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Zarrolou
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

Welcome to GMAT Club.

Note that if n=1 or -1, then n^2=1 and 1 is NOT a prime number.
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MacFauz
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.. :)

My answer is C
.
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 8-) ) 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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(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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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 !!
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manishuol
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.
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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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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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Yeah..You were correct ........ I was wrong . Thanks !!
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