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# If x and y are distinct positive integers

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If x and y are distinct positive integers [#permalink]  06 Nov 2012, 17:37
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If x and y are distinct positive integers, what is the value of x^4 - y^4?

(1) (y^2 + x^2)(y + x)(x - y) = 240
(2) x^y = y^x and x > y
[Reveal] Spoiler: OA

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Re: If x and y are distinct positive integers [#permalink]  06 Nov 2012, 17:50
1. sufficient b/c doing the algebra, comes out to be x^4 - y^4 = 240.
2. sufficient b/c only case this works is when x = 4 and y = 2
4^2 = 2^4

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Re: If x and y are distinct positive integers [#permalink]  07 Nov 2012, 02:40
Expert's post
If x and y are distinct positive integers, what is the value of x^4 - y^4?

(1) (y^2 + x^2)(y + x)(x - y) = 240 --> $$(x^2 +y^2)(x^2 - y^2) = 240$$ --> $$x^4-y^4= 240$$. Sufficient.

(2) x^y = y^x and x > y. With trial and error we can find that, since x and y are positive integers and x>y, then x=4 and y=2. Sufficient.

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Re: If x and y are distinct positive integers [#permalink]  27 Apr 2013, 01:04
bunuel

ok, 1 is pretty much understandable, but for 2, how do you get there? what is the mind process here, in similar problems on the actual test, how do you get there in under 2 minutes? do you start just plugging the numbers in?
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Re: If x and y are distinct positive integers [#permalink]  13 Aug 2013, 07:50
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lololol650 wrote:
bunuel

ok, 1 is pretty much understandable, but for 2, how do you get there? what is the mind process here, in similar problems on the actual test, how do you get there in under 2 minutes? do you start just plugging the numbers in?

Statement 2 is sufficient because, only one pair of integers that can satisfy the condition $$X^y = Y^x$$ is 4 and 2. $$(4^2 = 2^4)$$. So whenever you see such equation you can rest assured that one integer must be 4 and other one be 2.

Hope that helps.
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Re: If x and y are distinct positive integers [#permalink]  21 Jan 2014, 06:26
Narenn wrote:
lololol650 wrote:
bunuel

ok, 1 is pretty much understandable, but for 2, how do you get there? what is the mind process here, in similar problems on the actual test, how do you get there in under 2 minutes? do you start just plugging the numbers in?

Statement 2 is sufficient because, only one pair of integers that can satisfy the condition $$X^y = Y^x$$ is 4 and 2. $$(4^2 = 2^4)$$. So whenever you see such equation you can rest assured that one integer must be 4 and other one be 2.

Hope that helps.

Not really, they have to specify that such numbers must be positive integers otherwise one could have 4 combinations of 2 and 4
Re: If x and y are distinct positive integers   [#permalink] 21 Jan 2014, 06:26
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# If x and y are distinct positive integers

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