Prime factorization of $$$343$$$
Your Input
Find the prime factorization of $$$343$$$.
Solution
Start with the number $$$2$$$.
Determine whether $$$343$$$ is divisible by $$$2$$$.
Since it is not divisible, move to the next prime number.
The next prime number is $$$3$$$.
Determine whether $$$343$$$ is divisible by $$$3$$$.
Since it is not divisible, move to the next prime number.
The next prime number is $$$5$$$.
Determine whether $$$343$$$ is divisible by $$$5$$$.
Since it is not divisible, move to the next prime number.
The next prime number is $$$7$$$.
Determine whether $$$343$$$ is divisible by $$$7$$$.
It is divisible, thus, divide $$$343$$$ by $$${\color{green}7}$$$: $$$\frac{343}{7} = {\color{red}49}$$$.
Determine whether $$$49$$$ is divisible by $$$7$$$.
It is divisible, thus, divide $$$49$$$ by $$${\color{green}7}$$$: $$$\frac{49}{7} = {\color{red}7}$$$.
The prime number $$${\color{green}7}$$$ has no other factors then $$$1$$$ and $$${\color{green}7}$$$: $$$\frac{7}{7} = {\color{red}1}$$$.
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: $$$343 = 7^{3}$$$.
Answer
The prime factorization is $$$343 = 7^{3}$$$A.