Problem 1--(c)

Problem 1

05 October 2001

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.

Answer:	233168

code:
#include
int main(void)
{
int num,sum=0;
for(num=1;num<1000;num++)
{
if(num%3==0||num%5==0)
sum+=num;
}
printf("%d\n",sum);
return 0;
}