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 do

Answer 1

Using the stepping approach, you can make a version:

#include 

int main ( void ) {

    int sum = 0;

    for (int i = 0; i < 1000; i += 5) {
        sum += i;
    }
    for (int i = 0; i < 1000; i += 3) {
        if (i % 5) sum += i;  /* already counted */
    }
    printf("%d\n", sum);
    return 0;
}

which does a whole lot fewer modulo computations.

In fact, with a counter, you can make a version with none:

#include 

int main ( void ) {

    int sum = 0;
    int cnt = 6;

    for (int i = 0; i < 1000; i += 5) {
        sum += i;
    }
    for (int i = 0; i < 1000; i += 3) {
        if (--cnt == 0) cnt = 5;
        else sum += i;
    }
    printf("%d\n", sum);
    return 0;
}