Euler problem 1

Problem:

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

We just need to calculate the sum of all divisible by 3 then add the sum of all divisible by 5 and finally remove the sum of all divisible by 15 (both divisible by 3 and 5).

Code:

double gaussSum ( double number )
{
    return ( number * ( number + 1 ) ) / 2 ;
}
//Proyect euler 1
int SumMulti3and5 ( int numero )
{  
    int total = ( 3 * gaussSum ( numero / 3 ) ) + ( 5 * gaussSum ( numero / 5 ) ) ;
    int menos = 15 * gaussSum ( numero / 15 ) ;

    return total - menos ;
}