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# M26-32

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Math Expert
Joined: 02 Sep 2009
Posts: 47219

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16 Sep 2014, 01:26
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Difficulty:

85% (hard)

Question Stats:

35% (00:48) correct 65% (01:16) wrong based on 43 sessions

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If $$x$$ and $$y$$ are negative integers, then what is the value of $$xy$$?

(1) $$x^y=\frac{1}{81}$$

(2) $$y^x=-\frac{1}{64}$$

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16 Sep 2014, 01:26
1
Official Solution:

(1) $$x^y=\frac{1}{81}$$. As both $$x$$ and $$y$$ are negative integers then $$x^y=\frac{1}{81}=(-9)^{-2}=(-3)^{-4}$$ hence $$xy=18$$ or $$xy=12$$. Note that as negative integer ($$x$$) in negative integer power ($$y$$) gives positive number ($$\frac{1}{81}$$), then the power must be negative even number. Not sufficient.

(2) $$y^x=-\frac{1}{64}$$. As the result is negative then $$x$$ must be negative odd number, so: $$y^x=-\frac{1}{64}=(-4)^{-3}=(-64)^{-1}$$ hence $$xy=12$$ or $$xy=64$$. Not sufficient.

(1)+(2) Only one pair of negative integers $$x$$ and $$y$$ satisfies both statements $$x=-3$$ and $$y=-4$$. Therefore $$xy=12$$. Sufficient.

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13 Jul 2017, 07:54
Bunuel wrote:
Official Solution:

(1) $$x^y=\frac{1}{81}$$. As both $$x$$ and $$y$$ are negative integers then $$x^y=\frac{1}{81}=(-9)^{-2}=(-3)^{-4}$$ hence $$xy=18$$ or $$xy=12$$. Note that as negative integer ($$x$$) in negative integer power ($$y$$) gives positive number ($$\frac{1}{81}$$), then the power must be negative even number. Not sufficient.

(2) $$y^x=-\frac{1}{64}$$. As the result is negative then $$x$$ must be negative odd number, so: $$y^x=-\frac{1}{64}=(-4)^{-3}=(-64)^{-1}$$ hence $$xy=12$$ or $$xy=64$$. Not sufficient.

(1)+(2) Only one pair of negative integers $$x$$ and $$y$$ satisfies both statements $$x=-3$$ and $$y=-4$$. Therefore $$xy=12$$. Sufficient.

Very good question.
In statement 2, I forget the SPECIAL CASE in which x = -1.
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Joined: 13 May 2017
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12 Sep 2017, 06:17
bunuel ,

your questions are amazing and fantastic.
Re: M26-32 &nbs [#permalink] 12 Sep 2017, 06:17
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# M26-32

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