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Does anyone know of an easy way (besides using a calculator) to find the prime factorization of a number?

For example, 12347983?

2007-01-30 09:12:02 · 4 answers · asked by Napper 2 in Science & Mathematics Mathematics

PS. I know the answer to the example, i am just looking for the method.

2007-01-30 09:15:05 · update #1

note, i know how to turn a number into a prime factorization... i just need to know how the heck i'm supposed to know 281 is the first prime to divide into 12347983?

2007-01-30 09:21:48 · update #2

4 answers

Divide by two until it won't divide exactly and then move onto next prime number i.e. 3. Repeat until you have your answer.

2007-01-30 09:18:25 · answer #1 · answered by Anonymous · 0 0

I usually start with the smallest prime number and work my way up. So, if a number is even, divide by 2's. If odd, check to see if it divides by 3, then 5, then 7 and so on.

To check for 3, add the digits: 1+2+3+4+7+9+8+3=37

For 7 and up, simply divide with calculator. Unfortunately beyond 7, there are very few "easy" ways beyond using a calculator.

2007-01-30 17:44:30 · answer #2 · answered by danjlil_43515 4 · 0 0

only way I know is trial and error... if you're doing this for a class, this is the basis behind encryption.

It takes a heck of a lot longer to figure out what to factor something into if you don't already know one of the numbers.

The fastest way is to find a list of primes and just divide by those numbers. That's why prime number finding is such a wanted topic in mathematics.

2007-01-30 17:30:26 · answer #3 · answered by brothergoosetg 4 · 0 0

Start by trying to divide by 2. If you can divide the number by 2, do it, then test the quotient for divisibility by 2. Once you reach an odd number, test that number for divisibility by 3. Keep dividing by 3 until you can no longer do so, then move to 5, 7, 11, etc. until you reach a number you recognize as prime.

I was going to demonstrate your example, but I got tired of testing numbers once I reached 109.

2007-01-30 17:23:11 · answer #4 · answered by Chris S 5 · 0 0

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