Prime factorization of $$$4020$$$

Find the prime factorization of $$$4020$$$.

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

Determine whether $$$4020$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$4020$$$ by $$${\color{green}2}$$$: $$$\frac{4020}{2} = {\color{red}2010}$$$.

Determine whether $$$2010$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2010$$$ by $$${\color{green}2}$$$: $$$\frac{2010}{2} = {\color{red}1005}$$$.

Determine whether $$$1005$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1005$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$1005$$$ by $$${\color{green}3}$$$: $$$\frac{1005}{3} = {\color{red}335}$$$.

Determine whether $$$335$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$335$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$335$$$ by $$${\color{green}5}$$$: $$$\frac{335}{5} = {\color{red}67}$$$.

The prime number $$${\color{green}67}$$$ has no other factors then $$$1$$$ and $$${\color{green}67}$$$: $$$\frac{67}{67} = {\color{red}1}$$$.

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: $$$4020 = 2^{2} \cdot 3 \cdot 5 \cdot 67$$$.

The prime factorization is $$$4020 = 2^{2} \cdot 3 \cdot 5 \cdot 67$$$A.