Prime factorization of $$$445$$$

Find the prime factorization of $$$445$$$.

Start with the number $$$2$$$.

Determine whether $$$445$$$ is divisible by $$$2$$$.

Since it is not divisible, move to the next prime number.

The next prime number is $$$3$$$.

Determine whether $$$445$$$ is divisible by $$$3$$$.

Since it is not divisible, move to the next prime number.

The next prime number is $$$5$$$.

Determine whether $$$445$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$445$$$ by $$${\color{green}5}$$$: $$$\frac{445}{5} = {\color{red}89}$$$.

The prime number $$${\color{green}89}$$$ has no other factors then $$$1$$$ and $$${\color{green}89}$$$: $$$\frac{89}{89} = {\color{red}1}$$$.

Since we have obtained $$$1$$$, we are done.

Now, just count the number of occurences of the divisors (green numbers), and write down the prime factorization: $$$445 = 5 \cdot 89$$$.

The prime factorization is $$$445 = 5 \cdot 89$$$A.