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Simplying Exponents And Dividing

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by Anurag@Gurome » Sun Feb 26, 2012 7:57 pm
RyanP wrote:(8^2)(3^3)(2^4)/(96^2)=

A = 3
B = 6
C = 9
D = 12
E = 18

I tried to read through this post here: https://www.beatthegmat.com/can-someone- ... 16619.html but I just don't understand the math.
96 = 2^5 * 3

(8^2)(3^3)(2^4)/(96^2) = (2^3)^2 * (3^3) * (2^4)/(2^5 * 3)^2
= (2^6) * (3^3) * (2^4)/(2^10 * 3²)
= (2^10) * (3^3))/(2^10 * 3²)
= 3

The correct answer is A.
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by RyanP » Fri Mar 02, 2012 2:19 pm
Anurag@Gurome wrote:
(8^2)(3^3)(2^4)/(96^2) = (2^3)^2 * (3^3) * (2^4)/(2^5 * 3)^2
= (2^6) * (3^3) * (2^4)/(2^10 * 3²)
= (2^10) * (3^3))/(2^10 * 3²)
= 3

The correct answer is A.
I don't understand how you reduce 96^2 to (2^5*3)^2. 96*96 = 9216. (64*3)^2 = 36864. Am I missing something? How are those equivalent?
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by krusta80 » Fri Mar 02, 2012 4:52 pm
RyanP wrote:(8^2)(3^3)(2^4)/(96^2)=

A = 3
B = 6
C = 9
D = 12
E = 18

I tried to read through this post here: https://www.beatthegmat.com/can-someone- ... 16619.html but I just don't understand the math.
This is really just an exercise in breaking down 96 into a product of primes:

96 = 2^5 * 3^1

Let's also do the same for 8:

8 = 2^3

Applying this to the problem at hand...

(8^2)(3^3)(2^4)/(96^2)=((2^3)^2)(3^3)(2^4)/((2^5 * 3^1)^2)=(2^10)(3^3)/[(2^10)(3^2)]

Now we can start cancelling out terms:

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


Answer is A
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by krusta80 » Fri Mar 02, 2012 4:53 pm
RyanP wrote:
Anurag@Gurome wrote:
(8^2)(3^3)(2^4)/(96^2) = (2^3)^2 * (3^3) * (2^4)/(2^5 * 3)^2
= (2^6) * (3^3) * (2^4)/(2^10 * 3²)
= (2^10) * (3^3))/(2^10 * 3²)
= 3

The correct answer is A.
I don't understand how you reduce 96^2 to (2^5*3)^2. 96*96 = 9216. (64*3)^2 = 36864. Am I missing something? How are those equivalent?
2^5 is 32, not 64 :)
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