Posted by **Alex L.** on Wednesday, October 5, 2011 at 6:18pm.

I have to find the prime factorization of each number and write down the answers with exponents when repeated factors appear.

eighteen, four hundred eighty, six hundred forty

Next, I need to write this fraction in lowest terms.

one-hundred sixty-five over one-hundred eighty

- Math -
**Steve**, Wednesday, October 5, 2011 at 11:42pm
This is not hard. Just start with 2, and divide it into the number as many times as you can. Then try 3,5,7,... for all primes up to sqrt(N). Why stop there? because if one factis greater than sqrt(N), the other factor will be less.

18

2*9

2*3*3 = 2*3^2

480

2*240

2*2*120

2*2*2*60

2*2*2*2*30

2^5 * 15

2^5 * 3 * 5

640

2^6 * 10

2^7 * 5

165/180

3*5*11/2*2*3*3*5

11/2*2*3

11/12

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