Prime factorization of $$$1708$$$

Find the prime factorization of $$$1708$$$.

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

Determine whether $$$1708$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1708$$$ by $$${\color{green}2}$$$: $$$\frac{1708}{2} = {\color{red}854}$$$.

Determine whether $$$854$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$854$$$ by $$${\color{green}2}$$$: $$$\frac{854}{2} = {\color{red}427}$$$.

Determine whether $$$427$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$427$$$ is divisible by $$$3$$$.

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

The next prime number is $$$5$$$.

Determine whether $$$427$$$ is divisible by $$$5$$$.

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

The next prime number is $$$7$$$.

Determine whether $$$427$$$ is divisible by $$$7$$$.

It is divisible, thus, divide $$$427$$$ by $$${\color{green}7}$$$: $$$\frac{427}{7} = {\color{red}61}$$$.

The prime number $$${\color{green}61}$$$ has no other factors then $$$1$$$ and $$${\color{green}61}$$$: $$$\frac{61}{61} = {\color{red}1}$$$.

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: $$$1708 = 2^{2} \cdot 7 \cdot 61$$$.

The prime factorization is $$$1708 = 2^{2} \cdot 7 \cdot 61$$$A.