Prime factorization of $$$3810$$$

Find the prime factorization of $$$3810$$$.

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

Determine whether $$$3810$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$3810$$$ by $$${\color{green}2}$$$: $$$\frac{3810}{2} = {\color{red}1905}$$$.

Determine whether $$$1905$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1905$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$1905$$$ by $$${\color{green}3}$$$: $$$\frac{1905}{3} = {\color{red}635}$$$.

Determine whether $$$635$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$635$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$635$$$ by $$${\color{green}5}$$$: $$$\frac{635}{5} = {\color{red}127}$$$.

The prime number $$${\color{green}127}$$$ has no other factors then $$$1$$$ and $$${\color{green}127}$$$: $$$\frac{127}{127} = {\color{red}1}$$$.

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: $$$3810 = 2 \cdot 3 \cdot 5 \cdot 127$$$.

The prime factorization is $$$3810 = 2 \cdot 3 \cdot 5 \cdot 127$$$A.