# Prime factorization of $4120$

The calculator will find the prime factorization of $4120$, with steps shown.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

Find the prime factorization of $4120$.

### Solution

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.