Assignment - Perfect Numbers with Multiple functions

[Arslan7041]

I have to write a program involving multiple function, and im having a lot of trouble even starting it.

The assignment is the following:

A perfect number is one which is equal to the sum of its divisors (excluding itself). For example, 6 is a perfect number (the divisors of 6 are 1, 2, 3, 6; sum = 6).

Write a function sumdiv that accepts an integer and returns the sum of the divsors of the integer that was pass (excluding the integer itself). Thus a call of sumdiv(6) would return 6; a call to sumdiv(15) would return 9 (= 1 + 3 + 5).

I know this will involve at least one other function besides main, but im not sure how to set this up. How do I figure out the divisors of any integer I enter? And then how do I test to see if they add back to the original integer?

Can anyone at least help me with the setup?

Thank you in advance for any help.

[keskiverto]

There is operator %, modulus, that returns remainder of division. You have had loops already, have you?

[Arslan7041]

Ok, so far I've written this:

#include <iostream>
#include <cmath>

using namespace std;


int sumdiv (i);

int main()
{
    cout << sumdiv(6);
   return 0;
}

int sumdiv (int i){
int sum = 0

for(i = 1; i<=1,000); i++)
    for (int divisor = 1; divisor<i; divisor++)
        if(i%divisor==0)
        sum+= divisor;
        return sum;
}

But Im getting all sorts of error messages that I dont know what to do with. Whats wrong with the above program?

[Arslan7041]

Ok, revised code.

But still not compiling

#include <iostream>
#include <cmath>

using namespace std;


int sumdiv (i);

int main()
{
cout << sumdiv(6);
   return 0;
}

int sumdiv (int i){

int sum = 0;
int divisor;

for(i = 1; i<=1000; i++){
    {for(int divisor = 1; divisor<=i; divisor++)
        if(int divisor<i && i%divisor==0)
            sum+= divisor;
            return sum;}

}

Anyone can help?

[Arslan7041]

Anyone?!

[shadowCODE]

start by fixing line 7. you missed int

[Arslan7041]

start by fixing line 7. you missed int

Thank you. Revised code again:

#include <iostream>
#include <cmath>

using namespace std;


int sumdiv (int i){
{

    int sum = 0;
    int n, i = 1;

for(i=1; i>n; i++){
        for(n = 1; n<i; n++)
            if (i%n==0)
                sum+= n;}
       
        return sum;
    }
}


int main()
{
    cout << sumdiv(6) << "\n\n";
    return 0;
}

No more error messages but its still giving me the wrong output. I entered the number "6" into main, so the output should also be 6. But no matter what number I input, the output is always 0. Whats the problem?

[Arslan7041]

Nevermind. I figured it out. Here it is:

#include <iostream>
#include <cmath>

using namespace std;


int sumdiv (int i){

    int sum = 0;
    int n;

for(n = 1; n<i; n++)
            if (i%n==0)
                sum+= n;
                return sum;

    }

int main()
{
    cout << sumdiv(8) << "\n\n";
    return 0;
}

[keskiverto]

Two notes:

First on style. It is legal to have only one statement on a loop body or in if-clause, but it is more clear to use a brace-block.

Second: 1 divides every number. There is no need to calculate it, unless input is less than 1.

Therefore:

// Returns sum of divisors of param i
// The param i must be greater than 0
int sumdiv ( int i ) {
  int sum = 1;
  for ( int n = 2; n < i; ++n ) {
    if ( 0 == i % n ) {
      sum += n;
    }
  return sum;
}