Prime factorization of $$$1968$$$

Find the prime factorization of $$$1968$$$.

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

Determine whether $$$1968$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1968$$$ by $$${\color{green}2}$$$: $$$\frac{1968}{2} = {\color{red}984}$$$.

Determine whether $$$984$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$984$$$ by $$${\color{green}2}$$$: $$$\frac{984}{2} = {\color{red}492}$$$.

Determine whether $$$492$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$492$$$ by $$${\color{green}2}$$$: $$$\frac{492}{2} = {\color{red}246}$$$.

Determine whether $$$246$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$246$$$ by $$${\color{green}2}$$$: $$$\frac{246}{2} = {\color{red}123}$$$.

Determine whether $$$123$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$123$$$ is divisible by $$$3$$$.

It is divisible, thus, divide $$$123$$$ by $$${\color{green}3}$$$: $$$\frac{123}{3} = {\color{red}41}$$$.

The prime number $$${\color{green}41}$$$ has no other factors then $$$1$$$ and $$${\color{green}41}$$$: $$$\frac{41}{41} = {\color{red}1}$$$.

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: $$$1968 = 2^{4} \cdot 3 \cdot 41$$$.

The prime factorization is $$$1968 = 2^{4} \cdot 3 \cdot 41$$$A.