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Elementary arithmetic

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MateoLibre
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Elementary arithmetic [#permalink] New post   12 Nov 2010, 18:55
I stumbled across this problem recently and for the life of me couldn't solve it. My answer wasn't even close to any of the available answer choices. Can anyone break this one down for me?

\(\frac{(8^2)(3^3)(2^4)}{96^2}=?\)



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Re: Elementary arithmetic [#permalink] New post   12 Nov 2010, 19:06
Actually, I think I know what I did. I think I just found the prime factors of 96....not 96 SQUARED. So I was way off. I did it using 96 squared and the answer I got was 3. Can anyone else confirm this?
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Re: Elementary arithmetic [#permalink] New post   12 Nov 2010, 19:13
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MateoLibre wrote:
I stumbled across this problem recently and for the life of me couldn't solve it. My answer wasn't even close to any of the available answer choices. Can anyone break this one down for me?

\(\frac{(8^2)(3^3)(2^4)}{96^2}=?\)


In such questions, the best approach is to break it down into prime factors. The question testing whether you know your exponents well...

\(\frac{(8^2)(3^3)(2^4)}{96^2}=?\)

96 = 16 x 6 (Try and be comfortable with multiplication tables) = \(2^5 * 3\)
\(8 = 2^3\)

So, \(\frac{(8^2)(3^3)(2^4)}{96^2}=?\) = \(\frac{(2^3)^2.3^3.2^4}{(2^5 * 3)^2}\)

= \(\frac{2^{6 + 4} . 3^3}{(2^5* 3)^2}\)
\(=2^{10 - 10} .3^{3 - 2} = 3\)

If basics of exponents are an issue, check out my blog:
https://gmatquant.blogspot.com/2010/11/s ... tials.html
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Re: Elementary arithmetic [#permalink] New post   12 Nov 2010, 19:15
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MateoLibre wrote:
Actually, I think I know what I did. I think I just found the prime factors of 96....not 96 SQUARED. So I was way off. I did it using 96 squared and the answer I got was 3. Can anyone else confirm this?


Thats right! Looks like you are fine with exponents.. Just a silly mistake!



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Re: Elementary arithmetic [#permalink]
12 Nov 2010, 19:15
Moderator:
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