Sum of all the factors of a number

Difficulty Level : Medium
Last Updated : 25 Jul, 2020

Given a number n, the task is to find the sum of all the divisors.

Examples :

Input : n = 30
Output : 72
Dividers sum 1 + 2 + 3 + 5 + 6 + 
             10 + 15 + 30 = 72 

Input :  n = 15
Output : 24
Dividers sum 1 + 3 + 5 + 15 = 24

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

A simple solution is to traverse through all divisors and add them.

C++

// Simple 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  
//divisors of a given number 
int divSum(int n) 
{ 
    if(n == 1) 
      return 1; 
  
    // Sum of divisors 
    int result = 0; 
   
    // find all divisors which divides 'num' 
    for (int i = 2; i <= sqrt(n); i++) 
    { 
        // if 'i' is divisor of 'n' 
        if (n % i == 0) 
        { 
            // if both divisors are same 
            // then add it once else add 
            // both 
            if (i == (n / i)) 
                result += i; 
            else
                result += (i + n/i); 
        } 
    } 
   
    // Add 1 and n to result as above loop 
    // considers proper divisors greater  
    // than 1. 
    return (result + n + 1); 
} 
   
// Driver program to run the case 
int main() 
{ 
    int n = 30; 
    cout << divSum(n); 
    return 0; 
} 

Java

// Simple Java program to  
// find sum of all divisors  
// of a natural number 
import java.io.*;  
  
class GFG { 
  
    // Function to calculate sum of all  
    //divisors of a given number 
    static int divSum(int n) 
    { 
         if(n == 1) 
           return 1; 
        // Final result of summation  
        // of divisors 
        int result = 0; 
      
        // find all divisors which divides 'num' 
        for (int i = 2; i <= Math.sqrt(n); i++) 
        { 
            // if 'i' is divisor of 'n' 
            if (n % i == 0) 
            { 
                // if both divisors are same 
                // then add it once else add 
                // both 
                if (i == (n / i)) 
                    result += i; 
                else
                    result += (i + n / i); 
            } 
        } 
      
        // Add 1 and n to result as above loop 
        // considers proper divisors greater 
        // than 1. 
        return (result + n + 1); 
          
    } 
      
    // Driver program to run the case 
    public static void main(String[] args) 
    { 
        int n = 30; 
        System.out.println(divSum(n)); 
    } 
} 
  
// This code is contributed by Prerna Saini.  

Python3

# Simple Python 3 program to  
# find sum of all divisors of 
# a natural number 
import math   
  
# Function to calculate sum  
# of all divisors of given 
#  natural number 
def divSum(n) : 
    if(n == 1): 
       return 1
  
    # Final result of summation  
    # of divisors 
    result = 0
    
    # find all divisors which 
    # divides 'num' 
    for i in range(2,(int)(math.sqrt(n))+1) : 
  
        # if 'i' is divisor of 'n' 
        if (n % i == 0) : 
  
            # if both divisors are same  
            # then add it only once 
            # else add both 
            if (i == (n/i)) : 
                result = result + i 
            else : 
                result = result + (i + n//i) 
          
          
    # Add 1 and n to result as above  
    # loop considers proper divisors 
    # greater than 1. 
    return (result + n + 1) 
    
# Driver program to run the case 
n = 30
print(divSum(n)) 
  
# This code is contributed by Nikita Tiwari. 

C#

// Simple C# program to  
// find sum of all divisors  
// of a natural number 
using System; 
  
class GFG { 
  
    // Function to calculate sum of all  
    //divisors of a given number 
    static int divSum(int n) 
    { 
        if(n == 1) 
           return 1; 
  
        // Final result of summation  
        // of divisors 
        int result = 0; 
      
        // find all divisors which divides 'num' 
        for (int i = 2; i <= Math.Sqrt(n); i++) 
        { 
            // if 'i' is divisor of 'n' 
            if (n % i == 0) 
            { 
                // if both divisors are same 
                // then add it once else add 
                // both 
                if (i == (n / i)) 
                    result += i; 
                else
                    result += (i + n / i); 
            } 
        } 
      
        // Add 1 and n to result as above loop 
        // considers proper divisors greater 
        // than 1. 
        return (result + n + 1); 
    } 
      
