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## Main Question or Discussion Point

**Is this line of thought correct? Please correct me where I´m wrong.**

**Will this way of finding prime factors work when A is**

__any__integer?**Is there a proof for this or a proof that is closely related?**

**Is there a way to do it that requiers less iterations? It has to be a method that requires no intelligence ;)**

A = any integer

B = prime number

If A / B is evenly divisible then B is a prime factor of A

The first time A is devided by 2, then 3, 5, 7, 11....

Example:

2*5*5*5*5*7 = 8750 =

**A**

8750/2= 4375

4375/3 is not an integer

Now the prime number 3 is ruled out as a possible prime factor of 8750

4375/5 = 875

875/7 = 125

125/11 is not an integer

Now the prime number 11 is ruled out as a possible prime factor of 8750

125/13 is not an integer

Now the prime number 13 is ruled out as a possible prime factor of 8750

...............

The ratio is not an integer

Now the prime number n is ruled out as a possible prime factor of 8750

...............

The next prime, 67, is greater than 125/2.

**Can you still find the prime factors if you stop deviding before prime number**

__n__is greater than the last ratio that was an integer?We have found the greatest prime factor that 8750 is made up by and some of the rest but not all of them since the ratio isn´t 1. So far we have found 2,5 and 7

125/2 is not an integer

Now the prime number 2 is ruled out as a possible prime factor of 8750

We have already ruled out 3 and there is no need to devide by it again

125/5 = 25

25/7 is not an integer

Now the prime number 7 is ruled out as a possible prime factor of 8750

25/5 = 5

5/5 = 1

Now we have found all prime factors that 8750 is made up by. They are 2,5,5,5,5 and 7

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