Prime factorization of $$$2312$$$

Find the prime factorization of $$$2312$$$.

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

Determine whether $$$2312$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2312$$$ by $$${\color{green}2}$$$: $$$\frac{2312}{2} = {\color{red}1156}$$$.

Determine whether $$$1156$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1156$$$ by $$${\color{green}2}$$$: $$$\frac{1156}{2} = {\color{red}578}$$$.

Determine whether $$$578$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$578$$$ by $$${\color{green}2}$$$: $$$\frac{578}{2} = {\color{red}289}$$$.

Determine whether $$$289$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$289$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$289$$$ is divisible by $$$5$$$.

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

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

Determine whether $$$289$$$ is divisible by $$$7$$$.

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

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

Determine whether $$$289$$$ is divisible by $$$11$$$.

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

The next prime number is $$$13$$$.

Determine whether $$$289$$$ is divisible by $$$13$$$.

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

The next prime number is $$$17$$$.

Determine whether $$$289$$$ is divisible by $$$17$$$.

It is divisible, thus, divide $$$289$$$ by $$${\color{green}17}$$$: $$$\frac{289}{17} = {\color{red}17}$$$.

The prime number $$${\color{green}17}$$$ has no other factors then $$$1$$$ and $$${\color{green}17}$$$: $$$\frac{17}{17} = {\color{red}1}$$$.

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: $$$2312 = 2^{3} \cdot 17^{2}$$$.

The prime factorization is $$$2312 = 2^{3} \cdot 17^{2}$$$A.