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18 Jul 2010, 08:11
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (01:57) correct
40%
(01:53)
wrong
based on 552
sessions
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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 \(x > y\)
(1) \((y^2 + x^2)(y + x)(x - y) = 240\)
(2) \(x^y = y^x\) and \(x > y\)
18 Jul 2010, 11:05
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Suvorov
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 \(10^2<2^{10}\), ....
18 Jul 2010, 08:28
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ankitmania
Important property to know: \(x^2-y^2=(x+y)(x-y)\).
Given: \(x\) and \(y\) are distinct positive integers. Question: \(x^4-y^4=?\)
(1) \((y^2+x^2)(y+x)(x-y)=240\) --> \((y^2+x^2)(x^2-y^2)=240\) --> \(x^4-y^4=240\). Sufficient.
(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.
Answer: D.
