Prime factorization of $$$4263$$$

Find the prime factorization of $$$4263$$$.

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

Determine whether $$$4263$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$4263$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$4263$$$ by $$${\color{green}3}$$$: $$$\frac{4263}{3} = {\color{red}1421}$$$.

Determine whether $$$1421$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$1421$$$ is divisible by $$$5$$$.

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

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

Determine whether $$$1421$$$ is divisible by $$$7$$$.

It is divisible, thus, divide $$$1421$$$ by $$${\color{green}7}$$$: $$$\frac{1421}{7} = {\color{red}203}$$$.

Determine whether $$$203$$$ is divisible by $$$7$$$.

It is divisible, thus, divide $$$203$$$ by $$${\color{green}7}$$$: $$$\frac{203}{7} = {\color{red}29}$$$.

The prime number $$${\color{green}29}$$$ has no other factors then $$$1$$$ and $$${\color{green}29}$$$: $$$\frac{29}{29} = {\color{red}1}$$$.

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: $$$4263 = 3 \cdot 7^{2} \cdot 29$$$.

The prime factorization is $$$4263 = 3 \cdot 7^{2} \cdot 29$$$A.