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.

Examples:

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.

## C++

// C++ program to find sum of all divisors of // a natural number #include<bits/stdc++.h> using namespace std; // Function to calculate sum of all proper divisors // num --> given natural number int divSum(int num) { // Final result of summation of divisors int result = 0; // find all divisors which divides 'num' for (int i=2; i<=sqrt(num); i++) { // if 'i' is divisor of 'num' if (num%i==0) { // if both divisors are same then add // it only once else add both if (i==(num/i)) result += i; else result += (i + num/i); } } // Add 1 to the result as 1 is also a divisor return (result + 1); } // Driver program to run the case int main() { int num = 36; cout << divSum(num); return 0; }

## Java

// JAVA program to find sum of all divisors // of a natural number import java.math.*; class GFG { // Function to calculate sum of all proper // divisors num --> given natural number static int divSum(int num) { // Final result of summation of divisors int result = 0; // find all divisors which divides 'num' for (int i = 2; i <= Math.sqrt(num); i++) { // if 'i' is divisor of 'num' if (num % i == 0) { // if both divisors are same then // add it only once else add both if (i == (num / i)) result += i; else result += (i + num / i); } } // Add 1 to the result as 1 is also // a divisor return (result + 1); } // Driver program to run the case public static void main(String[] args) { int num = 36; System.out.println(divSum(num)); } } /*This code is contributed by Nikita Tiwari*/

## Python

# PYTHON program to find sum of all # divisors of a natural number import math # Function to calculate sum of all proper # divisors num --> given natural number def divSum(num) : # Final result of summation of divisors result = 0 # find all divisors which divides 'num' i = 2 while i<= (math.sqrt(num)) : # if 'i' is divisor of 'num' if (num % i == 0) : # if both divisors are same then # add it only once else add both if (i == (num / i)) : result = result + i; else : result = result + (i + num/i); i = i + 1 # Add 1 to the result as 1 is also # a divisor return (result + 1); # Driver program to run the case num = 36 print (divSum(num)) # This code is contributed by Nikita Tiwari

Output:

55

If you have another approach to solve this problem then please share.

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