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

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

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New post 09 Sep 2017, 23:53
hazelnut 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).

Answer: D.

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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110 = 1 + 1 + 0 = 2.
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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New post 10 Sep 2017, 21:07
hazelnut 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).

Answer: D.

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?


It's sum of the digits that you have to do.

110 = 1+1+0 = 2
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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New post 25 Mar 2018, 10:13
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

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

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New post 27 May 2019, 21:43
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
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Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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New post 27 May 2019, 22:08
24ymk96 wrote:
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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The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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New post 11 Aug 2019, 20:39
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


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


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

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New post 14 Sep 2019, 10:24
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


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

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

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New post 27 Jan 2020, 01:34
My solutions to the questions: summation of each value separately in their digits, tens and hundreds.

1. Digits:
1+2+3...+0=45
45 * 12 = 540
(we know that these digits appear 12 times in series of 10 numbers from 1 through 110 and. I.e 1-9, sum of digits is 45 appear once, 10-19, another series of 45...all the way till 101-110 which will be the 12th series)

2. Tens:
45 * 9= 405
405+1= 406
(we know that these tens appear 9 times. i.e 11-19 is once, 20-29 is twice...110-110 is 9th times)
405 plus 1 is to consider the tens value of 110
3. Hundreds:
Value is 11.
101, 102, 103...110 provides a hundreds value of 1+1+1+1...=11


4. sum of the digits used to write the sum = 540+406+11+1= 957 (D)
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The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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New post 01 Feb 2020, 23:27
My method (though the wording of the question is very poor and ambiguous):

We need to find 1234567891011......110 (the digit sum of this):
We know that the sum of all numbers from 1-9 is 45
Take the 2nd group into consideration: 10111213141516171819 : The digit sum will be 1*10 (for the 10 1s present in the tens places) + 45 = 55
Similarly for the group 20212223242526272829 : The digit sum will be 65
We can form an AP 45, 55, 65, 75, 85, 95, 105, 115, 125, 135 (which takes into account all the digits in 1234567891011...99
The sum of this AP is n/2[2a + (n-1)d] = 900
Then find the digit sum of 100101...110 = 57

So total sum = 957 (D)
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The sum of the digits used to write the sum 10 + 11 + 12 + 1   [#permalink] 01 Feb 2020, 23:27

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