Find the sum of all the multiples of 3 or 5 below 1000

Question

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. I have the following code but the answer do

Answer 1

I attempt the Project Euler #1: Multiples of 3 and 5. And got 60 points for that.

private static long euler1(long N) {
        long i, sum = 0;
        for (i = 3; i < N; i++) {
            if (i % 3 == 0 || i % 5 == 0) {
            sum += i;
            }
        }
        return sum;
    }

I use Java for it and you can use this logic for C. I worked hard and find an optimal solution.

private static void euler1(long n) {
        long a = 0, b = 0, d = 0;

        a = (n - 1) / 3;
        b = (n - 1) / 5;
        d = (n - 1) / 15;

        long sum3 = 3 * a * (a + 1) / 2;
        long sum5 = 5 * b * (b + 1) / 2;
        long sum15 = 15 * d * (d + 1) / 2;
        long c = sum3 + sum5 - sum15;
        System.out.println(c);
    }