Prime factorization of $$$3492$$$

Find the prime factorization of $$$3492$$$.

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

Determine whether $$$3492$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$3492$$$ by $$${\color{green}2}$$$: $$$\frac{3492}{2} = {\color{red}1746}$$$.

Determine whether $$$1746$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1746$$$ by $$${\color{green}2}$$$: $$$\frac{1746}{2} = {\color{red}873}$$$.

Determine whether $$$873$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$873$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$873$$$ by $$${\color{green}3}$$$: $$$\frac{873}{3} = {\color{red}291}$$$.

Determine whether $$$291$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$291$$$ by $$${\color{green}3}$$$: $$$\frac{291}{3} = {\color{red}97}$$$.

The prime number $$${\color{green}97}$$$ has no other factors then $$$1$$$ and $$${\color{green}97}$$$: $$$\frac{97}{97} = {\color{red}1}$$$.

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: $$$3492 = 2^{2} \cdot 3^{2} \cdot 97$$$.

The prime factorization is $$$3492 = 2^{2} \cdot 3^{2} \cdot 97$$$A.