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 ` |

## C#

`// C# program to find sum of all ` `// divisorsof a natural number ` `using` `System; ` ` ` `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 Code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `num = 36; ` ` ` `Console.Write(divSum(num)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Nitin Mittal. ` |

## PHP

`<?php ` `// PHP program to find sum of ` `// all divisors of a natural number ` ` ` `// Function to calculate sum of ` `// all proper divisors ` `// num --> given natural number ` `function` `divSum(` `$num` `) ` `{ ` ` ` `// Final result of ` ` ` `// summation of divisors ` ` ` `$result` `= 0; ` ` ` ` ` `// find all divisors ` ` ` `// which divides 'num' ` ` ` `for` `(` `$i` `= 2; ` `$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 Code ` `$num` `= 36; ` `echo` `(divSum(` `$num` `)); ` ` ` `// This code is contributed by Ajit. ` `?> ` |

**Output :**

55

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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