IMO E.

1. x is an integer. so, x = 0 or 1(regardless what n is) to satisfy x^n-x^-n = 0. Not Suff.
2. n <> 0. still we have the same solution as 1 . x can be 0 or x can be 1. Not Suff.

Please let me know the answer and FYI its not a simple question at least to me

Hey guys,

Yeah, I definitely wouldn't call this a simple question...this is one that isn't all that labor-intensive, but it definitely digs fairly deep into your knowledge of exponents, properties of numbers, etc.

I'd look at the given information this way:

x^n - x^-n = 0

To simplify, add x^-n to both sides to get:

x^n = x^-n

Or, to rephrase:

x^n = 1/(x^n)

Basically, x^n is the same as its own reciprocal.

Well, only two numbers are the same as their own reciprocal (remember, you can't divide by 0 so 0 doesn't count!). -1 = 1/-1 and 1 = 1/1. So either x is 1 or -1, or n is 0.


Statement 1 tells us that x is an integer, which isn't enough because as long as n is 0 x could be any integer.

Statement 2 tells us that n is not 0, but that still allows x to be 1 or -1.

Taken together, x could still be 1 or -1, so the answer is indeed E.

Keep in mind, though, that if statement 1 said that x is a positive integer, we'd be able to eliminate -1 from the mix, so this one is a lot closer to being sufficient than it might look at first glance! Definitely not a problem I'd call "simple"...