Project Euler #10 I made this method off the idea off of Project Euler's #10 of finding the sum of all primes up to two million, and made the method compatible with any limit you enter in the parameters. I started out doing the obvious thing and going through each number and finding if it was divisible by anything other than itself and 1, but ra…

SumOfPrimes.java

public class SumOfPrimes {
    
	public SumOfPrimes(){
		System.out.println(sum(2000000)); //running the sum of primes limit to 2,000,000 [For Project Euler's answer]
	}
	
	public long sum(int limit){ //Project Euler wanted it for 2,000,000 but this method can do it for any limit given
		boolean[] boolPrimes = new boolean[limit]; //initializing an array of booleans to the limit
		boolPrimes[0] = false; //I know 0 isn't a prime
		boolPrimes[1] = false;// I know 1 isn't a prime
		for(int i = 2; i < boolPrimes.length; ++i){ //setting every value after boolPrimes[0] & boolPrimes[1] equal to true
			boolPrimes[i] = true;
		}
		for(int i = 2; i <= Math.sqrt(limit); ++i){ //every non-prime number can be stated as i*i + ni
			int counter = 0; //takes place of n
			int j = i*i; // making j not a prime
			if(boolPrimes[i]){ //by doing this if-statement, it cancels out going through every single number from 2 -> limit
				while(j < limit){ //making sure not to go out of bounds in the boolPrime array
//					System.out.println("i = " + i);
//					System.out.println("Setting j to not prime = " + j);
					boolPrimes[j] = false; //setting non-prime to false in the array
					counter++;
					j = i + counter*i; //now setting j to the next non-prime number based off the prime number (i)
				}
			}
		}
		long sum = 0;
		for(int i = 0; i < limit; ++i){
			if(boolPrimes[i]){ //if this number is prime
//				System.out.println("i is prime = " + (i));
				sum += (i); //adds prime number to the sum
			}
		}
		return sum;
	}
	
	public static void main(String[] args){
		SumOfPrimes sum = new SumOfPrimes();
	}
}