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Math Expert V
Joined: 02 Sep 2009
Posts: 56277
Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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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).

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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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).

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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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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

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

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 56277
Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1  [#permalink]

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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.
_________________ Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1   [#permalink] 27 May 2019, 23:08

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