Prime factorization of $$$2820$$$

Find the prime factorization of $$$2820$$$.

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

Determine whether $$$2820$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2820$$$ by $$${\color{green}2}$$$: $$$\frac{2820}{2} = {\color{red}1410}$$$.

Determine whether $$$1410$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1410$$$ by $$${\color{green}2}$$$: $$$\frac{1410}{2} = {\color{red}705}$$$.

Determine whether $$$705$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$705$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$705$$$ by $$${\color{green}3}$$$: $$$\frac{705}{3} = {\color{red}235}$$$.

Determine whether $$$235$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$235$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$235$$$ by $$${\color{green}5}$$$: $$$\frac{235}{5} = {\color{red}47}$$$.

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: $$$2820 = 2^{2} \cdot 3 \cdot 5 \cdot 47$$$.

The prime factorization is $$$2820 = 2^{2} \cdot 3 \cdot 5 \cdot 47$$$A.