Prime factorization of $$$1888$$$

Find the prime factorization of $$$1888$$$.

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

Determine whether $$$1888$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1888$$$ by $$${\color{green}2}$$$: $$$\frac{1888}{2} = {\color{red}944}$$$.

Determine whether $$$944$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$944$$$ by $$${\color{green}2}$$$: $$$\frac{944}{2} = {\color{red}472}$$$.

Determine whether $$$472$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$472$$$ by $$${\color{green}2}$$$: $$$\frac{472}{2} = {\color{red}236}$$$.

Determine whether $$$236$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$236$$$ by $$${\color{green}2}$$$: $$$\frac{236}{2} = {\color{red}118}$$$.

Determine whether $$$118$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$118$$$ by $$${\color{green}2}$$$: $$$\frac{118}{2} = {\color{red}59}$$$.

The prime number $$${\color{green}59}$$$ has no other factors then $$$1$$$ and $$${\color{green}59}$$$: $$$\frac{59}{59} = {\color{red}1}$$$.

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: $$$1888 = 2^{5} \cdot 59$$$.

The prime factorization is $$$1888 = 2^{5} \cdot 59$$$A.