Prime factorization of $$$1940$$$

Find the prime factorization of $$$1940$$$.

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

Determine whether $$$1940$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$1940$$$ by $$${\color{green}2}$$$: $$$\frac{1940}{2} = {\color{red}970}$$$.

Determine whether $$$970$$$ is divisible by $$$2$$$.

It is divisible, thus, divide $$$970$$$ by $$${\color{green}2}$$$: $$$\frac{970}{2} = {\color{red}485}$$$.

Determine whether $$$485$$$ is divisible by $$$2$$$.

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

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

Determine whether $$$485$$$ is divisible by $$$3$$$.

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

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

Determine whether $$$485$$$ is divisible by $$$5$$$.

It is divisible, thus, divide $$$485$$$ by $$${\color{green}5}$$$: $$$\frac{485}{5} = {\color{red}97}$$$.

The prime number $$${\color{green}97}$$$ has no other factors then $$$1$$$ and $$${\color{green}97}$$$: $$$\frac{97}{97} = {\color{red}1}$$$.

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: $$$1940 = 2^{2} \cdot 5 \cdot 97$$$.

The prime factorization is $$$1940 = 2^{2} \cdot 5 \cdot 97$$$A.