If x and y are both integers, which is larger, x^x or y^y? : GMAT Data Sufficiency (DS)
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# If x and y are both integers, which is larger, x^x or y^y?

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If x and y are both integers, which is larger, x^x or y^y? [#permalink]

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23 Jul 2011, 01:25
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If x and y are both integers, which is larger, x^x or y^y?

(1) x = y + 1
(2) xy > x and x is positive.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-both-integers-which-is-larger-x-x-or-y-y-127756.html
[Reveal] Spoiler: OA

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Re: self exponential! [#permalink]

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23 Jul 2011, 01:44
(1)

x = y + 1

Let x = 2, y = 1

x^x > y

x = -1 and y = -2

x^x < y^y

Not Sufficient

The option here should say x and y != 0 as GMAT does not test 0^0.

(2)

xy > x and x is positive

=> y > 0

But x can be > y or y can be > x

Not Sufficient

(1) + (2)

x^x > y^y

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Re: self exponential! [#permalink]

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23 Jul 2011, 17:02
Hi Subashgosh,
are you sure about ur second example for statement 1?
(-1)^-1 should be greater than (-2)^-2 that is 1>(1/4)

I think statement 1 should be sufficient. A should be the answer.
Please correct me if i am wrong.
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Re: self exponential! [#permalink]

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23 Jul 2011, 21:07
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Subhashghosh is correct, as (-1)^-1 stays negative, while (-2)^-2 becomes positive.

-1 < 1/4

This is an odd problem, though. I agree that it should mention that x and y aren't 0. It also shouldn't ask which option is greater, as this rules out in advance the possibility that the two expressions are equal. A real GMAT question would typically ask if one or the other was greater, not *which* expression is greater.
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Re: self exponential! [#permalink]

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24 Jul 2011, 17:48
Thanks DmitryFarber and Subashgosh. Sometimes my mind stops working
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Re: self exponential! [#permalink]

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24 Jul 2011, 19:55
Stmt 1 : if y= -3 then x = -2 in which case x^2<y^2
if y =3 then x =4 ==> x^2> y^2 Hence Insuff

Stmt2: xy> x any positive values of x and y in which either x and y can be greater hence again Insuff

Combining 1 and 2 gives us that x is positive and therefore, x^2> y^2 . Hence C
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Re: self exponential! [#permalink]

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07 Sep 2013, 23:45
subhashghosh wrote:
(1)

x = y + 1

Let x = 2, y = 1

x^x > y

x = -1 and y = -2

x^x < y^y

Not Sufficient

The option here should say x and y != 0 as GMAT does not test 0^0.

(2)

xy > x and x is positive

=> y > 0

But x can be > y or y can be > x

Not Sufficient

(1) + (2)

x^x > y^y

doesnt this statement: "xy > x and x is positive" actually mean

==> y>1

if this is the case i am guessing statement 2 will be sufficient to solve the question. Am i making any mistake here?
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Re: self exponential! [#permalink]

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08 Sep 2013, 05:04
emailmkarthik wrote:
subhashghosh wrote:
(1)

x = y + 1

Let x = 2, y = 1

x^x > y

x = -1 and y = -2

x^x < y^y

Not Sufficient

The option here should say x and y != 0 as GMAT does not test 0^0.

(2)

xy > x and x is positive

=> y > 0

But x can be > y or y can be > x

Not Sufficient

(1) + (2)

x^x > y^y

doesnt this statement: "xy > x and x is positive" actually mean

==> y>1

if this is the case i am guessing statement 2 will be sufficient to solve the question. Am i making any mistake here?

If x and y are both integers, which is larger, x^x or y^y?

(1) x = y + 1 --> if $$y$$ is positive integer then $$x^x=(y+1)^{y+1}>y^y$$ but if $$y=-2$$ then $$x=-1$$ and $$x^x=-1<\frac{1}{4}=y^y$$

(2) x^y > x and x is positive --> since $$x$$ is positive then $$x^{y-1}>1$$ --> since $$x$$ and $$y$$ are integers then $$y>1$$. If $$x=1$$ and $$y=2$$ then $$x^x<y^y$$ but if $$x=3$$ and $$y=2$$ then $$x^x>y^y$$. Not sufficient.

(1)+(2) From (2) $$y>1$$, so it's a positive integer then from (1) $$x^x=(y+1)^{y+1}>y^y$$. Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-both-integers-which-is-larger-x-x-or-y-y-127756.html
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Re: self exponential!   [#permalink] 08 Sep 2013, 05:04
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# If x and y are both integers, which is larger, x^x or y^y?

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