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

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Manager
Joined: 03 Oct 2009
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If x and y are both integers, which is larger, x^x or y^y? [#permalink]

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18 Feb 2012, 08:33
2
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Difficulty:

65% (hard)

Question Stats:

54% (01:08) correct 46% (00:59) wrong based on 41 sessions

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

(1) x = y + 1
(2) x^y > x and x is positive.
[Reveal] Spoiler: OA

Kudos [?]: 150 [2], given: 8

Manager
Status: Employed
Joined: 17 Nov 2011
Posts: 97

Kudos [?]: 169 [2], given: 10

Location: Pakistan
GMAT 1: 720 Q49 V40
GPA: 3.2
WE: Business Development (Internet and New Media)
Re: Which is larger, x^x or y^y? [#permalink]

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18 Feb 2012, 10:38
2
KUDOS
Answer should be C. Here is how:

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

Statement A: $$x=y+1$$

So $$x$$ and $$y$$ are consecutive integers. Remember they can be positive, negative or $$0$$ (from the question stem).

Suppose $$y=1$$ and $$x=2$$ , then $$x^x$$ is larger , but suppose $$y=-2$$ and $$x =-1$$ then $$y^y$$ is larger.

Hence Insufficient.

Statement B: $$x^y>x$$ and $$x$$ is positive.

Knowing that $$x>0$$, $$x^y>x$$ is only possible if $$y>1$$ . Please note even when $$y=0$$ , $$x>0$$ and $$x$$ is an integer. So now we know that $$y$$ is positive and $$x$$ is positive but we do not know which is larger. Hence Insufficient

Combined:

From Statement 2 we know that $$x$$ and $$y$$ are both $$>0$$ and from Statement 1 we know that $$x$$ is bigger. So YES. $$x^x$$ is bigger than $$y^y$$

Sufficient. Hence Answer C . Seriously Kudos Hungry
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Kudos [?]: 169 [2], given: 10

Math Expert
Joined: 02 Sep 2009
Posts: 42669

Kudos [?]: 136009 [0], given: 12723

Re: Which is larger, x^x or y^y? [#permalink]

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18 Feb 2012, 11:08
omerrauf wrote:
Answer should be C. Here is how:

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

Statement A: $$x=y+1$$

So $$x$$ and $$y$$ are consecutive integers. Remember they can be positive, negative or $$0$$ (from the question stem).

Suppose $$y=1$$ and $$x=2$$ , then $$x^x$$ is larger , but suppose $$y=-2$$ and $$x =-1$$ then $$y^y$$ is larger.

Hence Insufficient.

Statement B: $$x^y>x$$ and $$x$$ is positive.

Let's re-arrage this a bit.

$$x^y>x$$ so $$x^y-x>0$$ so x*(y-1)>0 and we know that x is +ve
Now in order for $$x*(y-1)>0$$ the factor $$(y-1)$$ has to be $$>0$$ so we know that $$y>1$$ but we do not know which is bigger, $$x$$ or $$y$$.

Hence Insufficient

Combined:

From Statement 2 we know that $$x$$ and $$y$$ are both $$>0$$ and from Statement 1 we know that x is bigger. So YES. $$x^x$$ is bigger than $$y^y$$

Sufficient. Hence Answer C Seriously Kudos Hungry

The red part is not correct.

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.

P.S. Not a GMAT style question.
_________________

Kudos [?]: 136009 [0], given: 12723

Manager
Status: Employed
Joined: 17 Nov 2011
Posts: 97

Kudos [?]: 169 [0], given: 10

Location: Pakistan
GMAT 1: 720 Q49 V40
GPA: 3.2
WE: Business Development (Internet and New Media)
Re: If x and y are both integers, which is larger, x^x or y^y? [#permalink]

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18 Feb 2012, 20:05
Had already corrected that before you wrote bunuel. While writing the solution I mistakenly took $$x^y-x$$ for $$xy-x$$ but corrected it when i was reading my solution after I had posted. Thankyou anyways!
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"Nowadays, people know the price of everything, and the value of nothing." Oscar Wilde

Kudos [?]: 169 [0], given: 10

Manager
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04 Mar 2012, 09:15
If x and y are both integers, which is larger, x^x or y^y?

x = y + 1
x^y > x and x is positive.

Kudos [?]: 150 [0], given: 8

Senior Manager
Joined: 23 Mar 2011
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04 Mar 2012, 09:43
St 1:
x=y+1
eg; y=2; then x= 2+1 = 3
then 3^3 > 2^2

y=-3 then x= -3+1 = -2
then -2^-2 > -3^-3

Sufficient

St 2:
x^y>x

eg; x = 2 y = 3 then x^y>x

Thus, x^x < y^y

However, if x = 4 and y = 3 then also x^y> x
But, x^x > y^y

Not sufficient

ANS: A

I dont know if this is the best approach.
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04 Mar 2012, 12:34
http://www.platinumgmat.com/practice_gm ... on_id=2197
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Kudos [?]: 442 [0], given: 192

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Concentration: Finance
Re: Which is larger, x^x or y^y? [#permalink]

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27 Dec 2013, 09:09
Bunuel wrote:
omerrauf wrote:
Answer should be C. Here is how:

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

Statement A: $$x=y+1$$

So $$x$$ and $$y$$ are consecutive integers. Remember they can be positive, negative or $$0$$ (from the question stem).

Suppose $$y=1$$ and $$x=2$$ , then $$x^x$$ is larger , but suppose $$y=-2$$ and $$x =-1$$ then $$y^y$$ is larger.

Hence Insufficient.

Statement B: $$x^y>x$$ and $$x$$ is positive.

Let's re-arrage this a bit.

$$x^y>x$$ so $$x^y-x>0$$ so x*(y-1)>0 and we know that x is +ve
Now in order for $$x*(y-1)>0$$ the factor $$(y-1)$$ has to be $$>0$$ so we know that $$y>1$$ but we do not know which is bigger, $$x$$ or $$y$$.

Hence Insufficient

Combined:

From Statement 2 we know that $$x$$ and $$y$$ are both $$>0$$ and from Statement 1 we know that x is bigger. So YES. $$x^x$$ is bigger than $$y^y$$

Sufficient. Hence Answer C Seriously Kudos Hungry

The red part is not correct.

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.

P.S. Not a GMAT style question.

Just curious, why not a GMAT style question?

Thanks

Cheers!
J

Kudos [?]: 759 [0], given: 355

Re: Which is larger, x^x or y^y?   [#permalink] 27 Dec 2013, 09:09
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