simple DS

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by tdadic84 » Wed May 27, 2009 12:05 pm
is it E....?

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Re: simple DS

by dtweah » Wed May 27, 2009 1:14 pm
PAB2706 wrote:If n is an integer and x^n – x^-n = 0, what is the value of x ?
(1) x is an integer.
(2) n ≠ 0
Reorganize stem as

X^n -1/X^n=0

X^2n=1

X=1^1/2n.

For all altegral n except 0, X=1. But for n=0 the equation has an indeterminate form X=1^1/0=1^infinity. Since interger 0 leads to an indeterminate form 1) is Not SUFF.

Under 2)

We are told n is not equal to 0. Well then X=1 for all other integral n. For n=-2, X=1^-1/4=1/(1^1/4 )=1. -1 is not a solution because every power of the form 1/p with p even, can be reduced to a base of the square root of -1 which is imaginary. For example -1^1/18= (-1^1/2)^1/9 and this gives imaginary number. If p is odd, the idea of multiplying odd number of negative numbers to produce a positive one is out of question. Since n can only take integral values, 2 is SUFF and 1 is the only solution of the equation.

Choose B

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by 2010gmat » Wed May 27, 2009 9:04 pm
x^n - x^-n =0

---> x^2n - 1 = 0

1. x is an integer ... insuff... x can be 1 or -1; also n can take value = 0 then x can be anything

2. n is not 0, for any value of n, x can be 1 or -1...in suff

1. + 2. also doesn't help

E...