Prime factorization of $$$4392$$$

Find the prime factorization of $$$4392$$$.

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

Determine whether $$$4392$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$4392$$$ by $$${\color{green}2}$$$: $$$\frac{4392}{2} = {\color{red}2196}$$$.

Determine whether $$$2196$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2196$$$ by $$${\color{green}2}$$$: $$$\frac{2196}{2} = {\color{red}1098}$$$.

Determine whether $$$1098$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1098$$$ by $$${\color{green}2}$$$: $$$\frac{1098}{2} = {\color{red}549}$$$.

Determine whether $$$549$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$549$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$549$$$ by $$${\color{green}3}$$$: $$$\frac{549}{3} = {\color{red}183}$$$.

Determine whether $$$183$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$183$$$ by $$${\color{green}3}$$$: $$$\frac{183}{3} = {\color{red}61}$$$.

The prime number $$${\color{green}61}$$$ has no other factors then $$$1$$$ and $$${\color{green}61}$$$: $$$\frac{61}{61} = {\color{red}1}$$$.

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: $$$4392 = 2^{3} \cdot 3^{2} \cdot 61$$$.

The prime factorization is $$$4392 = 2^{3} \cdot 3^{2} \cdot 61$$$A.