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If X is an integer, X even? (1) X^{X} <= |X| (2) X^{3}

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If X is an integer, X even? (1) X^{X} <= |X| (2) X^{3} [#permalink]

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New post 03 Jun 2009, 00:26
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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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New post 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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New post 03 Jun 2009, 00:34
Nope, try again...
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Re: Tough DS 10 [#permalink]

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New post 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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New post 03 Jun 2009, 07:08
so is the answer E then ? :)
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Re: Tough DS 10 [#permalink]

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New post 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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New post 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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New post 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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New post 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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New post 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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New post 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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New post 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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Re: Tough DS 10   [#permalink] 24 Jul 2009, 09:43
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