Sum of all integer divisors of a number

Question

I want to find sum of all divisors of a number i.e. if the number is 6 i want to have 1+2+3+6=12. My attempt to approach it is:

#include <iostream>

using namespace std;

int divisorsSum(int n){
    int sum=0;
    for (int i=1; i<=n; i++){
        if(n%i==0)
            i=sum+i;

    }
    return sum;
}

int main()
{
    cout<<divisorsSum(6);
}

However surprisingly it does not work at all, it returns nothing and i am not able to figure out what is wrong with my code.

Thus the question is how to make it works? BTW: There is no point in immediately down voting everything I am not an expert and yes i do make mistakes.

`i=sum+i;` should be `sum=sum+i` or even better `sum += i;`. — NathanOliver, Nov 21 '16 at 18:59
pretty sure you mean `i=1` for your for loop and `sum += i;` after the if. Next time be more careful when copying the examples. — Michael Dorgan, Nov 21 '16 at 19:02

score 2 · Accepted Answer · answered Nov 21 '16 at 19:01

2

You have several issues in your code.

int i = i;

and i is still not defined. You probably wanted i = 1

i = sum + i;

sum is not updated above. You probably wanted sum += i

answered Nov 21 '16 at 19:01

user3437460

17,253
15
58
106

score 1 · Answer 2 · answered Nov 21 '16 at 19:02

1

You need to change your function divisorsSum to use the following code:

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

answered Nov 21 '16 at 19:02

Emmanuel Mathi-Amorim

382
5
20

score 1 · Answer 3 · answered Nov 21 '16 at 19:02

1

for (int i=i; i<=n; i++)

Change the i=i to i = 1

answered Nov 21 '16 at 19:02

Karan Nagpal · Answer 4 · 2016-11-21T19:07:04.813

0

int divisorsSum(int n){
    int sum=0;
    for (int i=1; i<=n; i++){
         if(n%i==0)
            sum+=i;

    }
return sum;
}

i starts from 1 and not i
sum+=i and not i=sum+i

edited Nov 21 '16 at 19:07

answered Nov 21 '16 at 19:01

Karan Nagpal

341
2
10

score 0 · Answer 5 · answered Nov 21 '16 at 19:08

Maybe there are better algorithms of finding divisors of a number, but here is the correct version of your code.

int divisorsSum(int n){
    int sum=0;
    for (int i = 1; i <= n; ++i){
       if(n % i == 0)
            sum += i;

    }
    return sum;
}

And here is a little optimized version, if a i is bigger than half of n then i cannot be a divisor of n.

int divisorsSum(int n) {
    int sum=0;
    for (int i = n / 2; i >= 1; --i){
       if(n % i == 0)
            sum += i;

    }
    return sum;
}

Sum of all integer divisors of a number

5 Answers5