Prime factorization of $$$4708$$$

Find the prime factorization of $$$4708$$$.

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

Determine whether $$$4708$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$4708$$$ by $$${\color{green}2}$$$: $$$\frac{4708}{2} = {\color{red}2354}$$$.

Determine whether $$$2354$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2354$$$ by $$${\color{green}2}$$$: $$$\frac{2354}{2} = {\color{red}1177}$$$.

Determine whether $$$1177$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1177$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$1177$$$ is divisible by $$$5$$$.

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

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

Determine whether $$$1177$$$ is divisible by $$$7$$$.

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

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

Determine whether $$$1177$$$ is divisible by $$$11$$$.

It is divisible, thus, divide $$$1177$$$ by $$${\color{green}11}$$$: $$$\frac{1177}{11} = {\color{red}107}$$$.

The prime number $$${\color{green}107}$$$ has no other factors then $$$1$$$ and $$${\color{green}107}$$$: $$$\frac{107}{107} = {\color{red}1}$$$.

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: $$$4708 = 2^{2} \cdot 11 \cdot 107$$$.

The prime factorization is $$$4708 = 2^{2} \cdot 11 \cdot 107$$$A.