scottaroner wrote:

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?

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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