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Program to check if N is a Nonagonal Number
  • Last Updated : 01 Apr, 2021

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

Nonagonal Number is a figurate number that extends the concept of triangular and square numbers to the Nonagon. Specifically, the nth Nonagonal Numbers count the number of dots in a pattern of n nested nonagons(9 sided polygon), all sharing a common corner, where the ith nonagon in the pattern has sides made of i dots spaced one unit apart from each other. The first few Nonagonal Numbers are 1, 9, 24, 46, 75, 111, 154, … 
 

Examples:  

Input: N = 9 
Output: Yes 
Explanation: 
Second Nonagonal Number is 9.

Input: N = 20 
Output: No 



Approach:  

1. The Kth term of the nonagonal number is given as
K^{th} Term = \frac{7*K^{2} - 5*K}{2}

2. As we have to check that the given number can be expressed as a Nonagonal Number or not. This can be checked as: 

=> N = \frac{7*K^{2} - 5*K}{2}
=> K = \frac{5 + \sqrt{56*N + 25}}{14}

3. If the value of K calculated using the above formula is an integer, then N is a Nonagonal Number.

4. Else N is not a Nonagonal Number.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if N is a
// is a Nonagonal Number
bool isnonagonal(int N)
{
    float n
        = (5 + sqrt(56 * N + 25))
          / 14;
 
    // Condition to check if the
    // number is a nonagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 9;
 
    // Function call
    if (isnonagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java program for the above approach
import java.lang.Math;
 
class GFG{
     
// Function to check if N is a
// nonagonal number
public static boolean isnonagonal(int N)
{
    double n = (5 + Math.sqrt(56 * N + 25)) / 14;
     
    // Condition to check if the
    // number is a nonagonal number
    return (n - (int)n) == 0;
}
 
// Driver code    
public static void main(String[] args)
{
         
    // Given number
    int N = 9;
     
    // Function call
    if (isnonagonal(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 program for the above approach
 
# Function to check if N is a
# nonagonal number
def isnonagonal(N):
    n = (5 + pow((56 * N + 25), 1 / 2)) / 14;
 
    # Condition to check if the
    # number is a nonagonal number
    return (n - int(n)) == 0;
 
# Driver code
if __name__ == '__main__':
 
    # Given number
    N = 9;
 
    # Function call
    if (isnonagonal(N)):
        print("Yes");
    else:
        print("No");
 
# This code is contributed by Rajput-Ji

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to check if N is a
// nonagonal number
public static bool isnonagonal(int N)
{
    double n = (5 + Math.Sqrt(56 * N + 25)) / 14;
     
    // Condition to check if the
    // number is a nonagonal number
    return (n - (int)n) == 0;
}
 
// Driver code    
public static void Main(string[] args)
{
         
    // Given number
    int N = 9;
     
    // Function call
    if (isnonagonal(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by rutvik_56

Javascript




<script>
// Javascript program for the above approach
 
// Function to check if N is a
// is a Nonagonal Number
function isnonagonal(N)
{
    let n
        = (5 + Math.sqrt(56 * N + 25))
          / 14;
 
    // Condition to check if the
    // number is a nonagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
 
// Given Number
let N = 9;
 
// Function call
if (isnonagonal(N))
{
    document.write("Yes");
}
else
{
    document.write("No");
}
 
// This code is contributed by subhammahato348.
</script>
Output: 
Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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