Sum of all proper divisors of a natural number
Given a natural number, calculate sum of all its proper divisors. A proper divisor of a natural number is the divisor that is strictly less than the number.
For example, number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
Input : num = 10 Output: 8 // proper divisors 1 + 2 + 5 = 8 Input : num = 36 Output: 55 // proper divisors 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 = 55
This problem has very simple solution, we all know that for any number ‘num’ all its divisors are always less than and equal to ‘num/2’ and all prime factors are always less than and equal to sqrt(num). So we iterate through ‘i’ till i<=sqrt(num) and for any ‘i’ if it divides ‘num’ , then we get two divisors ‘i’ and ‘num/i’ , continuously add these divisors but for some numbers divisors ‘i’ and ‘num/i’ will same in this case just add only one divisor , e.g; num=36 so for i=6 we will get (num/i)=6 , that’s why we will at 6 in the summation only once. Finally we add one as one is divisor of all natural numbers.
Please refer below post for an optimized solution and formula.
Efficient solution for sum of all the factors of a number
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