01_multiples_of_3_or_5.c

//https://projecteuler.net/problem=1
/*

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(void){
	
	int sum = 0;
	int i = 0;
	
	for(i;i < 1000; i++){
		
		if(i % 3 == 0 || i % 5 == 0 ){sum = sum + i; printf("i : %d\n", i);}
		
		
		}
	printf("Sum of all multiples of 3 or 5 below 1000 : %d", sum);
	
	
	return 0;
}