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# Suppose n is an integer, such that the sum of the digits of n is 2, and 1010 < n < 1011. The number of different values for n is: [CAT, 2014]

- May 14, 2018
- Posted by: allexamshelps
- Category: CAT

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**Solution:**

**Suppose n is an integer, such that the sum of the digits of n is 2:**

**The given question puts two conditions:**

**(1) There is a number n having m digits, and the sum of these m digits equals 2.**

**(2) The number n is between 10**^{10} < n < 10^{11}

^{10}< n < 10

^{11}

^{ }

^{ }

**(10**^{10} < n < 10^{11 }=^{ }n should be any number greater than 10 billion and less than 100 billion)

^{10}< n < 10

^{11 }=

^{ }n should be any number greater than 10 billion and less than 100 billion)

**That is, If we find 10**^{10}

^{10}

**10000000000 (ten zeroes), or 10,000,000,000 = 10 billion **

**and**

**10**^{11}= 100000000000 (eleven zeroes), or 100,000,000,000= 100 billion

^{11}= 100000000000 (eleven zeroes), or 100,000,000,000= 100 billion

**So, n can be from ten billion one dollars, 10,000,000,001 to twenty billion dollars to keep the sum of digits = 2.**

**1. $10,000,000,001**

**2. $10000000010**

**3. $10000000100**

**4. $10000001000**

**5. $10000010000**

**6. $10000100000**

**7. $10001000000**

**8. $10010000000**

**9. $10100000000**

**10. $11000000000**

**11. $20,000,000,000**

**There are 11 numbers which n can take.**

**Hence the option (a) is correct.**

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