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

Question

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. I have the following code but the answer does not match.

#include<stdio.h>
int main()
{
    long unsigned int i,sum=0;
    clrscr();
    for(i=0;i<=1000;i++)
    {
        if((i%5==0)||(i%3==0))
        {
            sum=sum+1;
        }
    }
    printf("%d\n",sum);
    getchar();
    return 0;
}

Answer 1

Perhaps you should do

sum += i // or sum = sum + i

instead of

sum = sum + 1

Additionally, be careful when printing long unsigned ints with printf. I guess the right specifier is %lu.

Answer 2

It should be sum = sum + i instead of 1.

Answer 3

Two things:

• you're including 1000 in the loop, and
• you're adding one to the sum each time, rather than the value itself.

Change the loop to

for(i=0;i<1000;i++)

And the sum line to

sum=sum+i;