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Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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02 Nov 2011, 15:42

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.

Last edited by kostyan5 on 02 Nov 2011, 17:43, edited 3 times in total.

Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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02 Nov 2011, 16:37

For B, I plugged in a few sets of numbers and all resulted in the original equation not holding true. Instead of wasting time trying to solve it, I decided to simulate the test conditions and simply assume that none of the numbers would work for the original equation. Therefore, B would be sufficient to say that the original equation will not hold true.

I will work out the actual solving at a later time.

Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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02 Nov 2011, 17:41

Continuing from previous post, the integer solutions to \(x^y=y^x\) are X=Y, (2,4), (4,2), (-2,-4), and (-4,-2).

(1) \(x^x>y\): It is possible to find solutions that both work for (1) and for original equation. It is also possible to find solutions that work for (1) but not for original equation. Therefore, (1) is insufficient to answer.

(2), \(x>y^y\): None of the solutions to the original equation work for (2). That means, given (2), there is no possible way to make the original equation work. Therefore, (2) is sufficient to answer whether \(x^y = y^x\), and that is NO.

So the answer to the question is B: (2) alone is sufficient and (1) is not sufficient.

Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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03 Nov 2011, 01:15

enigma123 wrote:

If x and y are integers, does x^y*y^-x=1

1. x^x>y 2. x>y^y

Guys - any idea the approach to solve this question please?

The above explanation is good enough. I'll just add some text to it.

Q: Is x^y=y^x

In other words: Is x=y OR Is (x,y) any of the pairs: (2, 4), (4, 2), (-2, -4), (-4, -2)

1. x^x>y Say (x,y)=(4,2) x^x=4^4>2; Good. Answer to the question=Yes, x^y is equal to y^x as (x, y) is one of the mentioned pairs.

But say (x,y)=(5,2) x^x=5^5>2; Good. Answer to the question=No

Not Sufficient.

2. x>y^y

Now, We can definitely say that x NOT equal to Y. Let's see whether they can be any of the mentioned pairs. (x,y)=(2,4); No; (x,y)=(4,2); No; (x,y)=(-2,-4); No; as -2 < (-4)^(-4) (x,y)=(-4,-2); No; as -4 < (-2)^(-2)

So, (x,y) is not one of the pairs that will make the expression true. So, we can definitely conclude that x^y*y^-x NOT equal to 1 A definite NO proves sufficiency. Sufficient.

Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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03 Nov 2011, 07:23

Thanks, fluke.

However, are you supposed to memorize solutions to that equation? I don't see how one could come up with those in a reasonable amount of time during the GMAT.

Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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15 Nov 2011, 15:10

I got B. Because x>y^y than 2 and 4 or are not an option. Therefore by the fraction made by the negative exponent will not = 1 whether it is postive or negative. Good question took me 1:47 seconds, a little longer than I would like to spend on 600-700. kudos

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.

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).

good Q.. thanks for the explanation...
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Best Vaibhav

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Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]

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