8^2)(3^3)(2^4)/96^2 A. 3 B. 6 C. 9 D. 12 E. 18 : Quant Question Archive [LOCKED]
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# 8^2)(3^3)(2^4)/96^2 A. 3 B. 6 C. 9 D. 12 E. 18

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8^2)(3^3)(2^4)/96^2 A. 3 B. 6 C. 9 D. 12 E. 18 [#permalink]

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09 Aug 2007, 10:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

(8^2)(3^3)(2^4)/96^2

A. 3
B. 6
C. 9
D. 12
E. 18
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Joined: 12 Jul 2007
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09 Aug 2007, 11:40
yup, I got A by breaking it all down into like terms

16*4*3*3*3*16

divided by

16*6*16*6

then just start dividing out like that. you're left with a 3 in the numerator and 1 in the bottom.
09 Aug 2007, 11:40
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