Prime factorization of $$$1160$$$

Find the prime factorization of $$$1160$$$.

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

Determine whether $$$1160$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1160$$$ by $$${\color{green}2}$$$: $$$\frac{1160}{2} = {\color{red}580}$$$.

Determine whether $$$580$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$580$$$ by $$${\color{green}2}$$$: $$$\frac{580}{2} = {\color{red}290}$$$.

Determine whether $$$290$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$290$$$ by $$${\color{green}2}$$$: $$$\frac{290}{2} = {\color{red}145}$$$.

Determine whether $$$145$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$145$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$145$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$145$$$ by $$${\color{green}5}$$$: $$$\frac{145}{5} = {\color{red}29}$$$.

The prime number $$${\color{green}29}$$$ has no other factors then $$$1$$$ and $$${\color{green}29}$$$: $$$\frac{29}{29} = {\color{red}1}$$$.

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: $$$1160 = 2^{3} \cdot 5 \cdot 29$$$.

The prime factorization is $$$1160 = 2^{3} \cdot 5 \cdot 29$$$A.