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If X is an integer, X even? (1) X^{X} <= X (2) X^{3} [#permalink]
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03 Jun 2009, 00:26
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This topic is locked. If you want to discuss this question please repost it in the respective forum. If \(X\) is an integer, \(X\) even? (1) \(X^{X} <= X\) (2) \(X^{3} < 0\)
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Hades
Last edited by Hades on 03 Jun 2009, 01:07, edited 1 time in total.



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Re: Tough DS 10 [#permalink]
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03 Jun 2009, 00:32
simple..answer E..
1, insufficient.. scenario possible for both X even or odd.. 2, infusfficent, scneario possible for both X even or odd...
combined .. dont provide much info abt Z being even or odd..
hence ansewr is E



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Re: Tough DS 10 [#permalink]
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03 Jun 2009, 00:34
Nope, try again...
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Re: Tough DS 10 [#permalink]
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03 Jun 2009, 01:07
My bad, yes you were correct, but I typed it in wrong try again
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Re: Tough DS 10 [#permalink]
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03 Jun 2009, 07:08
so is the answer E then ?



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Re: Tough DS 10 [#permalink]
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03 Jun 2009, 08:22
Hm, ok, lets try again From statement 2, you know that X is negative. Great, but no info on whether its odd or even. insuff From statement 1, X=1 and 1 both work, and they're both odd. Even numbers, both positive and negative, dont satisfy the inequality. So X must be odd. suff. Therefore, I now say A



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Re: Tough DS 10 [#permalink]
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03 Jun 2009, 15:22
IMO A x^x<=x for x^x=x x=1 for x^x<x x has negative odd values in both the cases x is odd
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Re: Tough DS 10 [#permalink]
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03 Jun 2009, 15:49
Hades wrote: If \(X\) is an integer, \(X\) even?
(1) \(X^{X} \leq X\)
(2) \(X^{3} < 0\) Question:(\(X\) EVEN)? (1) \(X^{X} \leq X\) Let's assume X is any integer for now, so \(X = ...,8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8,...\) Notice that \(2^{2} = 4 > 2 = 2\) which is contrary to fact #1, and likewise any number bigger than 2. So we're down to: \(X = ...,8,7,6,5,4,3,2,1,0,1\) Consider \(X=0\) \(0^0 = 1 > 0\), so \(X=0\) doesn't hold. So \(X = ...,8,7,6,5,4,3,2,1,1\) \(X=1\) holds as \((1)^{1} = 1 \leq 1 = 1\). \(X=2\) doesn't hold as raising a negative number to an even exponent makes it positive, and a positive integer raised to itself will always be bigger than itself, hence no negative even numbers will hold. And we're down to \(X = ...,7,5,3,1,1\) Which is always odd. \(\longrightarrow SUFFICIENT\). (2) \(X^{3} < 0\) This doesn't tell us much, just that X is negative X can still be even or odd. \(\longrightarrow INSUFFICIENT\). Final Answer, \(A\).
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Re: Tough DS 10 [#permalink]
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03 Jun 2009, 21:50
i differ...
X = 2 doesn't hold as raising a negative number to an even exponent makes it positive, and a positive integer raised to itself will always be bigger than itself, hence no negative even numbers will hold.
for X = 2, X^X = .25.. and .25 is less than 2.. it holds..
So still not defined..
I stick with my answer... E..



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Re: Tough DS 10 [#permalink]
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23 Jul 2009, 13:01
Quote: i differ...
X = 2 doesn't hold as raising a negative number to an even exponent makes it positive, and a positive integer raised to itself will always be bigger than itself, hence no negative even numbers will hold.
for X = 2, X^X = .25.. and .25 is less than 2.. it holds..
So still not defined..
I stick with my answer... E..
I know , Sorry to exhume this thread again !



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Re: Tough DS 10 [#permalink]
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24 Jul 2009, 01:46
Neochronic wrote: i differ...
X = 2 doesn't hold as raising a negative number to an even exponent makes it positive, and a positive integer raised to itself will always be bigger than itself, hence no negative even numbers will hold.
for X = 2, X^X = .25.. and .25 is less than 2.. it holds..
So still not defined..
I stick with my answer... E.. I agree. Let X be A (A>0 and A is an integer) (A) ^(A)= (1/A) ^A < 1 (A is an integer so 1/A <1 ) So X^X <1 if X <0 while X is always >= 1 (except X=0)



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Re: Tough DS 10 [#permalink]
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24 Jul 2009, 09:43
Hades wrote: Hades wrote: \(0^0 = 1 > 0\) Few things: 1. \(0^0\) is not defined. Quote: \(X=2\) doesn't hold as raising a negative number to an even exponent makes it positive, and a positive integer raised to itself will always be bigger than itself, hence no negative even numbers will hold. 2. If \(x=2\), \(x^x = (2)^(2) = 1/(2)^(2)= 1/4\) which is always < 2. OA should be E not A.
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