Prime factorization of $$$3938$$$

Find the prime factorization of $$$3938$$$.

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

Determine whether $$$3938$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$3938$$$ by $$${\color{green}2}$$$: $$$\frac{3938}{2} = {\color{red}1969}$$$.

Determine whether $$$1969$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1969$$$ is divisible by $$$3$$$.

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

The next prime number is $$$5$$$.

Determine whether $$$1969$$$ is divisible by $$$5$$$.

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

The next prime number is $$$7$$$.

Determine whether $$$1969$$$ is divisible by $$$7$$$.

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

The next prime number is $$$11$$$.

Determine whether $$$1969$$$ is divisible by $$$11$$$.

It is divisible, thus, divide $$$1969$$$ by $$${\color{green}11}$$$: $$$\frac{1969}{11} = {\color{red}179}$$$.

The prime number $$${\color{green}179}$$$ has no other factors then $$$1$$$ and $$${\color{green}179}$$$: $$$\frac{179}{179} = {\color{red}1}$$$.

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: $$$3938 = 2 \cdot 11 \cdot 179$$$.

The prime factorization is $$$3938 = 2 \cdot 11 \cdot 179$$$A.