# Sum of all perfect numbers present in an array

Given an array **arr[]** containing **N** positive integer. The task is to find the sum of all the perfect numbers from the array.

A number is perfect if is equal to sum of its proper divisors i.e. sum of its positive divisors excluding the number itself.

**Examples:**

Input:arr[] = {3, 6, 9}

Output:

Proper divisor sum of 3 = 1

Proper divisor sum of 6 = 1 + 2 + 3 = 6

Proper divisor sum of 9 = 1 + 3 = 4

Input:arr[] = {17, 6, 10, 6, 4}

Output:12

**Approach:** Initialise **sum = 0** and for every element of the array, find the sum of its proper divisors say **sumFactors**. If **arr[i] = sumFactors** then update the resultant sum as **sum = sum + arr[i]**. Print the **sum** in the end.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `// Function to return the sum of ` `// all the proper factors of n ` `int` `sumOfFactors(` `int` `n) ` `{ ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `f = 1; f <= n / 2; f++) ` ` ` `{ ` ` ` ` ` `// f is the factor of n ` ` ` `if` `(n % f == 0) ` ` ` `{ ` ` ` `sum += f; ` ` ` `} ` ` ` `} ` ` ` `return` `sum; ` `} ` ` ` `// Function to return the required sum ` `int` `getSum(` `int` `arr[], ` `int` `n) ` `{ ` ` ` ` ` `// To store the sum ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` ` ` `// If current element is non-zero and equal ` ` ` `// to the sum of proper factors of itself ` ` ` `if` `(arr[i] > 0 && ` ` ` `arr[i] == sumOfFactors(arr[i])) ` ` ` `{ ` ` ` `sum += arr[i]; ` ` ` `} ` ` ` `} ` ` ` `return` `sum; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[10] = { 17, 6, 10, 6, 4 }; ` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << (getSum(arr, n)); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the above approach ` `class` `GFG { ` ` ` ` ` `// Function to return the sum of ` ` ` `// all the proper factors of n ` ` ` `static` `int` `sumOfFactors(` `int` `n) ` ` ` `{ ` ` ` `int` `sum = ` `0` `; ` ` ` `for` `(` `int` `f = ` `1` `; f <= n / ` `2` `; f++) { ` ` ` ` ` `// f is the factor of n ` ` ` `if` `(n % f == ` `0` `) { ` ` ` `sum += f; ` ` ` `} ` ` ` `} ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Function to return the required sum ` ` ` `static` `int` `getSum(` `int` `[] arr, ` `int` `n) ` ` ` `{ ` ` ` ` ` `// To store the sum ` ` ` `int` `sum = ` `0` `; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) { ` ` ` ` ` `// If current element is non-zero and equal ` ` ` `// to the sum of proper factors of itself ` ` ` `if` `(arr[i] > ` `0` `&& arr[i] == sumOfFactors(arr[i])) { ` ` ` `sum += arr[i]; ` ` ` `} ` ` ` `} ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `[] arr = { ` `17` `, ` `6` `, ` `10` `, ` `6` `, ` `4` `}; ` ` ` `int` `n = arr.length; ` ` ` `System.out.print(getSum(arr, n)); ` ` ` `} ` `} ` |

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## Python3

`# Python3 implementation of the above approach ` ` ` `# Function to return the sum of ` `# all the proper factors of n ` `def` `sumOfFactors(n): ` ` ` ` ` `sum` `=` `0` ` ` `for` `f ` `in` `range` `(` `1` `, n ` `/` `/` `2` `+` `1` `): ` ` ` ` ` `# f is the factor of n ` ` ` `if` `(n ` `%` `f ` `=` `=` `0` `): ` ` ` `sum` `+` `=` `f ` ` ` ` ` `return` `sum` ` ` `# Function to return the required sum ` `def` `getSum(arr, n): ` ` ` ` ` `# To store the sum ` ` ` `sum` `=` `0` ` ` `for` `i ` `in` `range` `(n): ` ` ` ` ` `# If current element is non-zero and equal ` ` ` `# to the sum of proper factors of itself ` ` ` `if` `(arr[i] > ` `0` `and` ` ` `arr[i] ` `=` `=` `sumOfFactors(arr[i])) : ` ` ` `sum` `+` `=` `arr[i] ` ` ` ` ` `return` `sum` ` ` `# Driver code ` `arr ` `=` `[` `17` `, ` `6` `, ` `10` `, ` `6` `, ` `4` `] ` ` ` `n ` `=` `len` `(arr) ` `print` `(getSum(arr, n)) ` ` ` `# This code is contributed by Mohit Kumar ` |

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## C#

`// C# implementation of the above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the sum of ` ` ` `// all the proper factors of n ` ` ` `static` `int` `sumOfFactors(` `int` `n) ` ` ` `{ ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `f = 1; f <= n / 2; f++) ` ` ` `{ ` ` ` ` ` `// f is the factor of n ` ` ` `if` `(n % f == 0) ` ` ` `{ ` ` ` `sum += f; ` ` ` `} ` ` ` `} ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Function to return the required sum ` ` ` `static` `int` `getSum(` `int` `[] arr, ` `int` `n) ` ` ` `{ ` ` ` ` ` `// To store the sum ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` ` ` `// If current element is non-zero and equal ` ` ` `// to the sum of proper factors of itself ` ` ` `if` `(arr[i] > 0 && arr[i] == sumOfFactors(arr[i])) ` ` ` `{ ` ` ` `sum += arr[i]; ` ` ` `} ` ` ` `} ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `static` `public` `void` `Main () ` ` ` `{ ` ` ` `int` `[] arr = { 17, 6, 10, 6, 4 }; ` ` ` `int` `n = arr.Length; ` ` ` `Console.WriteLine(getSum(arr, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by @ajit_0023 ` |

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**Output:**

12

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