Prime factorization of a 4- digit number with at least three distinct factors Need two examples. SHOW ALL STEPS

We know that every number can be written as a product of prime numbers.

The method to find the factorized form of a number depends on the number, we just try to find the different factors by dividing by them, for example for the number 1000 we have:

1000 is an even number, then we can divide it by 2 (2 is a prime number)

1000 = 2*500   (so we already found a prime factor)

500 is also an even number, so we can divide it by 2

1000 = 2*500 = 2*2*250  (we found another prime factor)

dividing by 2 again we get:

1000 = 2*2*250 = 2*2*2*125

1000 = (2*2*2)*125

now we just need to factorize 125

we know that 125 is a multiple of 5, such that:

125 = 5*25 = 5*5*5

(5 is a prime number, so it is completely factorized).

Then the factorization of 1000 is:

1000 = (2*2*2)*(5*5*5) = 2^3*5^3

Now with another example, 1422

1422 is an even number, so we again start using the factor 2:

1422 = 2 = 711

then:

1422 = 2*711

we already found a factor.

711 is a multiple of 3 (the sum of its digits is a multiple of 3), then:

711/3 = 237

We can write our number as:

1422 = 2*3*237

237 is also a multiple of 3

237/3 = 79

then:

1422 = 2*3*3*79

and 79 is a prime number, so we already have 1422 completely factorized.