Prime factorization of $$$1917$$$

Find the prime factorization of $$$1917$$$.

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

Determine whether $$$1917$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1917$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$1917$$$ by $$${\color{green}3}$$$: $$$\frac{1917}{3} = {\color{red}639}$$$.

Determine whether $$$639$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$639$$$ by $$${\color{green}3}$$$: $$$\frac{639}{3} = {\color{red}213}$$$.

Determine whether $$$213$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$213$$$ by $$${\color{green}3}$$$: $$$\frac{213}{3} = {\color{red}71}$$$.

The prime number $$${\color{green}71}$$$ has no other factors then $$$1$$$ and $$${\color{green}71}$$$: $$$\frac{71}{71} = {\color{red}1}$$$.

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: $$$1917 = 3^{3} \cdot 71$$$.

The prime factorization is $$$1917 = 3^{3} \cdot 71$$$A.