Prime factorization of $$$3636$$$

Find the prime factorization of $$$3636$$$.

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

Determine whether $$$3636$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$3636$$$ by $$${\color{green}2}$$$: $$$\frac{3636}{2} = {\color{red}1818}$$$.

Determine whether $$$1818$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1818$$$ by $$${\color{green}2}$$$: $$$\frac{1818}{2} = {\color{red}909}$$$.

Determine whether $$$909$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$909$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$909$$$ by $$${\color{green}3}$$$: $$$\frac{909}{3} = {\color{red}303}$$$.

Determine whether $$$303$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$303$$$ by $$${\color{green}3}$$$: $$$\frac{303}{3} = {\color{red}101}$$$.

The prime number $$${\color{green}101}$$$ has no other factors then $$$1$$$ and $$${\color{green}101}$$$: $$$\frac{101}{101} = {\color{red}1}$$$.

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: $$$3636 = 2^{2} \cdot 3^{2} \cdot 101$$$.

The prime factorization is $$$3636 = 2^{2} \cdot 3^{2} \cdot 101$$$A.