Skip to content
Related Articles

Related Articles

Find sum of odd factors of a number
  • Difficulty Level : Medium
  • Last Updated : 23 Mar, 2021

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++




// 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;
}
Java



// 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.*/
Python3



# 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".
C#



// 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...*/
PHP



<?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.
?>
Javascript



<script>
 
// Formula based Javascript program
// to find sum of all
// divisors of n.
 
// Returns sum of all factors of n.
function sumofoddFactors(n)
{
    // Traversing through all
    // prime factors.
    let res = 1;
 
    // ignore even factors by
    // removing all powers of
    // 2
    while (n % 2 == 0)
        n = n / 2;
 
    for (let i = 3; i <= Math.sqrt(n); i++)
    {
 
        // While i divides n, print
        // i and divide n
        let count = 0;
        let curr_sum = 1;
        let 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
let n = 30;
document.write(sumofoddFactors(n));
 
// This code is contributed by _saurabh_jaiswal
 
</script>
Output : 
24
 
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.



My Personal Notes
arrow_drop_up
Previous
Find sum of even factors of a number
Next

Find largest prime factor of a number
Recommended Articles
Page :
Find number of factors of N when location of its two factors whose product is N is given
14, Aug 19
Maximum number of prime factors a number can have with exactly x factors
07, May 20
Check if a number exists having exactly N factors and K prime factors
10, May 20
Queries on sum of odd number digit sums of all the factors of a number
13, May 18
Print all numbers whose set of prime factors is a subset of the set of the prime factors of X
22, Apr 19
Expressing a number as sum of consecutive | Set 2 (Using odd factors)
18, Apr 18
Sum of all odd factors of numbers in the range [l, r]
05, Mar 19
Make all the elements of array odd by incrementing odd-indexed elements of odd-length subarrays
16, Dec 20
Number of elements with odd factors in given range
29, Aug 16
Check whether count of odd and even factors of a number are equal
03, Jul 20
Absolute difference between the count of odd and even factors of N
29, Apr 21
Find sum of even factors of a number
28, Dec 17
Find minimum sum of factors of number
08, Aug 17
Count of integers in a range which have even number of odd digits and odd number of even digits
21, Aug 19
Number of factors of very large number N modulo M where M is any prime number
31, Mar 20
Sum of M maximum distinct digit sum from 1 to N that are factors of K
04, Mar 20
Find four factors of N with maximum product and sum equal to N | Set-2
04, Oct 18
Find four factors of N with maximum product and sum equal to N
01, Oct 18
Find four factors of N with maximum product and sum equal to N | Set 3
10, Oct 18
Find prime factors of Array elements whose sum of exponents is divisible by K
15, Aug 20
Find sum of exponents of prime factors of numbers 1 to N
15, Jul 20
Sum of Factors of a Number using Prime Factorization
03, Oct 18
Sum of all the factors of a number
23, Dec 17
Sum of array elements which are prime factors of a given number
05, Mar 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
DevanshuAgarwal
@DevanshuAgarwal
Vote for difficulty
Current difficulty :
Medium





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox