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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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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

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

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New post 18 Feb 2012, 10:38
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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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Re: Which is larger, x^x or y^y? [#permalink]

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New post 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.

Answer: C.

P.S. Not a GMAT style question.
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Re: If x and y are both integers, which is larger, x^x or y^y? [#permalink]

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New post 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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Exponents [#permalink]

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New post 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.

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Re: Exponents [#permalink]

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New post 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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Re: Exponents [#permalink]

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New post 04 Mar 2012, 12:34
http://www.platinumgmat.com/practice_gm ... on_id=2197
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Re: Which is larger, x^x or y^y? [#permalink]

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New post 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.

Answer: C.

P.S. Not a GMAT style question.


Just curious, why not a GMAT style question?

Thanks

Cheers!
J :)

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Re: Which is larger, x^x or y^y?   [#permalink] 27 Dec 2013, 09:09
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If x and y are both integers, which is larger, x^x or y^y?

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