Python Program for Find Sum of Odd Factors of a Number (original) (raw)

Last Updated : 30 Oct, 2025

Given a number n, the task is to find the sum of all its odd factors. Odd factors are the divisors of n that are not divisible by 2. For example:

**Input: n = 30
**Output: 24
Odd divisors -> 1, 3, 5, 15 -> Sum = 24

**Input: n = 18
**Output: 13
Odd divisors -> 1, 3, 9 -> Sum = 13

Lets explore different methods to find sum of odd factors of a number in python.

Formula-based Approach

In this method, we use the prime factorization of n and ignore even factors. We repeatedly divide n by 2 to remove all even factors, then calculate the sum of odd divisors using powers of remaining primes.

python `

import math n = 30 res = 1

while n % 2 == 0: n //= 2

for i in range(3, int(math.sqrt(n)) + 1): curr_sum = 1 curr_term = 1 while n % i == 0: n //= i curr_term *= i curr_sum += curr_term res *= curr_sum

if n >= 2: res *= (1 + n)

print(res)

`

**Explanation:

• math.sqrt(n) gives upper limit for checking factors efficiently.
• while n % 2 == 0 removes all even factors from n.
• For each odd divisor i, multiply terms (1 + i + i² + ... ).
• Final product res gives the total odd factor sum.

Optimized Divisor Pairing

Here, we reduce the number of iterations by checking divisors only up to √n. For each divisor i, both i and n/i are considered if they are odd.

Python `

import math n = 30 odd_sum = 0

for i in range(1, int(math.sqrt(n)) + 1): if n % i == 0: if i % 2 == 1: odd_sum += i if (n // i) % 2 == 1 and n // i != i: odd_sum += n // i

print(odd_sum)

`

**Explanation:

• math.sqrt(n) limits loop to smallest half of divisors.
• For each divisor i, check both i and its pair n // i.
• Add them only if they are odd and avoids double-counting perfect squares.

Iterative Divisor Check

In this method, we loop through all numbers up to n and add only the odd divisors. It’s easy to understand but slower for large numbers.

Python `

n = 30 odd_sum = 0

for i in range(1, n + 1): if n % i == 0 and i % 2 == 1: odd_sum += i

print(odd_sum)

`

**Explanation:

• Loop from 1 to n to check all divisors.
• n % i == 0 ensures i divides n.
• i % 2 == 1 ensures i is odd.
• Adds only odd divisors to odd_sum.

Using List Comprehension

Here, we use list comprehension to quickly collect all odd divisors and sum them directly. It’s concise and Pythonic but not as efficient as formula-based or paired divisor methods.

Python `

n = 30 odd_sum = sum([i for i in range(1, n + 1) if n % i == 0 and i % 2 == 1]) print(odd_sum)

`

**Explanation:

• [i for i in range(1, n + 1) ...] builds a list of odd divisors.
• sum() adds all elements of that list.
• Simple one-line logic, easy to read for small values of n.

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