# Aliquot sum

In number theory, the aliquot sum s(n) of a positive integer n is the sum of all proper divisors of n, that is, all divisors of n other than n itself.
They are defined by the sums of their aliquot divisors. The aliquot divisors of a number are all of its divisors except the number itself. The aliquot sum is the sum of the aliquot divisors so, for example, the aliquot divisors of 12 are 1, 2, 3, 4, and 6 and it’s aliquot sum is 16.
A number whose aliquot sum equals its value is a PERFECT number (6 for example).
Examples :

```Input : 12
Output : 16
Explanation :
Proper divisors of 12 is = 1, 2, 3, 4, 6
and sum 1 + 2 + 3 + 4 + 6 = 16

Input : 15
Output : 9
Explanation :
Proper divisors of 15 is 1, 3, 5
and sum 1 + 3 + 5 = 9```

A simple solution is to traverse through all numbers smaller than n. For every number i, check if i divides n. If yes, we add it to result.

## C++

 `// CPP program for aliquot sum` `#include ` `using` `namespace` `std;`   `// Function to calculate sum of ` `// all proper divisors` `int` `aliquotSum(``int` `n)` `{` `    ``int` `sum = 0;` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``if` `(n % i == 0) ` `            ``sum += i;        ` `    `  `    ``return` `sum; ` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 12;` `    ``cout << aliquotSum(n); ` `    ``return` `0;` `}`

## Java

 `// Java program for aliquot sum` `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Function to calculate sum of ` `    ``// all proper divisors` `    ``static` `int` `aliquotSum(``int` `n)` `    ``{` `        ``int` `sum = ``0``;` `        ``for` `(``int` `i = ``1``; i < n; i++) ` `            ``if` `(n % i == ``0``) ` `                ``sum += i;` `                `  `        ``return` `sum; ` `    ``}` `    `  `    ``// Driver Code` `    ``public` `static` `void` `main(String args[])` `                           ``throws` `IOException` `    ``{` `        ``int` `n = ``12``;` `        ``System.out.println(aliquotSum(n));` `    ``}` `}`   `/* This code is contributed by Nikita Tiwari.*/`

## Python3

 `# Python 3 program for aliquot sum`   `# Function to calculate sum of ` `# all proper divisors` `def` `aliquotSum(n) :` `    ``sm ``=` `0` `    ``for` `i ``in` `range``(``1``,n) :` `        ``if` `(n ``%` `i ``=``=` `0``) :` `            ``sm ``=` `sm ``+` `i     ` `    `  `    ``return` `sm ``# return sum`     `# Driver Code` `n ``=` `12` `print``(aliquotSum(n))`   `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program for aliquot sum` `using` `System;`   `class` `GFG {` `    `  `    ``// Function to calculate sum of ` `    ``// all proper divisors` `    ``static` `int` `aliquotSum(``int` `n)` `    ``{` `        ``int` `sum = 0;` `        ``for` `(``int` `i = 1; i < n; i++) ` `            ``if` `(n % i == 0) ` `                ``sum += i;` `                `  `        ``return` `sum; ` `    ``}` `    `  `    ``// Driver Code` `    ``public` `static` `void` `Main()` `                        `  `    ``{` `        ``int` `n = 12;` `        ``Console.WriteLine(aliquotSum(n));` `    ``}` `}`   `/* This code is contributed by vt_m.*/`

## PHP

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## Javascript

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Output :

`16`

Time Complexity: O(n)
Auxiliary Space: O(1)

Efficient Solutions :
Sum of all proper divisors of a natural number
Sum of all the factors of a number

Please suggest if someone has a better solution which is more efficient in terms of space and time.