits numbers
90 numbers
On average, sum of digits is 9,5
=855

3 digit numbers
11 numbers x 1 (hundredth)
+9x5
+1 for the last non accounted 1 in (110)
=57

Sum is 957
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The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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16 Aug 2015, 03:34
1
honchos wrote:
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

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
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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25 Jun 2016, 10:44
2
honchos wrote:
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

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.
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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25 Jun 2016, 23:33
Bunuel wrote:
honchos wrote:
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

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).

Hope it's clear.

This is the credited response.

I think the question was made unnecessarily complicated by including the numbers from 101 to 110. Solving the same problem of numbers from 1 to 100 is plenty challenging IMO.
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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 _________________ Test confidently with gmatprepnow.com RSM Erasmus Moderator Joined: 26 Mar 2013 Posts: 2367 Concentration: Operations, Strategy Schools: Erasmus RSM "22 Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1 [#permalink] ### Show Tags 29 Jul 2016, 07:33 GMATPrepNow wrote: honchos wrote: 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 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 GMAT Club Legend Joined: 11 Sep 2015 Posts: 4567 Location: Canada GMAT 1: 770 Q49 V46 Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1 [#permalink] ### Show Tags 29 Jul 2016, 09:37 3 Top Contributor Mo2men wrote: 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 _________________ Test confidently with gmatprepnow.com Director Joined: 28 Nov 2014 Posts: 812 Concentration: Strategy Schools: Fisher '19 (M$)
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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26 Oct 2016, 01:26
Bunuel wrote:
honchos wrote:
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

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).

Hope it's clear.

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!
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The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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26 Oct 2016, 03:59
GMATPrepNow wrote:
honchos wrote:
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

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

Keats wrote:
Bunuel wrote:
honchos wrote:
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

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).

Hope it's clear.

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!

Refer to the highlighted text in quote from Brent, I hope that answers your question.

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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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26 Oct 2016, 07:11
1
1
Keats wrote:
Bunuel wrote:
honchos wrote:
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

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).

Hope it's clear.

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

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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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01 Aug 2017, 20:23
HanoiGMATtutor wrote:
Bunuel wrote:
honchos wrote:
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

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).

Hope it's clear.

This is the credited response.

I think the question was made unnecessarily complicated by including the numbers from 101 to 110. Solving the same problem of numbers from 1 to 100 is plenty challenging IMO.

Agreed. I don't even get the point of the question. I knew how to do it but all the solutions above are too slow.
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The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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01 Aug 2017, 20:31
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.
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The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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09 Sep 2017, 19:17
Bunuel wrote:
honchos wrote:
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

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).

Hope it's clear.

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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The sum of the digits used to write the sum 10 + 11 + 12 + 1   [#permalink] 09 Sep 2017, 19:17

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