Question

How to sum the proper divisors of a number in C++

Answer 1

#include <iostream>
 
unsigned SumNumberProperDivisors(const unsigned num) {
    unsigned sum = 0;
   
    for (unsigned i = 1; i <= num / 2; i++) {
        if (num % i == 0) {
            printf("%u, ", i);
            sum += i;
        }
    }
     
    return sum;
}
   
int main()
{
    int sum = SumNumberProperDivisors(220);
  
    std::cout << "\n" << sum;
}

   
   
/*
run:
   
1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 
284
 
*/

 

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