problem1/problem1.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() {
	unsigned int sum;
	for (int i = 1; i < 1000; i++) {
		if (i % 3 == 0 || i % 5 == 0) {
			sum += i;
		}
	}
	printf("%u\n", sum);
}