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Is x an odd number? [#permalink]
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28 Feb 2013, 07:23
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Is x an odd number? 1) \(n^x = 1\) 2) \(n^2\) is a prime number. My own question. So, no OA. Comments and feedback recommended.. .
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Re: Is x an odd number? [#permalink]
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28 Feb 2013, 08:51
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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. Fame
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Re: Is x an odd number? [#permalink]
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28 Feb 2013, 09:09
if n^2 is prime, then n=1 or n=1 n^(x)=1 => if n=1, is true for any x ; => if n=1, is true only if x is odd; I cannot say if x is odd... for example n=1 n^2=1^2=1 OK n^(odd x)=1 OK n^(even x)=1 OK IMO E
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Re: Is x an odd number? [#permalink]
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Re: Is x an odd number? [#permalink]
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28 Feb 2013, 10:10
MacFauz wrote: Is x an odd number? 1) \(n^x = 1\) 2) \(n^2\) is a prime number. My own question. So, no OA. Comments and feedback recommended.. . 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



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Re: Is x an odd number? [#permalink]
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28 Feb 2013, 16:11
(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



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Re: Is x an odd number? [#permalink]
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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, noninterger, 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 !!
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Re: Is x an odd number? [#permalink]
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02 Apr 2013, 03:54
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, noninterger, 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.
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Last edited by mau5 on 02 Apr 2013, 04:10, edited 1 time in total.



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Re: Is x an odd number? [#permalink]
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Oh!! dear....... if 1 raise to the power 0 then also the result will be 1 & therefore x=0 i.e; even .. .. .. .. hence, A.
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Re: Is x an odd number? [#permalink]
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02 Apr 2013, 04:09
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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Re: Is x an odd number? [#permalink]
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19 Apr 2013, 21:48
Yeah..You were correct ........ I was wrong . Thanks !!
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