Prime factorization of $$$1745$$$

Find the prime factorization of $$$1745$$$.

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

Determine whether $$$1745$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$1745$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$1745$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$1745$$$ by $$${\color{green}5}$$$: $$$\frac{1745}{5} = {\color{red}349}$$$.

The prime number $$${\color{green}349}$$$ has no other factors then $$$1$$$ and $$${\color{green}349}$$$: $$$\frac{349}{349} = {\color{red}1}$$$.

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: $$$1745 = 5 \cdot 349$$$.

The prime factorization is $$$1745 = 5 \cdot 349$$$A.