# Refactorable number

Given an integer n. Check whether the number is refactorable or not. A refactorable number is an integer n that is divisible by count of all it’s divisors.
Example :

```Input:  n = 8
Output: yes

Explanation:
8 has 4 divisors: 1, 2, 4, 8
Since 8 is divisible by 4 therefore 8 is
refactorable number.

Input : n = 4
Output: no```

This solution is pretty straightforward. The idea is to iterate from 1 to sqrt(n) and count all the divisors of a number. After that we just need to check whether the number n is divisible by it’s total count or not.

## C++

 `// C++ program to check whether number is  ``// refactorable or not ``#include `` ` `// Function to count all divisors ``bool` `isRefactorableNumber(``int` `n) ``{  ``    ``// Initialize result ``    ``int` `divCount = 0;  ``     ` `    ``for` `(``int` `i = 1; i <= ``sqrt``(n); ++i) ``    ``{ ``        ``if` `(n % i==0) ``        ``{ ``            ``// If divisors are equal, count  ``            ``// only one. ``            ``if` `(n / i == i) ``                ``++divCount; `` ` `            ``// Otherwise count both ``            ``else` `                ``divCount += 2; ``        ``} ``    ``} `` ` `    ``return` `n % divCount == 0; ``} `` ` `//Driver Code ``int` `main() ``{ ``    ``int` `n = 8; ``    ``if` `(isRefactorableNumber(n)) ``        ``puts``(``"yes"``); ``    ``else``        ``puts``(``"no"``); `` ` `    ``n = 14; ``    ``if` `(isRefactorableNumber(n)) ``        ``puts``(``"yes"``); ``    ``else``        ``puts``(``"no"``); `` ` `    ``return` `0; ``} `

## Java

 `// Java program to check whether number is  ``// refactorable or not `` ` `class` `GFG ``{ ``    ``// Function to count all divisors ``    ``static` `boolean` `isRefactorableNumber(``int` `n) ``    ``{ ``        ``// Initialize result ``        ``int` `divCount = ``0``;  ``         ` `        ``for` `(``int` `i = ``1``; i <= Math.sqrt(n); ++i) ``        ``{ ``            ``if` `(n % i==``0``) ``            ``{ ``                ``// If divisors are equal, count  ``                ``// only one. ``                ``if` `(n / i == i) ``                    ``++divCount; ``                     ` `                ``// Otherwise count both ``                ``else` `                    ``divCount += ``2``; ``            ``} ``        ``} ``        ``return` `n % divCount == ``0``; ``    ``} ``     ` `    ``public` `static` `void` `main (String[] args) ``    ``{ ``        ``int` `n = ``8``; ``        ``if` `(isRefactorableNumber(n)) ``            ``System.out.println(``"yes"``); ``        ``else``            ``System.out.println(``"no"``); ``             ` `        ``n = ``14``; ``        ``if` `(isRefactorableNumber(n)) ``            ``System.out.println(``"yes"``); ``        ``else``            ``System.out.println(``"no"``); ``    ``} ``} `` ` `// This code is contributed by Saket Kumar `

## Python3

 `# Python program to check whether number is ``# refactorable or not ``import` `math `` ` `def` `isRefactorableNumber(n): `` ` `    ``# Initialize result ``    ``divCount ``=` `0`` ` `    ``for` `i ``in` `range``(``1``,``int``(math.sqrt(n))``+``1``): `` ` `        ``if` `n ``%` `i ``=``=` `0``: `` ` `            ``# If divisors are equal, count only one ``            ``if` `n``/``i ``=``=` `i: ``                ``divCount ``+``=` `1`` ` `            ``else``:  ``# Otherwise count both ``                ``divCount ``+``=` `2`` ` `    ``return` `n ``%` `divCount ``=``=` `0`` ` ` ` `# Driver Code ``n ``=` `8``if` `isRefactorableNumber(n): ``    ``print` `(``"yes"``) ``else``: ``    ``print` `(``"no"``) `` ` `n ``=` `14``if` `(isRefactorableNumber(n)): ``    ``print` `(``"yes"``) ``else``: ``    ``print` `(``"no"``) `

## C#

 `// C# program to check whether number is  ``// refactorable or not ``using` `System; `` ` `class` `GFG ``{ ``    ``// Function to count all divisors ``    ``static` `bool` `isRefactorableNumber(``int` `n) ``    ``{ ``         ` `        ``// Initialize result ``        ``int` `divCount = 0;  ``         ` `        ``for` `(``int` `i = 1; i <= Math.Sqrt(n); ++i) ``        ``{ ``            ``if` `(n % i==0) ``            ``{ ``                ``// If divisors are equal, count  ``                ``// only one. ``                ``if` `(n / i == i) ``                    ``++divCount; ``                 ` `                ``// Otherwise count both ``                ``else` `                    ``divCount += 2; ``            ``} ``        ``} ``        ``return` `n % divCount == 0; ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `Main () ``    ``{ ``        ``int` `n = 8; ``        ``if` `(isRefactorableNumber(n)) ``            ``Console.WriteLine(``"yes"``); ``        ``else``            ``Console.Write(``"no"``); ``             ` `        ``n = 14; ``        ``if` `(isRefactorableNumber(n)) ``            ``Console.Write(``"yes"``); ``        ``else``            ``Console.Write(``"no"``); ``    ``} ``} `` ` `// This code is contributed by nitin mittal. `

## PHP

 ` `

## Javascript

 ` `

```Output:
yes
no```

Time complexity: O(sqrt(n))
Auxiliary space: O(1)
Facts about refactorable number

1. There is no refactorable number which is perfect

2. There is no three consecutive integers can all be refactorable.

3. Refactorable number have natural density zero.

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