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If n is an integer such that the sum of the digits of n is 2, and 10^1 28 Jan 2020, 03:24
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B
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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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B
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If n is an integer such that the sum of the digits of n is 2, and 10^1 28 Jan 2020, 04:10
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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 —???

10,000,000,000 < n < 100,000,000,000

—> 10,000,000,001
—> 10,000,000,010
—> 10,000,000,100
—> 10,000,001,000
—> 10,000,010,000
—> 10,000,100,000
—> 10,001,000,000
—> 10,010,000,000
—> 10,100,000,000
—> 11,000,000,000

Also, we got 20,000,000,000
In total, there are 11 values of n between \(10^{10}\) and \(10^{11}\).

The answer is B

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If n is an integer such that the sum of the digits of n is 2, and 10^1 28 Jan 2020, 08:15
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Bunuel
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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10^10=10,000,000,000
10,000,000,001
10,000,000,010
10,000,000,100
…
11,000,000,000
= 10 cases (10 zeros to substitute with one)
2*10^10 = 1 case
= 11 cases total

Ans (B)
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If n is an integer such that the sum of the digits of n is 2, and 10^1 28 Jan 2020, 09:03
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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
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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If n is an integer such that the sum of the digits of n is 2, and 10^1 29 Jan 2020, 03:23
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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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If n is an integer such that the sum of the digits of n is 2, and 10^1 12 Aug 2025, 16:13
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