Prime factorization of $$$1833$$$

Find the prime factorization of $$$1833$$$.

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

Determine whether $$$1833$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1833$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$1833$$$ by $$${\color{green}3}$$$: $$$\frac{1833}{3} = {\color{red}611}$$$.

Determine whether $$$611$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$611$$$ is divisible by $$$5$$$.

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

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

Determine whether $$$611$$$ is divisible by $$$7$$$.

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

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

Determine whether $$$611$$$ is divisible by $$$11$$$.

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

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

Determine whether $$$611$$$ is divisible by $$$13$$$.

It is divisible, thus, divide $$$611$$$ by $$${\color{green}13}$$$: $$$\frac{611}{13} = {\color{red}47}$$$.

The prime number $$${\color{green}47}$$$ has no other factors then $$$1$$$ and $$${\color{green}47}$$$: $$$\frac{47}{47} = {\color{red}1}$$$.

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: $$$1833 = 3 \cdot 13 \cdot 47$$$.

The prime factorization is $$$1833 = 3 \cdot 13 \cdot 47$$$A.