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Sum of all proper divisors of a natural number

Given a natural number, calculate sum of all its proper divisors. A proper divisor of a natural number is the divisor that is strictly less than the number.

For example, number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

Examples :

Input : num = 10
Output: 8
// proper divisors 1 + 2 + 5 = 8 

Input : num = 36
Output: 55
// proper divisors 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 = 55 


This problem has very simple solution, we all know that for any number ‘num’ all its divisors are always less than and equal to ‘num/2’ and all prime factors are always less than and equal to sqrt(num). So we iterate through ‘i’ till i<=sqrt(num) and for any 'i' if it divides 'num' , then we get two divisors 'i' and 'num/i' , continuously add these divisors but for some numbers divisors 'i' and 'num/i' will same in this case just add only one divisor , e.g; num=36 so for i=6 we will get (num/i)=6 , that's why we will at 6 in the summation only once. Finally we add one as one is divisor of all natural numbers.



C++

// C++ program to find sum of all divisors of
// a natural number
#include<bits/stdc++.h>
using namespace std;
  
// Function to calculate sum of all proper divisors
// num --> given natural number
int divSum(int num)
{
    // Final result of summation of divisors
    int result = 0;
  
    // find all divisors which divides 'num'
    for (int i=2; i<=sqrt(num); i++)
    {
        // if 'i' is divisor of 'num'
        if (num%i==0)
        {
            // if both divisors are same then add
            // it only once else add both
            if (i==(num/i))
                result += i;
            else
                result += (i + num/i);
        }
    }
  
    // Add 1 to the result as 1 is also a divisor
    return (result + 1);
}
  
// Driver program to run the case
int main()
{
    int num = 36;
    cout << divSum(num);
    return 0;
}

Java

// JAVA program to find sum of all divisors
// of a natural number
import java.math.*;
  
class GFG {
      
    // Function to calculate sum of all proper
    // divisors num --> given natural number
    static int divSum(int num)
    {
        // Final result of summation of divisors
        int result = 0;
       
        // find all divisors which divides 'num'
        for (int i = 2; i <= Math.sqrt(num); i++)
        {
            // if 'i' is divisor of 'num'
            if (num % i == 0)
            {
                // if both divisors are same then 
                // add it only once else add both
                if (i == (num / i))
                    result += i;
                else
                    result += (i + num / i);
            }
        }
       
        // Add 1 to the result as 1 is also
        // a divisor
        return (result + 1);
    }
       
    // Driver program to run the case
    public static void main(String[] args)
    {
        int num = 36;
        System.out.println(divSum(num));
    }
}
  
/*This code is contributed by Nikita Tiwari*/

Python

# PYTHON program to find sum of all 
# divisors of a natural number
import math
      
# Function to calculate sum of all proper
# divisors num --> given natural number
def divSum(num) :
      
    # Final result of summation of divisors
    result = 0
      
    # find all divisors which divides 'num'
    i = 2
    while i<= (math.sqrt(num)) :
        
        # if 'i' is divisor of 'num'
        if (num % i == 0) :
        
            # if both divisors are same then
            # add it only once else add both
            if (i == (num / i)) :
                result = result + i;
            else :
                result = result +  (i + num/i);
        i = i + 1
          
    # Add 1 to the result as 1 is also 
    # a divisor
    return (result + 1);
   
# Driver program to run the case
num = 36
print (divSum(num))
  
# This code is contributed by Nikita Tiwari

C#

// C# program to find sum of all 
// divisorsof a natural number
using System;
  
class GFG {
      
    // Function to calculate sum of all proper
    // divisors num --> given natural number
    static int divSum(int num)
    {
          
        // Final result of summation of divisors
        int result = 0;
      
        // find all divisors which divides 'num'
        for (int i = 2; i <= Math.Sqrt(num); i++)
        {
              
            // if 'i' is divisor of 'num'
            if (num % i == 0)
            {
                  
                // if both divisors are same then 
                // add it only once else add both
                if (i == (num / i))
                    result += i;
                else
                    result += (i + num / i);
            }
        }
      
        // Add 1 to the result as 1 
        // is also a divisor
        return (result + 1);
    }
      
    // Driver Code
    public static void Main()
    {
        int num = 36;
        Console.Write(divSum(num));
    }
}
  
// This code is contributed by Nitin Mittal.

PHP

<?php
// PHP program to find sum of 
// all divisors of a natural number
  
// Function to calculate sum of 
// all proper divisors 
// num --> given natural number
function divSum($num)
{
    // Final result of 
    // summation of divisors
    $result = 0;
  
    // find all divisors 
    // which divides 'num'
    for ($i = 2; $i <= sqrt($num); 
                 $i++)
    {
        // if 'i' is divisor of 'num'
        if ($num % $i == 0)
        {
            // if both divisors are 
            // same then add it only
            // once else add both
            if ($i == ($num / $i))
                $result += $i;
            else
                $result += ($i + $num / $i);
        }
    }
  
    // Add 1 to the result as
    // 1 is also a divisor
    return ($result + 1);
}
  
// Driver Code
$num = 36;
echo(divSum($num));
  
// This code is contributed by Ajit.
?>


Output :

55

Please refer below post for an optimized solution and formula.
Efficient solution for sum of all the factors of a number

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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