# 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 ``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

 `= 2)``        ``\$res` `*= (1 + ``\$n``);` `    ``return` `\$res``;``}` `// Driver code``\$n` `= 30;``echo` `sumofoddFactors(``\$n``);` `// This code is contributed``// by nitin mittal.``?>`

## Javascript

 ``

Output :

`24`

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