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# M26-32

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16 Sep 2014, 00:26
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Difficulty:

85% (hard)

Question Stats:

43% (01:06) correct 57% (01:13) wrong based on 255 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, 00:26
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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, 06:54
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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12 Sep 2017, 05:17
bunuel ,

your questions are amazing and fantastic.
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30 Aug 2018, 06:14
In st 2, can we also take 8 as value of y? (extremely tricky question)
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30 Aug 2018, 06:20
nigina93 wrote:
In st 2, can we also take 8 as value of y? (extremely tricky question)

No. First of all, we are told that x and y are negative integers, plus 8^x=-1/64 (8^(x + 2) = -1) does not have real solution for x.
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31 Aug 2018, 08:59
as x and y both are negative,

1) prime factorization of $$81=$$ $$3*3*3*3$$= $$3^4$$ or $$9^2$$ or $$81^1$$
so x could be -3 and y could be -4 OR x could be -9 and y could be -2. INSUFFICIENT

2) prime factorization of $$64 =$$ $$2*2*2*2*2*2$$ = $$2^6$$ or $$4^3$$ or $$8^2$$ or $$64^1$$
so y could be -4 and x could be -3 OR y could be -64 and x could be -1. INSUFFICIENT

by combining both:
there is only one common pair of x and y ($$x=-3 & y=-4$$)
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Hasnain Afzal

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31 Aug 2018, 15:28
Answered B - completely disregarded (-64)^-1
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31 Aug 2018, 18:02
hanamana wrote:
Answered B - completely disregarded (-64)^-1

its insufficient as the x, y pair could also be -4 and -3. i.e (-4)^-3
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Hasnain Afzal

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31 Aug 2018, 18:26
hasnain3047 wrote:
hanamana wrote:
Answered B - completely disregarded (-64)^-1

its insufficient as the x, y pair could also be -4 and -3. i.e (-4)^-3

yep, i was only able to come up with -4 and -3 and didn't think of -64 and -1 so ended up choosing B at first.
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31 Aug 2018, 18:32
hanamana wrote:
hasnain3047 wrote:
hanamana wrote:
Answered B - completely disregarded (-64)^-1

its insufficient as the x, y pair could also be -4 and -3. i.e (-4)^-3

yep, i was only able to come up with -4 and -3 and didn't think of -64 and -1 so ended up choosing B at first.

that's great...hope you got understand now...
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Hasnain Afzal

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05 Sep 2018, 02:45
Pattern - Option A gives 2 pairs, Option B gives 2 pairs. Together both point to one pair. Hence (C)
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18 Sep 2018, 06:39
Another great question. Forgot about the case where the exponent is 1 on statement 2.
Re: M26-32 &nbs [#permalink] 18 Sep 2018, 06:39
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# M26-32

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