Prime factorization of $$$4120$$$

Find the prime factorization of $$$4120$$$.

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

Determine whether $$$4120$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$4120$$$ by $$${\color{green}2}$$$: $$$\frac{4120}{2} = {\color{red}2060}$$$.

Determine whether $$$2060$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$2060$$$ by $$${\color{green}2}$$$: $$$\frac{2060}{2} = {\color{red}1030}$$$.

Determine whether $$$1030$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1030$$$ by $$${\color{green}2}$$$: $$$\frac{1030}{2} = {\color{red}515}$$$.

Determine whether $$$515$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$515$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$515$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$515$$$ by $$${\color{green}5}$$$: $$$\frac{515}{5} = {\color{red}103}$$$.

The prime number $$${\color{green}103}$$$ has no other factors then $$$1$$$ and $$${\color{green}103}$$$: $$$\frac{103}{103} = {\color{red}1}$$$.

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: $$$4120 = 2^{3} \cdot 5 \cdot 103$$$.

The prime factorization is $$$4120 = 2^{3} \cdot 5 \cdot 103$$$A.