Explain how to find the prime factorization of any whole number?

Answer 1

If you are in school and get this type of question, you pretty much can handle them by try & error. Simply divide your number by trying prime numbers one by one, from smallest to larger, until you get a prime number.

Assume you get 1750.

First try 2, yes, it is a prime factor. The reminder is 875.
Next, go for 3. No it is not a prime factor.
Next, try 5. Yes, it is, you get reminder 175.
Ter 5 again. Yes, it is still a prime factor. The remainder is 35.
Try 5 again, you get 7, and 7 is a prime number, you stop.
So, 1750 = 2 * 5 * 5 * 5 * 7.

If you are a programmer, then you may choose a different approach to write a program.

Answer 2

Here's how I'd do it. Let's use an example like 78192.

I do three things first. (a) Is it even? If so, it's divisible by 2. (b) Does it end in 5 or 0? If so, 5 or 10 goes into it. And (c) are the "sum of digits" divisible by 3? If so, the number itself is divisible by 3.

Those three things take care of the first three primes -- 2, 3, and 5 ... and those are the three that divide most frequently.

After taking care of 2, 3, and 5, you can start going up the list -- 7, 11, 13, and so on, up as far as the square root of the number you're working on. (Because if nothing divides up to the square root, you have a prime number.)

Let's apply these steps to 78192. It's even. The sum of digits 7+8+1+9+2 = 27, which is divisible by 3. I also note that 5 does not go into the number. (But 6 does.)

Divide out the 6 to get 13032. Repeat the process. 6 goes into 13032 also. Divide it out to get 2172. 6 goes into that as well. Divide it out to get 362. This is even, but 3 will not go into it. Divide out the 2 to get 181.

At this point, you have 78192 = 6 x 6 x 6 x 2 x 181. Only now are you ready to try the higher primes. (You no longer need be concerned about 2, 3, and 5.) Since sqrt(181) is between 13 and 14 (13^2 = 169 and 14^2 = 196), you only need to check 7, 11, and 13. None of these divide, so 181 is prime, and 78192 = 2^4 x 3^3 x 181.

Here's a last example -- 27300. The zeroes mean 10^2 divides it, and the sum of digits is 12 (divisible by 3). So 27300 = 10^2 x 3 x 91.

Neither 2, 3, or 5 divide 77, so you look at primes up to sqrt(91). Well, 10^2 is too high, so you only have to look at 7. As it turns out, 91 = 7 x 13, and the prime factorization of 27300 is 2^2 x 3 x 5^2 x 7 x 13.

Answer 3

An algorithm would look something like this.

do
└ if( N % Pn == 0)
─┼ Pn is a prime factor
─┼ print Pn
─┴ N = N / Pn
└ else
─┴ Next Pn
while N != 1

The % sign is the modulus sign in most programming languages. Modulus finds the remainder of one number divided by another.
eg.
5 % 5 = 0
5 % 4 = 1
5 % 3 = 2
5 % 2 = 1
5 % 1 = 0

This algorithm loops through a list of primes. It is simple enough that you can do it in your head or write it down on paper.

Let's use the number 28 as an example

28 % 2 = 0
─ print 2
─ N → 14 (N = 28 / 2)

N != 1 so start over

14 % 2 = 0
─ print 2
─ N → 7 (N = 14 / 2)

N != 1 so start over

7 % 2 = 1
─ Next Pn

N != 1 so start over

7 % 3 = 1
─ Next Pn

N != 1 so start over

7 % 5 = 2
─ Next Pn

N != 1 so start over

7 % 7
─ Print Pn
─ N → 1 (N = 7 / 7)

N = 1
─ Stop

Answer 4

you have to make a prime factorization tree

like this maybe for 40

40
^
10 4
^ ^
2 5 2 2

the prime factorization of 40 is 2x5x2x2

if it was 50 you would have to find two numbers to multiply that equal 50 if any of those numbers are not prime you break it down to two more numbers that equal that then at the end you keep the prime numbers like this


50
^
10 5 ( Keep the five because it's prime)
^
2 5 Keep both because they are both prime.

The prime factorization is 5x5x2

Answer 5

This would best be shown by example

Say 60

60 = 15 * 4

4 = 2 * 2
15 = 3 * 5

60 = 2 * 2 * 3 * 5

Basically, find any 2 smaller factors and then factor each of those factors into smaller factors until all the factors are prime numbers.

Answer 6

just keep dividing the whole number by a prime number

The most common prime divisors are 2, 3, 5, and 7

64 = 2 * 2 * 2 * 2 * 2 * 2 or 2^6
20 = 2 * 2 * 5 or 2^2 * 5
34 = 2 * 17
28 = 2 * 2 * 7 or 2^2 * 7
etc...

Answer 7

You don't.
You use the diagram and you walk through it several times with your students. Then you let them try it on their own. Then you go over it. They practice some more while you monitor their progress. Assign homework.