Find sum of odd factors of a number

Given a number n, the task is to find the odd factor sum.

Examples :

Input : n = 30
Output : 24
Odd dividers sum 1 + 3 + 5 + 15 = 24 

Input : 18
Output : 13
Odd dividers sum 1 + 3 + 9 = 13

Prerequisite : Sum of all the factors of a number

As discussed in above mentioned previous post, sum of factors of a number is



Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
                  (1 + p2 + p22 ... p2a2) *
                  .............................................
                  (1 + pk + pk2 ... pkak) 

To find sum of odd factors, we simply need to ignore even factors and their powers. For example, consider n = 18. It can be written as 2132 and sun of all factors is (1)*(1 + 2)*(1 + 3 + 32). Sum of odd factors (1)*(1+3+32) = 13.

To remove all even factors, we repeatedly divide n while it is divisible by 2. After this step, we only get odd factors. Note that 2 is the only even prime.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 sumofoddFactors(int n)
{
    // Traversing through all
    // prime factors.
    int res = 1;
  
    // ignore even factors by 
    // removing all powers of
    // 2
    while (n % 2 == 0)
        n = n / 2;
  
    for (int i = 3; i <= sqrt(n); i++) 
    {
  
        // While i divides n, print
        // i and divide n
        int count = 0, curr_sum = 1;
        int curr_term = 1;
        while (n % i == 0) {
            count++;
  
            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.
    if (n >= 2)
        res *= (1 + n);
  
    return res;
}
  
// Driver code
int main()
{
    int n = 30;
    cout << sumofoddFactors(n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Formula based Java program 
// to find sum of all divisors
// of n.
import java.io.*;
import java.math.*;
  
class GFG {
      
    // Returns sum of all
    // factors of n.
    static int sumofoddFactors(int n)
    {
        // Traversing through 
        // all prime factors.
        int res = 1;
      
        // ignore even factors by
        // removing all powers
        // of 2
        while (n % 2 == 0)
            n = n / 2;
      
        for (int i = 3; i <= Math.sqrt(n); i++)
        {
      
            // While i divides n, print i 
            // and divide n
            int count = 0, curr_sum = 1;
            int curr_term = 1;
            while (n % i == 0)
            {
                count++;
      
                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.
        if (n >= 2)
            res *= (1 + n);
      
        return res;
    }
      
    // Driver code
    public static void main(String args[])
                        throws IOException
    {
        int n = 30;
        System.out.println(sumofoddFactors(n));
    }
}
  
/* This code is contributed by Nikita Tiwari.*/

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Formula based Python3 program 
# to find sum of all divisors
# of n.
import math
  
# Returns sum of all factors
# of n.
def sumofoddFactors( n ):
      
    # Traversing through all 
    # prime factors.
    res = 1
      
    # ignore even factors by 
    # of 2
    while n % 2 == 0:
        n = n // 2
      
    for i in range(3, int(math.sqrt(n) + 1)):
          
        # While i divides n, print
        # i and divide n
        count = 0
        curr_sum = 1
        curr_term = 1
        while n % i == 0:
            count+=1
              
            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.
    if n >= 2:
        res *= (1 + n)
      
    return res
  
# Driver code
n = 30
print(sumofoddFactors(n))
  
# This code is contributed by "Sharad_Bhardwaj".

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// Formula based C# program to
// find sum of all divisors of n.
using System;
  
class GFG {
      
    // Returns sum of all
    // factors of n.
    static int sumofoddFactors(int n)
    {
        // Traversing through 
        // all prime factors.
        int res = 1;
      
        // ignore even factors by
        // removing all powers
        // of 2
        while (n % 2 == 0)
            n = n / 2;
      
        for (int i = 3; i <= Math.Sqrt(n); i++)
        {
            // While i divides n, print i 
            // and divide n
            int count = 0, curr_sum = 1;
            int curr_term = 1;
            while (n % i == 0)
            {
                count++;
                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.
        if (n >= 2)
            res *= (1 + n);
      
        return res;
    }
      
    // Driver code
    public static void Main(String[] argc)
    {
        int n = 30;
        Console.Write(sumofoddFactors(n));
    }
}
  
/* This code is contributed by parashar...*/

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// Formula based PHP program 
// to find sum of all 
// divisors of n.
  
// Returns sum of all factors of n.
function sumofoddFactors($n)
{
    // Traversing through all
    // prime factors.
    $res = 1;
  
    // ignore even factors by 
    // removing all powers of
    // 2
    while ($n % 2 == 0)
        $n = $n / 2;
  
    for ($i = 3; $i <= sqrt($n); $i++) 
    {
  
        // While i divides n, print
        // i and divide n
        $count = 0; $curr_sum = 1;
        $curr_term = 1;
        while ($n % $i == 0) 
        {
            $count++;
  
            $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.
    if ($n >= 2)
        $res *= (1 + $n);
  
    return $res;
}
  
// Driver code
$n = 30;
echo sumofoddFactors($n);
  
// This code is contributed
// by nitin mittal. 
?>

chevron_right



Output :

24

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.