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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] New post 18 Feb 2012, 08:33
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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
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Re: Which is larger, x^x or y^y? [#permalink] 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] New post 18 Feb 2012, 11:08
Expert's post
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] 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] 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] 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] 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] 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 :)
Re: Which is larger, x^x or y^y?   [#permalink] 27 Dec 2013, 09:09
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