Given a number n, the task is to find the odd factor sum.
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.
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- Product of factors of number
- Prime factors of a big number
- Check whether a number has exactly three distinct factors or not
- Sum of Factors of a Number using Prime Factorization
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