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Difficulty:
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Question Stats:
27% (02:34) correct
73%
(02:22)
wrong
based on 483
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If x and y are integers, does x^y * y^(-x) = 1?
(1) x^x > y
(2) x > y^y
(1) x^x > y
(2) x > y^y
06 Jun 2012, 23:44
Kudos
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If x and y are integers, does x^y * y^(-x) = 1?
(1) x^x > y
(2) x > y^y
SOLUTION:
Let's re-arrange the question first:
Is \((x)^y * (y)^{-x} = 1\)?
Is \((x)^y = (y)^x\)?
Check this post for a detailed discussion on this: try-this-one-700-level-number-properties-103461.html#p805817
So, \((x)^y = (y)^x\) when x = y or x and y take values 2,4 or -2,-4
Look at the statements now:
(1) \((x)^x > y\)
We know this relation is true for many random values of x and y e.g. x = 4, y = 5 etc. So the answer to the question is NO in this case. \((x)^y\) is not equal to \((y)^x\).
But does it hold for any values which will make \((x)^y = (y)^x\)?
Yes it does! If x = y, x^x > y is true for say, x = y = 3. 3^3 is greater than 3. So x and y can take values which will give the answer YES.
Not sufficient.
(2) \(x > (y)^y\)
Again, it holds for many random values of x and y e.g. x = 10, y = 2 etc. So the answer to the question is NO in this case.
But does it hold for any values which will make \((x)^y = (y)^x\)?
Let's see. If x = y, x cannot be greater than \(y^y\). Check for a few values to figure out the pattern.
If x = 4 and y = 2, x is not greater than \(y^y\).
Similarly, it doesn't work for x = -2, y = -4 and x = -4 and y = -2 since x will be negative while y^y will be positive.
Therefore, if \(x > (y)^y\), \((x)^y = (y)^x\) cannot hold for any values of x and y. Hence answer to the question stays NO.
Sufficient.
Answer (B).
(1) x^x > y
(2) x > y^y
SOLUTION:
Let's re-arrange the question first:
Is \((x)^y * (y)^{-x} = 1\)?
Is \((x)^y = (y)^x\)?
Check this post for a detailed discussion on this: try-this-one-700-level-number-properties-103461.html#p805817
So, \((x)^y = (y)^x\) when x = y or x and y take values 2,4 or -2,-4
Look at the statements now:
(1) \((x)^x > y\)
We know this relation is true for many random values of x and y e.g. x = 4, y = 5 etc. So the answer to the question is NO in this case. \((x)^y\) is not equal to \((y)^x\).
But does it hold for any values which will make \((x)^y = (y)^x\)?
Yes it does! If x = y, x^x > y is true for say, x = y = 3. 3^3 is greater than 3. So x and y can take values which will give the answer YES.
Not sufficient.
(2) \(x > (y)^y\)
Again, it holds for many random values of x and y e.g. x = 10, y = 2 etc. So the answer to the question is NO in this case.
But does it hold for any values which will make \((x)^y = (y)^x\)?
Let's see. If x = y, x cannot be greater than \(y^y\). Check for a few values to figure out the pattern.
If x = 4 and y = 2, x is not greater than \(y^y\).
Similarly, it doesn't work for x = -2, y = -4 and x = -4 and y = -2 since x will be negative while y^y will be positive.
Therefore, if \(x > (y)^y\), \((x)^y = (y)^x\) cannot hold for any values of x and y. Hence answer to the question stays NO.
Sufficient.
Answer (B).
General Discussion
kostyan5
Joined: 29 Oct 2011
Last visit: 27 Apr 2021
Posts: 114
Own Kudos:
Given Kudos: 19
Concentration: General Management, Technology
Schools: Sloan '16 (D)
GMAT 1: 760 Q49 V44
GPA: 3.76
Original attempt (or as I would have done it on a GMAT never having seen this question before). For a more thorough solution see below.
First, rewrite the original equation. \(x^y*y^{-x}=1 -> (x^y)/(y^x)=1 -> x^y = y^x\)
I worked it using sample numbers.
Case (1): x=2, y=2 works in original equation
x = 2, y=1 does not work
Insufficient
Case (2): I couldn't quickly come up with 2 numbers that would satisfy the original equation so I assumed that it would be sufficient to say that the original equation will not be equal to 1. This is how I would guess on a GMAT.
I'd say B.
First, rewrite the original equation. \(x^y*y^{-x}=1 -> (x^y)/(y^x)=1 -> x^y = y^x\)
I worked it using sample numbers.
Case (1): x=2, y=2 works in original equation
x = 2, y=1 does not work
Insufficient
Case (2): I couldn't quickly come up with 2 numbers that would satisfy the original equation so I assumed that it would be sufficient to say that the original equation will not be equal to 1. This is how I would guess on a GMAT.
I'd say B.