    // Driver program to run the case 
    public static void Main() 
    { 
          
        int n = 30; 
          
        Console.WriteLine(divSum(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// Find sum of all divisors  
// of a natural number 
  
// Function to calculate sum of all  
//divisors of a given number 
function divSum($n) 
{ 
    if($n == 1) 
      return 1; 
  
    // Sum of divisors 
    $result = 0; 
  
    // find all divisors  
    // which divides 'num' 
    for ( $i = 2; $i <= sqrt($n); $i++) 
    { 
        // if 'i' is divisor of 'n' 
        if ($n % $i == 0) 
        { 
            // if both divisors are same 
            // then add it once else add 
            // both 
            if ($i == ($n / $i)) 
                $result += $i; 
            else
                $result += ($i + $n / $i); 
        } 
    } 
  
    // Add 1 and n to result as 
    // above loop considers proper  
    // divisors greater than 1. 
    return ($result + $n + 1); 
} 
  
// Driver Code 
$n = 30; 
echo divSum($n); 
  
// This code is contributed by SanjuTomar. 
?> 

Output :

An efficient solution is to use below formula.
Let p₁, p₂, … p_k be prime factors of n. Let a₁, a₂, .. a_k be highest powers of p₁, p₂, .. p_k respectively that divide n, i.e., we can write n as n = (p₁^a₁)*(p₂^a₂)* … (p_k^a_k).

Sum of divisors = (1 + p₁ + p₁² ... p₁^a₁) * 
                  (1 + p₂ + p₂² ... p₂^a₂) *
                  .............................................
                  (1 + p_k + p_k² ... p_k^a_k) 

We can notice that individual terms of above 
formula are Geometric Progressions (GP). We
can rewrite the formula as.

Sum of divisors = (p₁^a₁+1 - 1)/(p₁ -1) * 
                  (p₂^a₂+1 - 1)/(p₂ -1) *
                  ..................................
                  (p_k^a_k+1 - 1)/(p_k -1)

How does above formula work?

Consider the number 18.  

Sum of factors = 1 + 2 + 3 + 6 + 9 + 18

Writing divisors as powers of prime factors.
Sum of factors = (2⁰)(3⁰) + (2¹)(3⁰) + (2^0)(3¹) +
                 (2¹)(3¹) + (2⁰)(3^²) + (2^1)(3²)
               = (2⁰)(3⁰) + (2^0)(3¹) + (2^0)(3²) +
                 (2¹)(3^0) + (2¹)(3¹) + (2¹)(3²)
               = (2⁰)(3⁰ + 3¹ + 3²) + 
                 (2¹)(3⁰ + 3¹ + 3²)
               = (2⁰ + 2¹)(3⁰ + 3¹ + 3²)
If we ta_ke a closer look, we can notice that the
above expression is in the form.
(1 + p₁) * (1 + p₂ + p₂²)
Where p₁ = 2 and p₂ = 3 and 18 = 2¹3²

So the task reduces to finding all prime factors and their powers.

C++

// Formula based CPP program to 
// find sum of all  divisors of n. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns sum of all factors of n. 
int sumofFactors(int n) 
{ 
    // Traversing through all prime factors. 
    int res = 1; 
    for (int i = 2; i <= sqrt(n); i++) 
    { 
  
          
        int curr_sum = 1; 
        int curr_term = 1; 
        while (n % i == 0) { 
  
            // THE BELOW STATEMENT MAKES 
            // IT BETTER THAN ABOVE METHOD  
            //  AS WE REDUCE VALUE OF n. 
            n = n / i; 
  
            curr_term *= i; 
            curr_sum += curr_term; 
        } 
  
        res *= curr_sum; 
    } 
  
    // This condition is to handle  
    // the case when n is a prime 
    // number greater than 2. 
    if (n >= 2) 
        res *= (1 + n); 
  
    return res; 
} 
  
// Driver code 
int main() 
{ 
    int n = 30; 
    cout << sumofFactors(n); 
    return 0; 
} 

Java

// Formula based Java program to  
// find sum of all divisors of n. 
  
import java.io.*; 
import java.math.*; 
public class GFG{ 
      
    // Returns sum of all factors of n. 
    static int sumofFactors(int n) 
    { 
        // Traversing through all prime factors. 
        int res = 1; 
        for (int i = 2; i <= Math.sqrt(n); i++) 
        { 
      
