Prime factorization of $$$4936$$$

Find the prime factorization of $$$4936$$$.

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

Determine whether $$$4936$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$4936$$$ by $$${\color{green}2}$$$: $$$\frac{4936}{2} = {\color{red}2468}$$$.

Determine whether $$$2468$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2468$$$ by $$${\color{green}2}$$$: $$$\frac{2468}{2} = {\color{red}1234}$$$.

Determine whether $$$1234$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1234$$$ by $$${\color{green}2}$$$: $$$\frac{1234}{2} = {\color{red}617}$$$.

The prime number $$${\color{green}617}$$$ has no other factors then $$$1$$$ and $$${\color{green}617}$$$: $$$\frac{617}{617} = {\color{red}1}$$$.

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: $$$4936 = 2^{3} \cdot 617$$$.

The prime factorization is $$$4936 = 2^{3} \cdot 617$$$A.