# Prime Factorization

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## What is Prime Factorization?

Prime factorization allows us to write any number as a product of prime factors.

For example, consider the number 360. Let us write this number as follows:

$\begin{array}{l}360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5\\\;\;\;\;\;\, = {2^3} \times {3^2} \times {5^1}\end{array}$

Basically, we have expressed 360 as a product of Prime Numbers. Let us take some other (composite) numbers and express them as products of prime numbers:

$\begin{array}{l}\;\,80 = 2 \times 2 \times 2 \times 2 \times 5\\\;\;\;\;\;\,\,\, = {2^4} \times {5^1}\\144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3\\\;\;\;\;\;\,\,\, = {2^4} \times {3^2}\\600 = 2 \times 2 \times 2 \times 3 \times 3 \times 5\\\;\;\;\;\;\,\,\,= {2^3} \times {3^2} \times {5^1}\end{array}$

Expressing a number this way (as a product of primes) is called the prime factorization of that number. Note that the prime factorization of a prime number is trivial, since the only non-unity divisor of any prime number is that number itself.

✍Note: if a number is not prime, it has a Prime Numbers as its factor.

## Can every number be prime-factorized in a unique way?

Two questions now arise:

1. Can every composite number be expressed as a product of primes, that is, can every composite number be prime factorized?

2. If we prime factorize a composite number, is that factorization unique?

The answer is YES for both the above questions. The uniqueness of prime factorization is an incredibly important result, thus earning the name of

## Solved Examples:

Example 1: What are the prime factors of 12?

Solution: The factors of 12 are 1, 2, 3, 4, 6, and 12. The prime factors are 2 and 3.

Example 2: If $$N = a \times b \times c$$ such that $$a$$, $$b$$ and $$c$$ are three different prime numbers, how many positive divisors does $$N$$ have excluding 1 and itself.

Solution: Here, $$N = a \times b \times c$$, we can conclude that $$a$$, $$b$$ and $$c$$ are the factors of $$N$$. Since $$a$$, $$b$$ and $$c$$ are also prime numbers, we can't factor them to get any other number, so that gives us a total of 3 numbers.

But we know that if $$a$$ and $$b$$ are factors of $$N$$, then $$a \times b$$ is also a factor of $$N$$. So a combination of two factors out of these three factors is also a divisor of $$N$$. In other words, we have $$a \times b$$, $$b \times c$$, $$c \times a$$ as factors of $$N$$, which are another 3 in addition to the 3 above.

Note that $$a \times b \times c$$ is also a combination that is a factor of $$N$$, but it equals the number itself and is therefore omitted.

So we have a total of 6 divisors, excluding 1 and the number itself.

Challenge: Write all the factors of 120 and separate prime factors from them.

⚡Tip: Similar to example 1.

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