If X is an integer, X even? (1) X^{X}

Question

If  X  is an integer,  X  even?

(1)  X^{X} <= |X|

(2)  X^{3} < 0

Neochronic · Answer

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

Hades · Answer

Nope, try again...

Hades · Answer

My bad, yes you were correct, but I typed it in wrong-- try again

pmenon · Answer

so is the answer E then ?

pmenon · Answer

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

lahoosaher · Answer

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

Hades · Answer

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 .

Neochronic · Answer

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..

skpMatcha · Answer

I know , Sorry to exhume this thread again !

DavidArchuleta · Answer

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)

GMAT TIGER · Answer

Few things:

1.  0^0  is not defined.

2. If  x=-2 ,  x^x = (-2)^(-2) = 1/(-2)^(2)= 1/4  which is always < 2.

OA should be E not A.

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

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