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.

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

The problem is very easy. We are checking division of the numbers 3 and/or 5, the end number is not high and an idea of the sum is very clear. Despite the fact we need to brute-forcefully iterate all over the first 1000 numbers, I did not think much about any other complicated algorithm. The brute-force can be enough in this case.

PHP:

<?php 
$sum = 0;
for ($i=0;$i<1000;$i++) {
	if(($i%3 == 0) OR ($i%5 == 0)){
		$sum += $i;
	}
}
echo $sum;
?>

JAVA:

public class Project_Euler_Problem_001 {

	public static void main(String[] args) {
		
		System.out.println("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.");
		System.out.println("Find the sum of all the multiples of 3 or 5 below 1000.");

		int sum = 0;
		for (int i = 0; i<1000; i++) {
			if((i%3 == 0) || (i%5 == 0)){
				sum += i;
			}
		}
		System.out.println("\n"+ "Sum is: " + sum);;
	}
}