# Prime factorization of $4140$

The calculator will find the prime factorization of $4140$, with steps shown.

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Find the prime factorization of $4140$.

### Solution

Start with the number $2$.

Determine whether $4140$ is divisible by $2$.

It is divisible, thus, divide $4140$ by ${\color{green}2}$: $\frac{4140}{2} = {\color{red}2070}$.

Determine whether $2070$ is divisible by $2$.

It is divisible, thus, divide $2070$ by ${\color{green}2}$: $\frac{2070}{2} = {\color{red}1035}$.

Determine whether $1035$ is divisible by $2$.

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

The next prime number is $3$.

Determine whether $1035$ is divisible by $3$.

It is divisible, thus, divide $1035$ by ${\color{green}3}$: $\frac{1035}{3} = {\color{red}345}$.

Determine whether $345$ is divisible by $3$.

It is divisible, thus, divide $345$ by ${\color{green}3}$: $\frac{345}{3} = {\color{red}115}$.

Determine whether $115$ is divisible by $3$.

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

The next prime number is $5$.

Determine whether $115$ is divisible by $5$.

It is divisible, thus, divide $115$ by ${\color{green}5}$: $\frac{115}{5} = {\color{red}23}$.

The prime number ${\color{green}23}$ has no other factors then $1$ and ${\color{green}23}$: $\frac{23}{23} = {\color{red}1}$.

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: $4140 = 2^{2} \cdot 3^{2} \cdot 5 \cdot 23$.

The prime factorization is $4140 = 2^{2} \cdot 3^{2} \cdot 5 \cdot 23$A.