Prime factorization of $$$1204$$$

Find the prime factorization of $$$1204$$$.

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

Determine whether $$$1204$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1204$$$ by $$${\color{green}2}$$$: $$$\frac{1204}{2} = {\color{red}602}$$$.

Determine whether $$$602$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$602$$$ by $$${\color{green}2}$$$: $$$\frac{602}{2} = {\color{red}301}$$$.

Determine whether $$$301$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$301$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$301$$$ is divisible by $$$5$$$.

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

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

Determine whether $$$301$$$ is divisible by $$$7$$$.

It is divisible, thus, divide $$$301$$$ by $$${\color{green}7}$$$: $$$\frac{301}{7} = {\color{red}43}$$$.

The prime number $$${\color{green}43}$$$ has no other factors then $$$1$$$ and $$${\color{green}43}$$$: $$$\frac{43}{43} = {\color{red}1}$$$.

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: $$$1204 = 2^{2} \cdot 7 \cdot 43$$$.

The prime factorization is $$$1204 = 2^{2} \cdot 7 \cdot 43$$$A.