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# If x and y are integers, does x^y * y^(-x) = 1?

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If x and y are integers, does x^y * y^(-x) = 1? [#permalink]  06 Jun 2012, 22:44
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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:
[Reveal] Spoiler:
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.

[Reveal] Spoiler: OA

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Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]  06 Jun 2012, 22:58
Thank you Karishma for such a detailed reply.

Thanks.
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Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]  06 Jun 2012, 23:28
VeritasPrepKarishma wrote:
Responding to a pm:

Question: If x and y are integers, does (x)^y * (y)^-x = 1?

(1) (x)^x > y
(2) x > (y)^y

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.

good Q.. thanks for the explanation...
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Re: If x and y are integers, does x^y * y^(-x) = 1? [#permalink]  31 Jan 2015, 16:47
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Re: If x and y are integers, does x^y * y^(-x) = 1?   [#permalink] 31 Jan 2015, 16:47
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# If x and y are integers, does x^y * y^(-x) = 1?

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