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29 Dec 2015, 21:07
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Below is the question - there is a gap in the simplification for me:
What is the greatest prime factor of \(12^3 – 96\)?
Pulling out the factors is easy enough:
\((2^2*3)^3-(2^5*3)\)
\(2^6*3^3-2^5*3\)
Then the solution shows this simplification and I get stuck following:
\(2^5*3*(2*3^2-1)\)
Would someone break down these steps for me?
Once you get to this part is it straightforward, just do the math & you get:
\(2^5*3*17\)
Thank you in advance
What is the greatest prime factor of \(12^3 – 96\)?
Pulling out the factors is easy enough:
\((2^2*3)^3-(2^5*3)\)
\(2^6*3^3-2^5*3\)
Then the solution shows this simplification and I get stuck following:
\(2^5*3*(2*3^2-1)\)
Would someone break down these steps for me?
Once you get to this part is it straightforward, just do the math & you get:
\(2^5*3*17\)
Thank you in advance
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29 Dec 2015, 21:16
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elvk11
There are 2 ways to look at it: realize that 96 is also a multiple of 12,
\(12^3-96 = 12^3-12*8 = 12*(12^2-8) = 12*(136) = 12*8*17\) , now as 12 and 8 are NOT prime, 17 will be the greatest prime number factor of 12^3-96.
Alternately, once you get \(2^5*3*(2*3^2-1)\) ---> \(2^5*3*(2*9-1)\) ---> \(2^5*3*(18-1)\) ---> \(2^5*3*17\) , as all 2,3,17 are all prime numbers with 17 being the greatest you get the same answer using both approaches.
Hope this helps.
29 Dec 2015, 21:34
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Thank you - I didn't think of the first way you showed.
My OP was a bit ambiguous I believe, I get stuck doing the simplification from prime factors of 12^3-prime factors of 96 into the format where you are able to solve
My OP was a bit ambiguous I believe, I get stuck doing the simplification from prime factors of 12^3-prime factors of 96 into the format where you are able to solve
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
