Prime factorization of $$$260$$$

Find the prime factorization of $$$260$$$.

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

Determine whether $$$260$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$260$$$ by $$${\color{green}2}$$$: $$$\frac{260}{2} = {\color{red}130}$$$.

Determine whether $$$130$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$130$$$ by $$${\color{green}2}$$$: $$$\frac{130}{2} = {\color{red}65}$$$.

Determine whether $$$65$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$65$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$65$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$65$$$ by $$${\color{green}5}$$$: $$$\frac{65}{5} = {\color{red}13}$$$.

The prime number $$${\color{green}13}$$$ has no other factors then $$$1$$$ and $$${\color{green}13}$$$: $$$\frac{13}{13} = {\color{red}1}$$$.

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: $$$260 = 2^{2} \cdot 5 \cdot 13$$$.

The prime factorization is $$$260 = 2^{2} \cdot 5 \cdot 13$$$A.