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

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Joined: 02 Sep 2009
Posts: 41871

Kudos [?]: 128551 [0], given: 12180

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, 00:53
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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).

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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Kudos [?]: 128551 [0], given: 12180

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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, 22:07
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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Kudos [?]: 2273 [0], given: 51

Re: The sum of the digits used to write the sum 10 + 11 + 12 + 1   [#permalink] 10 Sep 2017, 22:07

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