Python Program for Find Sum of Odd Factors of a Number

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.

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)

24

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.

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)

24

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.

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)

24

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.

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

24

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!