Prime factorization of $$$1896$$$
Your Input
Find the prime factorization of $$$1896$$$.
Solution
Start with the number $$$2$$$.
Determine whether $$$1896$$$ is divisible by $$$2$$$.
It is divisible, thus, divide $$$1896$$$ by $$${\color{green}2}$$$: $$$\frac{1896}{2} = {\color{red}948}$$$.
Determine whether $$$948$$$ is divisible by $$$2$$$.
It is divisible, thus, divide $$$948$$$ by $$${\color{green}2}$$$: $$$\frac{948}{2} = {\color{red}474}$$$.
Determine whether $$$474$$$ is divisible by $$$2$$$.
It is divisible, thus, divide $$$474$$$ by $$${\color{green}2}$$$: $$$\frac{474}{2} = {\color{red}237}$$$.
Determine whether $$$237$$$ is divisible by $$$2$$$.
Since it is not divisible, move to the next prime number.
The next prime number is $$$3$$$.
Determine whether $$$237$$$ is divisible by $$$3$$$.
It is divisible, thus, divide $$$237$$$ by $$${\color{green}3}$$$: $$$\frac{237}{3} = {\color{red}79}$$$.
The prime number $$${\color{green}79}$$$ has no other factors then $$$1$$$ and $$${\color{green}79}$$$: $$$\frac{79}{79} = {\color{red}1}$$$.
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: $$$1896 = 2^{3} \cdot 3 \cdot 79$$$.
Answer
The prime factorization is $$$1896 = 2^{3} \cdot 3 \cdot 79$$$A.