Prime factorization of $$$4112$$$

Find the prime factorization of $$$4112$$$.

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

Determine whether $$$4112$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$4112$$$ by $$${\color{green}2}$$$: $$$\frac{4112}{2} = {\color{red}2056}$$$.

Determine whether $$$2056$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2056$$$ by $$${\color{green}2}$$$: $$$\frac{2056}{2} = {\color{red}1028}$$$.

Determine whether $$$1028$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1028$$$ by $$${\color{green}2}$$$: $$$\frac{1028}{2} = {\color{red}514}$$$.

Determine whether $$$514$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$514$$$ by $$${\color{green}2}$$$: $$$\frac{514}{2} = {\color{red}257}$$$.

The prime number $$${\color{green}257}$$$ has no other factors then $$$1$$$ and $$${\color{green}257}$$$: $$$\frac{257}{257} = {\color{red}1}$$$.

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: $$$4112 = 2^{4} \cdot 257$$$.

The prime factorization is $$$4112 = 2^{4} \cdot 257$$$A.