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M26-32

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M26-32  [#permalink]

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New post 16 Sep 2014, 01:26
11
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

43% (01:02) correct 57% (01:10) wrong based on 242 sessions

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Re M26-32  [#permalink]

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New post 16 Sep 2014, 01:26
3
2
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.


Answer: C
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Re: M26-32  [#permalink]

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New post 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.


Answer: C


Very good question.
In statement 2, I forget the SPECIAL CASE in which x = -1.
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Re: M26-32  [#permalink]

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New post 12 Sep 2017, 06:17
bunuel ,

your questions are amazing and fantastic.
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Re: M26-32  [#permalink]

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New post 30 Aug 2018, 07:14
In st 2, can we also take 8 as value of y? (extremely tricky question)
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Re: M26-32  [#permalink]

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New post 30 Aug 2018, 07:20
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Re: M26-32  [#permalink]

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New post 31 Aug 2018, 09:59
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 (\(x=-3 & y=-4\))
Answer is C
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Re: M26-32  [#permalink]

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New post 31 Aug 2018, 16:28
Answered B - completely disregarded (-64)^-1
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Re: M26-32  [#permalink]

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New post 31 Aug 2018, 19:02
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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Re: M26-32  [#permalink]

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New post 31 Aug 2018, 19:26
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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Re: M26-32  [#permalink]

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New post 31 Aug 2018, 19:32
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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Re: M26-32  [#permalink]

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New post 05 Sep 2018, 03:45
Pattern - Option A gives 2 pairs, Option B gives 2 pairs. Together both point to one pair. Hence (C)
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Re: M26-32  [#permalink]

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New post 18 Sep 2018, 07:39
Another great question. Forgot about the case where the exponent is 1 on statement 2.
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Re: M26-32 &nbs [#permalink] 18 Sep 2018, 07:39
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