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27 Mar 2012, 16:29
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Hi, I was wondering if there is a quick way to find the prime factors of a number like 720. How would I know that 720 = 2^4 x3^2 x5 without a lot of trial and error. Thanks?
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27 Mar 2012, 17:11
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The best suggestion I can give is to do LOTS of prime factorization - get fast.
This is how I'd go at 720.
\(720\)
\(72 * 10\)
\(8 * 9 * 5 * 2\)
\(2^3 * 3^2 * 5 * 2\)
\(2^4 * 3^2 * 5\)
This is how I'd go at 720.
\(720\)
\(72 * 10\)
\(8 * 9 * 5 * 2\)
\(2^3 * 3^2 * 5 * 2\)
\(2^4 * 3^2 * 5\)
29 Mar 2012, 00:53
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scottaroner
Start by breaking down the number into its components. Use whatever comes to your mind first.
e.g. \(720 = 72*10 = (8*9)*(2*5) = 2^3*3^2*2*5 = 2^4*3^2*5\)
\(365 = 5*73\) (when I see 365, the first thing that comes to mind is that it is divisible by 5. Don't know which other factors it has so simply divide it by 5 to break it down. 73 is prime.)
\(1064 = 8*133 = 2^3 * 19*7\) (Since last 3 digits of 1064 i.e. '064' is divisible by 8, 1064 is divisible by 8. I know the multiplication table of 19 so 133 was easy. You could have started by looking for a factor of 133 if you don't the tables)
etc
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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!