              
            int  curr_sum = 1; 
            int curr_term = 1; 
              
            while (n % i == 0)  
            { 
      
                // THE BELOW STATEMENT MAKES 
                // IT BETTER THAN ABOVE METHOD  
                // AS WE REDUCE VALUE OF n. 
                n = n / i; 
      
                curr_term *= i; 
                curr_sum += curr_term; 
            } 
      
            res *= curr_sum; 
        } 
      
        // This condition is to handle  
        // the case when n is a prime  
        // number greater than 2 
        if (n > 2) 
            res *= (1 + n); 
      
        return res; 
    } 
      
    // Driver code 
    public static void main(String args[]) 
    { 
        int n = 30; 
        System.out.println(sumofFactors(n)); 
    } 
} 
  
/*This code is contributed by Nikita Tiwari.*/

Python3

# Formula based Python3 code to find  
# sum of all divisors of n. 
import math as m 
  
# Returns sum of all factors of n. 
def sumofFactors(n): 
      
    # Traversing through all 
    # prime factors 
    res = 1
    for i in range(2, int(m.sqrt(n) + 1)): 
          
        curr_sum = 1
        curr_term = 1
          
        while n % i == 0: 
              
            n = n / i; 
  
            curr_term = curr_term * i; 
            curr_sum += curr_term; 
              
        res = res * curr_sum 
      
    # This condition is to handle the  
    # case when n is a prime number  
    # greater than 2 
    if n > 2: 
        res = res * (1 + n) 
  
    return res; 
  
# driver code     
sum = sumofFactors(30) 
print ("Sum of all divisors is: ",sum) 
  
# This code is contributed by Saloni Gupta 

C#

// Formula based Java program to  
// find sum of all divisors of n. 
using System; 
  
public class GFG { 
      
    // Returns sum of all factors of n. 
    static int sumofFactors(int n) 
    { 
          
        // Traversing through all prime factors. 
        int res = 1; 
        for (int i = 2; i <= Math.Sqrt(n); i++) 
        { 
      
              
            int curr_sum = 1; 
            int curr_term = 1; 
              
            while (n % i == 0)  
            { 
      
                // THE BELOW STATEMENT MAKES 
                // IT BETTER THAN ABOVE METHOD  
                // AS WE REDUCE VALUE OF n. 
                n = n / i; 
      
                curr_term *= i; 
                curr_sum += curr_term; 
            } 
      
            res *= curr_sum; 
        } 
      
        // This condition is to handle  
        // the case when n is a prime  
        // number greater than 2 
        if (n > 2) 
            res *= (1 + n); 
      
        return res; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
          
        int n = 30; 
          
        Console.WriteLine(sumofFactors(n)); 
    } 
} 
  
/*This code is contributed by vt_m.*/

PHP

<?php 
// Formula based PHP program to  
// find sum of all divisors of n. 
  
// Returns sum of all factors of n. 
function sumofFactors($n) 
{ 
      
    // Traversing through  
    // all prime factors. 
    $res = 1; 
    for ($i = 2; $i <= sqrt($n); $i++) 
    { 
  
        $curr_sum = 1; 
        $curr_term = 1; 
          
        while ($n % $i == 0)  
        { 
  
            // THE BELOW STATEMENT MAKES 
            // IT BETTER THAN ABOVE METHOD  
            // AS WE REDUCE VALUE OF n. 
            $n = $n / $i; 
  
            $curr_term *= $i; 
            $curr_sum += $curr_term; 
        } 
  
        $res *= $curr_sum; 
    } 
  
    // This condition is to handle  
    // the case when n is a prime  
    // number greater than 2 
    if ($n > 2) 
        $res *= (1 + $n); 
  
    return $res; 
} 
  
// Driver Code 
$n = 30; 
echo sumofFactors($n); 
  
// This code is contributed by Anuj_67. 
?> 

Output :

Further Optimization.
If there are multiple queries, we can use Sieve to find prime factors and their powers.

References:
https://www.math.upenn.edu/~deturck/m170/wk3/lecture/sumdiv.html
http://mathforum.org/library/drmath/view/71550.html

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Recommended: Please solve it on PRACTICE first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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Java

Python3

C#

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