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package ProjectEuler;
 * 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.
 * <p>Find the sum of all the multiples of 3 or 5 below 1000.
 * <p>Link: https://projecteuler.net/problem=1
public class Problem01 {
  public static void main(String[] args) {
    int[][] testNumber = {
      {3, 0},
      {4, 3},
      {10, 23},
      {-1, 0}
    for (int[] ints : testNumber) {
      assert solution1(ints[0]) == ints[1];
      assert solution2(ints[0]) == ints[1];
  private static int solution1(int n) {
    int sum = 0;
    for (int i = 3; i < n; ++i) {
      if (i % 3 == 0 || i % 5 == 0) {
        sum += i;
    return sum;
  private static int solution2(int n) {
    int sum = 0;
    int terms = (n - 1) / 3;
    sum += terms * (6 + (terms - 1) * 3) / 2;
    terms = (n - 1) / 5;
    sum += terms * (10 + (terms - 1) * 5) / 2;
    terms = (n - 1) / 15;
    sum -= terms * (30 + (terms - 1) * 15) / 2;
    return sum;

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