Prime factorization of $$$3105$$$

Find the prime factorization of $$$3105$$$.

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

Determine whether $$$3105$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$3105$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$3105$$$ by $$${\color{green}3}$$$: $$$\frac{3105}{3} = {\color{red}1035}$$$.

Determine whether $$$1035$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$1035$$$ by $$${\color{green}3}$$$: $$$\frac{1035}{3} = {\color{red}345}$$$.

Determine whether $$$345$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$345$$$ by $$${\color{green}3}$$$: $$$\frac{345}{3} = {\color{red}115}$$$.

Determine whether $$$115$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$115$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$115$$$ by $$${\color{green}5}$$$: $$$\frac{115}{5} = {\color{red}23}$$$.

The prime number $$${\color{green}23}$$$ has no other factors then $$$1$$$ and $$${\color{green}23}$$$: $$$\frac{23}{23} = {\color{red}1}$$$.

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: $$$3105 = 3^{3} \cdot 5 \cdot 23$$$.

The prime factorization is $$$3105 = 3^{3} \cdot 5 \cdot 23$$$A.