i have the following code, but the answer is coming out to be 2.

public class Main {

    public static void main(String[] args) {
        int sum = 0;
        int count = 0;
        for(int i = 2; i <= 50; i  ) {
            for(int j = 1; j <= i; j  ) {
                System.out.println(j);
                if(i % j == 0) {

                    count  ;
                }
                if(count == 2) {
                    System.out.println(i);
                    sum = sum   i;
                }
            }

        }
        System.out.println("The sum: "   sum);
    }
}

CodePudding user response：

You were close. But here is all you need.

• first, set sum to 2. It is the only even prime and doing this allows a better increment below
• now increment the test cases starting at 3 and incrementing by 2. Skip even values.
• now divide by each test case by numbers from 2 to the sqrt(i). No need to go beyond that.
• as soon as you have a successful division, end the inner loop and start the next of the outer loop.
• otherwise, if the inner loop completes, add i to sum.

int sum = 2;
outer:
for(int i = 3; i <= 50; i = 2) {
    for(int j = 2; j <= Math.sqrt(i); j  ) {
        if(i % j == 0) {
            continue outer;
        }
    }
    sum  = i;
}
System.out.println("The sum: "   sum);

An improvement would be to store the primes as they are found and then only divide by those instead of arbitrary numbers.

CodePudding user response：

Disclaimer: this code was written by Github Copilot with minimal supervision.

class Scratch {

    // find the sum of all prime numbers between 2 and n
    public static int sumPrimes(int n) {
        int sum = 0;
        for (int i = 2; i < n; i  ) {
            if (isPrime(i)) {
                sum  = i;
            }
        }
        return sum;
    }

    // return true if n is prime
    public static boolean isPrime(int n) {
        if (n == 2) {
            return true;
        }
        if (n % 2 == 0) {
            return false;
        }
        for (int i = 3; i * i <= n; i  = 2) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }


    public static void main(String[] args) {

        // print sum of primes between 2 and 50
        System.out.println("The sum: "   sumPrimes(50));

    }
}

Prints:

The sum: 328

CodePudding user response：

Maybe a minimal version with recursive method based on Aivean response:

class Scratch {
    // find the sum of all prime numbers between 2 and n
    public static int sumPrimes(int n) {
        if (n >= 2) {
            return (isPrime(n) ? n : 0)   sumPrimes(n-1);
        }
        
        return 0;
    }

    // return true if n is prime
    public static boolean isPrime(int n) {
        for (int i = 2; i < n; i  ) {
            if (n % i == 0) {
                return false;
            }
        }
        
        return true;
    }


    public static void main(String[] args) {

        // print sum of primes between 2 and 50
        System.out.println("The sum: "   sumPrimes(50));

    }
}