Prime factorization of $$$2212$$$

Find the prime factorization of $$$2212$$$.

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

Determine whether $$$2212$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2212$$$ by $$${\color{green}2}$$$: $$$\frac{2212}{2} = {\color{red}1106}$$$.

Determine whether $$$1106$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1106$$$ by $$${\color{green}2}$$$: $$$\frac{1106}{2} = {\color{red}553}$$$.

Determine whether $$$553$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$553$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$553$$$ is divisible by $$$5$$$.

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

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

Determine whether $$$553$$$ is divisible by $$$7$$$.

It is divisible, thus, divide $$$553$$$ by $$${\color{green}7}$$$: $$$\frac{553}{7} = {\color{red}79}$$$.

The prime number $$${\color{green}79}$$$ has no other factors then $$$1$$$ and $$${\color{green}79}$$$: $$$\frac{79}{79} = {\color{red}1}$$$.

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: $$$2212 = 2^{2} \cdot 7 \cdot 79$$$.

The prime factorization is $$$2212 = 2^{2} \cdot 7 \cdot 79$$$A.