Help! Ron!?!

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Help! Ron!?!

by san2009 » Thu Jun 03, 2010 12:29 am
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


the way that i approached this problem is:

n=0, n=1, n=-1 all satisfy the equation above, i.e. with each of those values of n, you get result zero

statement 1) x is an integer.
x=1, satisfies eqt when n=0
x=-1, satisfies eqt when n=0
since we can have two values of x, insuff.

statement 2) n cannot be zero
n can still be 1 or -1
if we take n=1, then we can simplify the expression to x=1/x
which gives us x^2=1, and that results in 2 values of x, specifically x=1, x=-1
thus insuff

statement 1 + statement 2)
n cannot be zero and x must be an integer
taking the same example that i used for statement 2, it can be proven that you get two values for x
thus this is also insufficient

thus answer E

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by jube » Thu Jun 03, 2010 12:40 am
Hi,

You might want to look at it in another way:

the statement can be re-written as: x^n - (1/x^n) = 0 (since x^-a = 1/x^a)
thus, (x^n.x^n - 1)/x^n = 0
or x^2n - 1 = 0
or x^2n = 1
which is possible only if 2n=0 or x=1

St. 1 -- doesn't tell us whether x is 1
St. 2 - explicitly says n is not 0 which implies x is 1. --sufficient.

B

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by san2009 » Thu Jun 03, 2010 12:43 am
OA is E

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by jube » Thu Jun 03, 2010 12:49 am
san2009 wrote:OA is E
Ok, then I guess I must be doing something wrong though I'm not able to figure out what.

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by kstv » Thu Jun 03, 2010 1:40 am
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

x^n - 1/x^n = 0
((x^n)² -1 /x^n = 0 so the numerator is = 0
(x^n)² - 1 = 0 (x^n+1)(x^n-1) = 0

x^n = 1 , & if n≠ 0 then x = 1.
x^n = -1 & if n ≠ 0 then x is -ve and n is a odd no.
(2) is not sufficent.
If x is an integer then x = 1 or - 1 if n is odd. Not sufficient.

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by Haaress » Thu Jun 03, 2010 9:09 am
Jube, you left out a tiny piece, which is a third option in which X = -1. This is true because the power of X will always even in the equation X^2n = 1, meaning (-1) ^even number = 1. This undermines the sufficiency of statement 2. Thus E.

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by jube » Fri Jun 04, 2010 2:03 am
Haaress wrote:Jube, you left out a tiny piece, which is a third option in which X = -1. This is true because the power of X will always even in the equation X^2n = 1, meaning (-1) ^even number = 1. This undermines the sufficiency of statement 2. Thus E.
ahh, yes!

thanks :)

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by boazkhan » Fri Jun 04, 2010 9:31 am
problem can be reworded as X^2n = 1, which basically means is X=1 or n=0

Statement 1 - no information about n - Insuff

Statement 2 - no information about X - Insuff

together we get both a yes and a no...hence E.

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by Haaress » Fri Jun 04, 2010 9:51 am
Yes, the problem can be reworded as X^2n = 1, which basically means is X=1 , n=0 and , X= (-1), , without which the answer would have been B.