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# (A^x)(A^y)(A^z) = A^(-21). If A > 0, and x, y and z are each different

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Math Expert
Joined: 02 Sep 2009
Posts: 43787
(A^x)(A^y)(A^z) = A^(-21). If A > 0, and x, y and z are each different [#permalink]

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21 May 2017, 08:34
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55% (hard)

Question Stats:

56% (00:48) correct 44% (01:13) wrong based on 51 sessions

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$$(A^x)(A^y)(A^z) = A^{(-21)}$$. If A > 0, and x, y and z are each different negative integers, what is the smallest that x could be?

A. -21
B. -19
C. -18
D. -11
E. -1
[Reveal] Spoiler: OA

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(A^x)(A^y)(A^z) = A^(-21). If A > 0, and x, y and z are each different [#permalink]

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21 May 2017, 08:48
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Bunuel wrote:
$$(A^x)(A^y)(A^z) = A^{(-21)}$$. If A > 0, and x, y and z are each different negative integers, what is the smallest that x could be?

A. -21
B. -19
C. -18
D. -11
E. -1

$$(A^x)(A^y)(A^z) = A^{(-21)}$$

$$A^{(x+y+z)} = A^{(-21)}$$

x+y+z = -21

x, y and z are each different negative integers. Therefore, x,y,z could be (-18,-1,-2)

x+y+z = (-18) + (-1) + (-2) = -18-1-2 = -21

Therefore Smallest value of x could be -18

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Re: (A^x)(A^y)(A^z) = A^(-21). If A > 0, and x, y and z are each different [#permalink]

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21 May 2017, 09:44
X+Y+z=-21,
Minimum x
So take y & z maximum
Y=-1, z=-2
X= -21+3=-18
Option C

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Re: (A^x)(A^y)(A^z) = A^(-21). If A > 0, and x, y and z are each different   [#permalink] 21 May 2017, 09:44
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