What is the best way to find factors?

Answer 1

There are a few ways to go about factoring. You can decide what works best for you. I always find the prime factorization first. Let's look at a random number: 108

The prime factorization can be found by using a factor tree.

108

54,2

27,2,2

9,3,2,2

3,3,3,2,2

2^2 x 3^3 = 108

Half of the factors will be less than the square root, half greater. If the number is a perfect square, there will be an equal number of factors on either side of the square root. In this case, the square root is between 10 and 11.

Adding one to the exponents of the prime factorization and multiplying them will tell you how many factors there are. In this case, the exponents are 2 and 3. Add one to each. 3 x 4 = 12

108 has 12 factors. Six of them are 10 or less, six of them are 11 or greater. All we have to do is divide the numbers one through ten into 108. If the result (quotient) turns out to be an integer, you've found a factor pair. Knowing the rules of divisibility will make that even easier.

108 is divisible by...

1 because everything is.

2 because it's even.

3 because its digits add up to a multiple of 3.

4 because its last two digits are a multiple of 4.

6 because it's a multiple of 2 and 3.

9 because its digits add up to a multiple of 9.

That's six factors less than 10. Divide them into 108. That's the rest of them.

(108,1)(54,2)(36,3)(27,4)(18,6)(12,9)

1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108

Notice that all of those numbers, except for 1, can also be found in the prime factorization.

Answer 2

The first step is to find the prime factorisation for the number. The best way for doing this depends partly on your arithmetic skills and partly on the size of the number that you are trying to factorise.

Divisibility algorithm: check if the smallest prime, 2, is a factor. If so, divided the number by 2: this is the number that you need to factorise now. Check divisibility by 2 again (and again). If (when) it is no longer a factor, check divisibility by the next prime (3). Continue through the primes until you get to the square root of the number you are trying to factorise. At that stage the number that you were trying to factorise is the last prime factor.

Factor Tree: this method is useful is you can spot a factor easily. Divide the number by the factor to give you two numbers. Each of these is smaller than the original and so, to that extent, the problem has been reduced. The next step is to factorise the two factors and continue doing so until you are left with only prime numbers.

WARNING: There are numbers, particularly those used in cryptography, which have factors that are numbers with a hundred or more digits. You will not be able to factorise these without some very sophisticated software.

if a number n can be expressed as = p1^a1 * p2^a2 * ... * pk^ak where p1, p2, ... pk are prime numbers and a1, a2, ... ak are integers.

Then the factors of n are of the form p1^b1 * p2^b2 * ... * pk^bk where b1, b2, ... bk are integers and for each i 0 <= bi <=ai (i = 1, 2, ...k).