Prime factorization of $$$992$$$

Find the prime factorization of $$$992$$$.

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

Determine whether $$$992$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$992$$$ by $$${\color{green}2}$$$: $$$\frac{992}{2} = {\color{red}496}$$$.

Determine whether $$$496$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$496$$$ by $$${\color{green}2}$$$: $$$\frac{496}{2} = {\color{red}248}$$$.

Determine whether $$$248$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$248$$$ by $$${\color{green}2}$$$: $$$\frac{248}{2} = {\color{red}124}$$$.

Determine whether $$$124$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$124$$$ by $$${\color{green}2}$$$: $$$\frac{124}{2} = {\color{red}62}$$$.

Determine whether $$$62$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$62$$$ by $$${\color{green}2}$$$: $$$\frac{62}{2} = {\color{red}31}$$$.

The prime number $$${\color{green}31}$$$ has no other factors then $$$1$$$ and $$${\color{green}31}$$$: $$$\frac{31}{31} = {\color{red}1}$$$.

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: $$$992 = 2^{5} \cdot 31$$$.

The prime factorization is $$$992 = 2^{5} \cdot 31$$$A.