Project Euler Problem 1

 Add all the natural numbers below one thousand that are multiples of 3 or 5.

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.

Solution#

This problem calls for the sum of all natural numbers that are multiples of 3 or 5 below 1000. The smallest number which is a multiple of 3 or 5 is 3. So my starting number is 3. The terminating condition is also given in the problem as below 1000. Consider the following things while programming

1. Check whether the number is a multiple of 3.

2. Check if the number is multiple of 5 and not a multiple of 3.(This step is to avoid adding duplicate numbers).

Finding the sum of all multiples of 3 or 5 below 20.

Initially sum = 0

3 is the smallest number that is a multiple of 3 or 5.

Implementation

int main()
{
    int sum, number;
    sum = 0;
    for (number = 3; number < 1000; number++)
    {
        if ((number % 3) == 0)
            sum = sum + number;
        else if ((number % 5) == 0)
            sum = sum + number;
    }
    printf("sum :%d", sum);
    return 0;

}