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++

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// 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;
}

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Java














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// 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.*/



















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Python3














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# 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". 



















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C#














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// 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...*/



















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PHP














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<?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.  


?> 



















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Output :


24





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