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algebra tricky question, 600 level

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algebra tricky question, 600 level

by Night reader » Sat Oct 30, 2010 3:15 pm
(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) - 11
(B) -18
(C) -1
(D) -19
(E) -21
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by pesfunk » Sat Oct 30, 2010 8:47 pm
(A^x)(A^y)(A^z) = A^(X+Y+Z)

A^(X+Y+Z) = Z^-21

X+Y+Z = -21

So, smallest value of X can be -19

Hence, Answer D
Night reader 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) - 11
(B) -18
(C) -1
(D) -19
(E) -21
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by limestone » Sat Oct 30, 2010 9:01 pm
(A^x)(A^y)(A^z) = A^(x+y+z) = A^-21

Thus x+y+z = -21.
x,y,z are three different negative integers, so, if we want to have the smallest x, we must assign the largest y,z possible.

The largest different negative y,z are : -1 and -2. Thus the smallest x must be : - 18

So pick B.
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by pesfunk » Sat Oct 30, 2010 9:06 pm
Ahh...i missed the word "different negative integers".


limestone wrote:(A^x)(A^y)(A^z) = A^(x+y+z) = A^-21

Thus x+y+z = -21.
x,y,z are three different negative integers, so, if we want to have the smallest x, we must assign the largest y,z possible.

The largest different negative y,z are : -1 and -2. Thus the smallest x must be : - 18

So pick B.
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