Prime factorization of $$$4796$$$

Find the prime factorization of $$$4796$$$.

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

Determine whether $$$4796$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$4796$$$ by $$${\color{green}2}$$$: $$$\frac{4796}{2} = {\color{red}2398}$$$.

Determine whether $$$2398$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2398$$$ by $$${\color{green}2}$$$: $$$\frac{2398}{2} = {\color{red}1199}$$$.

Determine whether $$$1199$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1199$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$1199$$$ is divisible by $$$5$$$.

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

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

Determine whether $$$1199$$$ is divisible by $$$7$$$.

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

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

Determine whether $$$1199$$$ is divisible by $$$11$$$.

It is divisible, thus, divide $$$1199$$$ by $$${\color{green}11}$$$: $$$\frac{1199}{11} = {\color{red}109}$$$.

The prime number $$${\color{green}109}$$$ has no other factors then $$$1$$$ and $$${\color{green}109}$$$: $$$\frac{109}{109} = {\color{red}1}$$$.

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: $$$4796 = 2^{2} \cdot 11 \cdot 109$$$.

The prime factorization is $$$4796 = 2^{2} \cdot 11 \cdot 109$$$A.