Prime factorization of $$$3932$$$

Find the prime factorization of $$$3932$$$.

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

Determine whether $$$3932$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$3932$$$ by $$${\color{green}2}$$$: $$$\frac{3932}{2} = {\color{red}1966}$$$.

Determine whether $$$1966$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1966$$$ by $$${\color{green}2}$$$: $$$\frac{1966}{2} = {\color{red}983}$$$.

The prime number $$${\color{green}983}$$$ has no other factors then $$$1$$$ and $$${\color{green}983}$$$: $$$\frac{983}{983} = {\color{red}1}$$$.

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: $$$3932 = 2^{2} \cdot 983$$$.

The prime factorization is $$$3932 = 2^{2} \cdot 983$$$A.