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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: 52164
(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
00:00

Difficulty:

45% (medium)

Question Stats:

60% (01:10) correct 40% (01:35) wrong based on 68 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

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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
1
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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Joined: 15 Feb 2017
Posts: 11
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 &nbs [#permalink] 21 May 2017, 09:44
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