How would I speed up this program?

Question

I am currently attempting to solve a ProjectEuler problem and I have got everything down, except the speed. I am almost certain the reason the program executes so slowly is due to the nested loops. I would love some advice on how to speed this up. I am a novice programmer, so I am not familiar with a lot of the more advanced methods/topics.

public class Problem12 {

    public static void main(String[] args) {
        int num;

        for (int i = 1; i < 15000; i++) {
            num = i * (i + 1) / 2;
            int counter = 0;

            for (int x = 1; x <= num; x++) {
                if (num % x == 0) {
                    counter++;
                }
            }
            System.out.println("[" + i + "] - " + num + " is divisible by " + counter + " numbers.");
        }
    }
}

EDIT : Below is the new code that is exponentially faster. Removed the constant line printing as well to speed it up even more.

public class Problem12 {

    public static void main(String[] args) {
        int num;

        outerloop:
        for (int i = 1; i < 25000; i++) {
            num = i * (i + 1) / 2;
            int counter = 0;

            double root = Math.sqrt(num);
            for (int x = 1; x < root; x++) {
                if (num % x == 0) {
                    counter += 2;
                    if (counter >= 500) {
                        System.out.println("[" + i + "] - " + num + " is divisible by " + counter + " numbers.");
                        break outerloop;
                    }
                }
            }
        }
    }
}

Answer 1

For starters, when looking at divisors, you never need to go further than the root square of the number, because each divisor below the square root has an equivalent above.

n = a * b => a <= sqrt(n) or b <= sqrt(n)

Then you need to count the other side of the division:

double root = Math.sqrt(num);
for (int x = 1; x < root; x++) {
    if (num % x == 0) {
        counter += 2;
    }
}

The square root is special because it counts only once if it is integer:

if ((double) ((int) root) == root) {
    counter += 1;
}

Answer 2

You just need to factorize the number. p^a * q^b * r^c has (a+1)*(b+1)*(c+1) divisors. Here is some basic implementation using this idea:

static int Divisors(int num) {
    if (num == 1) {
        return 1;
    }

    int root = (int) Math.sqrt(num);
    for (int x = 2; x <= root; x++) {
        if (num % x == 0) {
            int c = 0;
            do {
                ++c;
                num /= x;
            } while (num % x == 0);
            return (c + 1) * Divisors(num);
        }
    }

    return 2;
}

public static void test500() {
    int i = 1, num = 1;
    while (Divisors(num) <= 500) {
        num += ++i;
    }

    System.out.println("\nFound: [" + i + "] - " + num);
}