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Re: Is x an odd number? [#permalink]
28 Feb 2013, 07:51

2

This post received KUDOS

Is x an odd number? 1) \(n^x = 1\) 2) \(n^2\) is a prime number.

Statement 1) - n = 1, x = any number - Insufficient Statement 2) - \(n^2\) = Prime number, so "n" must be square root of a prime number. No info about x- Thus Insufficient Statement 1& 2) - Because "n" is a square root of a prime number, the only condition when \(n^x = 1\) is if x= 0. As we know "Zero (0)" is an Even Integer so the answer is "NO" - Thus sufficient

Hence Answer is C

Hope it helps.

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Statement 1: Clearly insufficient. for n=1, x can be any value. Statement 2: n2 is prime. Now there is no integer n for which n2 is prime (1 is not prime ) Nothing is mentioned about x. Insufficient Now for a non integer value of n such that n2 is prime like \(\sqrt{2}\) means that n cannot be 1 thus x must be equal to 0 which is not odd. Sufficient Answer C

Re: Is x an odd number? [#permalink]
28 Feb 2013, 15:11

(1)n^x = 1: insufficient, because x = 0, any n -> n^x = 1, n = 1, x = 2 --> 1^2 = 1; n = -1, x= 2 --> (-1)^2 = 1 (2) n^2 is a prime number: insufficient Prime number should be 2, 3, 5, 7, 11, 13, 17, 19, ... --> n^2 is a prime number if n = sqrt(2), sqrt(3),... (1) + (2): sufficient, x must be 0 Answer is C

Re: Is x an odd number? [#permalink]
02 Apr 2013, 02:10

2

This post received KUDOS

okay... correct me if I'm wrong. pls...

Is x an odd number?

1) \(n^x = 1\) 2) \(n^2\) is a prime number.

so as per the question, there is nothing like n okay. now anything raise to the power 0 makes that integer 1. so from stmt 1 we don't know what is n & we also don't want to know about n coz . from this \(n^x = 1\) if we take n as any integer, non-interger, anything & raise to the power 0 then, it satisfies the stmt 1 completely coz 0 is neither -ve nor +ve but an even integer, this means that x=0 that means even, straight No... & then there is no need of stmt. 2 . Hence the answer must be A. Please corect me if I'm wrong. Thanks !! _________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Re: Is x an odd number? [#permalink]
02 Apr 2013, 02:54

Expert's post

manishuol wrote:

okay... correct me if I'm wrong. pls...

Is x an odd number?

1) \(n^x = 1\) 2) \(n^2\) is a prime number.

so as per the question, there is nothing like n okay. now anything raise to the power 0 makes that integer 1. so from stmt 1 we don't know what is n & we also don't want to know about n coz . from this \(n^x = 1\) if we take n as any integer, non-interger, anything & raise to the power 0 then, it satisfies the stmt 1 completely coz 0 is neither -ve nor +ve but an even integer, this means that x=0 that means even, straight No... & then there is no need of stmt. 2 . Hence the answer must be A. Please corect me if I'm wrong. Thanks !!

What happens when n itself is 1? Then, 1^3 is also 1 and 1^4 is also 1. Thus, you wouldn't know whether x is an odd number or not. _________________

Re: Is x an odd number? [#permalink]
02 Apr 2013, 03:09

Expert's post

manishuol wrote:

Oh!! dear....... if 1 raise to the power 0 then also the result will be 1 & therefore x=0 i.e; even .. .. .. .. hence, A.

What part of the question has told you that x can ONLY be 0? x can take any value,as long as it satisfies the condition mentioned in the problem,the condition in this case being n^x = 1. _________________

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