## What is the fastest way to find if a number is prime?

Quoting wikipedia,

A

prime number(or aprime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.

So, at times you might find a need to check if a number is prime. And if the numbers to be checked for primality, you might need the fastest implementation. In this post, I am sharing the implementation as suggested by the algorithm here.

```
boolean isPrime(long number) {
if (number == 2) {
return true; // 2 is a prime
}
// if the number is even
if (number % 2 == 0 || number < 1) {
return false; // not a prime.
}
// a number would be prime if it is not divisible by a number less than or equal to its square root.
double root = Math.sqrt(number);
//we start the loop from 3 (2 is already checked above)
//Additionally, we will increment i by 2 as checking for even divisor is not required at this point.
for (double i = 3; i <= root; i = i + 2) {
if (number % i == 0) {
return false;
}
}
return true;
}
```