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# If n is an integer such that the sum of the digits of n is 2, and 10^1

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Re: If n is an integer such that the sum of the digits of n is 2, and 10^1 [#permalink]
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for every place value of the integers we will have to move the value of 1 starting from 1,00,00,00,00,01 so on till 11,00,00,00,00,0
IMO B ; 11 values

Bunuel wrote:
If n is an integer such that the sum of the digits of n is 2, and $$10^{10} < n < 10 ^ {11}$$. The number of different values of n is

A. 12
B. 11
C. 10
D. 8
E. 9

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Re: If n is an integer such that the sum of the digits of n is 2, and 10^1 [#permalink]
Since 10^10 < n< 10^11,
n is a 11 digit integer (10 zeroes after 1)

Now, for sum of digits to be 2, either there can be two 1s or one 2.

Case 1: Two 1s
First is fixed at the beginning i.e. 10000000000
The second can be in place of any of the ten zeroes. So, the possibility is 10C1 = 10.

Case 2: One 2.
This can only be fixed at the beginning i.e. 20000000000. So, only 1 possibility.

Total possibilities are 10+1=11.
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Re: If n is an integer such that the sum of the digits of n is 2, and 10^1 [#permalink]
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Re: If n is an integer such that the sum of the digits of n is 2, and 10^1 [#permalink]
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