Prime factorization of $$$1836$$$
Your Input
Find the prime factorization of $$$1836$$$.
Solution
Start with the number $$$2$$$.
Determine whether $$$1836$$$ is divisible by $$$2$$$.
It is divisible, thus, divide $$$1836$$$ by $$${\color{green}2}$$$: $$$\frac{1836}{2} = {\color{red}918}$$$.
Determine whether $$$918$$$ is divisible by $$$2$$$.
It is divisible, thus, divide $$$918$$$ by $$${\color{green}2}$$$: $$$\frac{918}{2} = {\color{red}459}$$$.
Determine whether $$$459$$$ is divisible by $$$2$$$.
Since it is not divisible, move to the next prime number.
The next prime number is $$$3$$$.
Determine whether $$$459$$$ is divisible by $$$3$$$.
It is divisible, thus, divide $$$459$$$ by $$${\color{green}3}$$$: $$$\frac{459}{3} = {\color{red}153}$$$.
Determine whether $$$153$$$ is divisible by $$$3$$$.
It is divisible, thus, divide $$$153$$$ by $$${\color{green}3}$$$: $$$\frac{153}{3} = {\color{red}51}$$$.
Determine whether $$$51$$$ is divisible by $$$3$$$.
It is divisible, thus, divide $$$51$$$ by $$${\color{green}3}$$$: $$$\frac{51}{3} = {\color{red}17}$$$.
The prime number $$${\color{green}17}$$$ has no other factors then $$$1$$$ and $$${\color{green}17}$$$: $$$\frac{17}{17} = {\color{red}1}$$$.
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: $$$1836 = 2^{2} \cdot 3^{3} \cdot 17$$$.
Answer
The prime factorization is $$$1836 = 2^{2} \cdot 3^{3} \cdot 17$$$A.