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Can someone explain the process here?

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Can someone explain the process here?

by blaque831 » Tue Aug 19, 2008 1:43 pm
#163 of the Official Quant Review (green):

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

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

Can someone explain how I could figure out that 96^2 is equal to (8^2)(3^2)(2^4)? I understand how to work the problem but not how I can figure out that such a large number (96^2) is broken down into smaller powers. Thanks in advance!
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by pepeprepa » Tue Aug 19, 2008 2:02 pm
The principle is that you want to delete factors which are both at the denominator and at the numerator so you will try to break into pieces 96²

How to arrive to that: 96² = 8² * 3² * 2^4

96=2*48=2*2*24=2*2*2*12=2*2*2*2*6=2^5*3
so
96²=2^10*3²=2^4*2^6*3²=2^4*3²*8²
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by parallel_chase » Tue Aug 19, 2008 2:04 pm
Whenever you see exponents like these always prime factorize.

You should remember all the powers of 2 till 10

2^1 =2
2^2 =4
2^3 =8 so on so forth.


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


96 = 3 * 2^5
96^2 = (3^2) (2^10)

8 = 2^3
8^2 = 2^6

Numerator (8^2)(3^3)(2^4) = (2^6)(3^3)(2^4) = (2^10)(3^3)
Denominator (3^2) (2^10)

(2^10)(3^3) / (3^2) (2^10) =3

Hope this helps.
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by blaque831 » Tue Aug 19, 2008 2:43 pm
Parallel that was the perfect explanation for my thought process. Thanks!
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