Prime factorization of $$$4068$$$

Find the prime factorization of $$$4068$$$.

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

Determine whether $$$4068$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$4068$$$ by $$${\color{green}2}$$$: $$$\frac{4068}{2} = {\color{red}2034}$$$.

Determine whether $$$2034$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2034$$$ by $$${\color{green}2}$$$: $$$\frac{2034}{2} = {\color{red}1017}$$$.

Determine whether $$$1017$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1017$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$1017$$$ by $$${\color{green}3}$$$: $$$\frac{1017}{3} = {\color{red}339}$$$.

Determine whether $$$339$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$339$$$ by $$${\color{green}3}$$$: $$$\frac{339}{3} = {\color{red}113}$$$.

The prime number $$${\color{green}113}$$$ has no other factors then $$$1$$$ and $$${\color{green}113}$$$: $$$\frac{113}{113} = {\color{red}1}$$$.

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: $$$4068 = 2^{2} \cdot 3^{2} \cdot 113$$$.

The prime factorization is $$$4068 = 2^{2} \cdot 3^{2} \cdot 113$$$A.