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# Amenable numbers

• Last Updated : 13 Jul, 2021

Given a number N, the task is to check if N is an Amenable Number or not. If N is an Amenable number then print “Yes” else print “No”.

Amenable numbers are numbers if there exists a multiset of integers whose size, sum and product equal to the number.
For example, 8 is an Amenable Number because there is a multiset of integers {-1, -1, 1, 1, 1, 1, 2, 4} whose size, sum and product is 8.

Examples:

Input: N = 8
Output: Yes
Explanation:
8 is an Amenable Number because there is a multiset of integers {-1, -1, 1, 1, 1, 1, 2, 4} whose size, sum and product is 8.
Input: N = 30
Output: No
Explanation:
30 is not an Amenable Number because there doesn’t exists a multiset of integers whose size, sum and product is 30.

Approach:
The first few Amenable Numbers are 1, 5, 8, 9, 12, 13, 16, 17, 20, 21….. which can be represented as the form 4K or 4K + 1
Therefore, any number N of the form 4K or 4K + 1 is an Amenable Numbers.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to check if N``// is Amenable number``bool` `isAmenableNum(``int` `N)``{` `    ``// Return true if N is of the form``    ``// 4K or 4K + 1``    ``return` `(N % 4 == 0``            ``|| (N - 1) % 4 == 0);``}` `// Driver Code``int` `main()``{``    ``// Given Number``    ``int` `n = 8;` `    ``// Function Call``    ``if` `(isAmenableNum(n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;``    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{` `// Function to check if N``// is Amenable number``static` `boolean` `isAmenableNum(``int` `N)``{``    ` `    ``// Return true if N is of the form``    ``// 4K or 4K + 1``    ``return` `(N % ``4` `== ``0` `|| (N - ``1``) % ``4` `== ``0``);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``8``;``    ` `    ``if` `(isAmenableNum(n))``    ``{``        ``System.out.println(``"Yes"``);``    ``}``    ``else``    ``{``        ``System.out.println(``"No"``);``    ``}``}``}` `// This code is contributed by shubham`

## Python3

 `# Python3 program for the above approach``import` `math` `# Function to check if N``# is Amenable number``def` `isAmenableNum(N):` `    ``# Return true if N is of the``    ``# form 4K or 4K + 1``    ``return` `(N ``%` `4` `=``=` `0` `or``           ``(N ``-` `1``) ``%` `4` `=``=` `0``);` `# Driver code` `# Given number``N ``=` `8``;` `# Function call``if` `(isAmenableNum(N)):``    ``print``(``"Yes"``);``else``:``    ``print``(``"No"``);` `# This code is contributed by rock_cool`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG{`` ` `// Function to check if N``// is Amenable number``static` `bool` `isAmenableNum(``int` `N)``{``     ` `    ``// Return true if N is of the form``    ``// 4K or 4K + 1``    ``return` `(N % 4 == 0 || (N - 1) % 4 == 0);``}`` ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 8;``     ` `    ``if` `(isAmenableNum(n))``    ``{``        ``Console.WriteLine(``"Yes"``);``    ``}``    ``else``    ``{``        ``Console.WriteLine(``"No"``);``    ``}``}``}` `// This code is contributed by amal kumar choubey`

## Javascript

 ``
Output:
`Yes`

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

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