Prime factorization of $$$3580$$$

Find the prime factorization of $$$3580$$$.

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

Determine whether $$$3580$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$3580$$$ by $$${\color{green}2}$$$: $$$\frac{3580}{2} = {\color{red}1790}$$$.

Determine whether $$$1790$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1790$$$ by $$${\color{green}2}$$$: $$$\frac{1790}{2} = {\color{red}895}$$$.

Determine whether $$$895$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$895$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$895$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$895$$$ by $$${\color{green}5}$$$: $$$\frac{895}{5} = {\color{red}179}$$$.

The prime number $$${\color{green}179}$$$ has no other factors then $$$1$$$ and $$${\color{green}179}$$$: $$$\frac{179}{179} = {\color{red}1}$$$.

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: $$$3580 = 2^{2} \cdot 5 \cdot 179$$$.

The prime factorization is $$$3580 = 2^{2} \cdot 5 \cdot 179$$$A.