# Prime factorization of $1833$

The calculator will find the prime factorization of $1833$, with steps shown.

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Find the prime factorization of $1833$.

### Solution

Start with the number $2$.

Determine whether $1833$ is divisible by $2$.

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

The next prime number is $3$.

Determine whether $1833$ is divisible by $3$.

It is divisible, thus, divide $1833$ by ${\color{green}3}$: $\frac{1833}{3} = {\color{red}611}$.

Determine whether $611$ is divisible by $3$.

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

The next prime number is $5$.

Determine whether $611$ is divisible by $5$.

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

The next prime number is $7$.

Determine whether $611$ is divisible by $7$.

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

The next prime number is $11$.

Determine whether $611$ is divisible by $11$.

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

The next prime number is $13$.

Determine whether $611$ is divisible by $13$.

It is divisible, thus, divide $611$ by ${\color{green}13}$: $\frac{611}{13} = {\color{red}47}$.

The prime number ${\color{green}47}$ has no other factors then $1$ and ${\color{green}47}$: $\frac{47}{47} = {\color{red}1}$.

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: $1833 = 3 \cdot 13 \cdot 47$.

The prime factorization is $1833 = 3 \cdot 13 \cdot 47$A.