# Python Program for 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
```

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.

## 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". `

Output:

```24
```

Please refer complete article on Find sum of odd factors of a number for more details!

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