Prime factorization of $$$3295$$$

Find the prime factorization of $$$3295$$$.

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

Determine whether $$$3295$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$3295$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$3295$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$3295$$$ by $$${\color{green}5}$$$: $$$\frac{3295}{5} = {\color{red}659}$$$.

The prime number $$${\color{green}659}$$$ has no other factors then $$$1$$$ and $$${\color{green}659}$$$: $$$\frac{659}{659} = {\color{red}1}$$$.

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: $$$3295 = 5 \cdot 659$$$.

The prime factorization is $$$3295 = 5 \cdot 659$$$A.