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 sum 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 = 30print(sumofoddFactors(n)) # This code is contributed by “Sharad_Bhardwaj”. |
Output:
24
Time complexity: O(sqrt(n))
Auxiliary Space: O(1)
Please refer complete article on Find sum of odd factors of a number for more details!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
