p1/main.c


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/*
 * 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.
 */

#include <stdio.h>

int main(int argc, char **argv)
{

	int sum = 0;
	int i;

	for (i = 1; i < 1000; ++i)
	{
		if (i % 3 == 0 || i % 5 == 0)
		{
			sum += i;
		}
	}

	printf("sum: %d\n", sum);

	return 0;
}