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(2) \(x^y = y^x\) and \(x>y\), also \(x\) and \(y\) are distinct positive integers --> only one such pair is possible \(x=4>y=2\): \(x^y=4^2=16=2^4=y^x\) --> \(x^4-y^4=240\). Sufficient.

only one such pair is possible \(x=4>y=2\): \(x^y=4^2=16=2^4=y^x\) --> \(x^4-y^4=240\). Sufficient.

How do you come to this conclusion? Just picking numbers/knowing or is there some mathematical rule for this? I understand that it revolves around the power of 2, but can't put my finger on it.

only one such pair is possible \(x=4>y=2\): \(x^y=4^2=16=2^4=y^x\) --> \(x^4-y^4=240\). Sufficient.

How do you come to this conclusion? Just picking numbers/knowing or is there some mathematical rule for this? I understand that it revolves around the power of 2, but can't put my finger on it.

I think it's worth remembering that \(4^2=16=2^4\), I've seen several GMAT questions on number properties using this (another useful property \(8^2=4^3=2^6=64\)).

But if you don't know this property:

Given: \(x^y = y^x\) and \(x>y\), also \(x\) and \(y\) are distinct positive integers. Couple of things: \(x\) and \(y\) must be either distinct positive odd integers or distinct positive even integers (as odd in ANY positive integer power is odd and even in ANY positive integer power is even).

After testing several options you'll see that \(x=4\) and \(y=2\) is the only possible scenario: because, when \(y\geq{2}\) and \(x>{4}\), then \(x^y\) (bigger value in smaller power) will be always less than \(y^x\) (smaller value in bigger power): \(5^3<3^5\), or \(6^2<2^6\), or \(8^2<2^8\), or \(10^2<2^{10}\), ....
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Re: If x and y are distinct positive integers, what is the value [#permalink]

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15 Oct 2013, 06:20

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Re: If x and y are distinct positive integers, what is the value [#permalink]

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20 Nov 2014, 13:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x and y are distinct positive integers, what is the value [#permalink]

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10 Mar 2015, 02:46

Bunuel wrote:

Suvorov wrote:

Bunuel wrote:

only one such pair is possible \(x=4>y=2\): \(x^y=4^2=16=2^4=y^x\) --> \(x^4-y^4=240\). Sufficient.

How do you come to this conclusion? Just picking numbers/knowing or is there some mathematical rule for this? I understand that it revolves around the power of 2, but can't put my finger on it.

I think it's worth remembering that \(4^2=16=2^4\), I've seen several GMAT questions on number properties using this (another useful property \(8^2=4^3=2^6=64\)).

But if you don't know this property:

Given: \(x^y = y^x\) and \(x>y\), also \(x\) and \(y\) are distinct positive integers. Couple of things: \(x\) and \(y\) must be either distinct positive odd integers or distinct positive even integers (as odd in ANY positive integer power is odd and even in ANY positive integer power is even).

After testing several options you'll see that \(x=4\) and \(y=2\) is the only possible scenario: because, when \(y\geq{2}\) and \(x>{4}\), then \(x^y\) (bigger value in smaller power) will be always less than \(y^x\) (smaller value in bigger power): \(5^3<3^5\), or \(6^2<2^6\), or \(8^2<2^8\), or \(10^2<2^{10}\), ....

Want to solve a GMAT like problem using this property 8^2=4^3=2^6=64 ??? Do you have any in your arsenal??Pls share
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Re: If x and y are distinct positive integers, what is the value [#permalink]

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19 Apr 2016, 05:06

Hello from the GMAT Club BumpBot!

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\(x^4 - y^4\) i started by factoring out... this basically equals: \((y^2 + x^2)(y + x)(x - y)\)

1. we are given the exact answer...sufficient.

2. we must have multiples of 2..otherwise it would not work.. i started with x=4 and y=2 4^2 = 2^4 2^4 = 2^4. works tried few more options, don't work.