# Program to check if N is a Icositetragonal number

Given an integer N, the task is to check if it is a icositetragonal number or not.

icositetragonal number:
s a class of figurate number. It has a 24-sided polygon called Icositetragon. The N-th Icositetragonal number count’s the number of dots and all others dots are surrounding with a common sharing corner and make a pattern
The first few icositetragonal numbers are 1, 24, 69, 136, 225, 336, …

Examples:

Input: N = 24
Output: Yes
Explanation:
Second icositetragonal number is 24.

Input: N = 30
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. The Kth term of the icositetragonal number is given as 2. As we have to check that the given number can be expressed as a icositetragonal number or not. This can be checked as follows –

=> => 3. Finally, check the value of computed using this formulae is an integer, which means that N is a icositetragonal number.

Below is the implementation of the above approach:

## C++

 // C++ implementation to check that  // a number is icositetragonal number or not     #include     using namespace std;     // Function to check that the  // number is a icositetragonal number  bool isicositetragonal(int N)  {      float n          = (10 + sqrt(44 * N + 100))            / 22;         // Condition to check if the      // number is a icositetragonal number      return (n - (int)n) == 0;  }     // Driver Code  int main()  {      int i = 24;         // Function call      if (isicositetragonal(i)) {          cout << "Yes";      }      else {          cout << "No";      }      return 0;  }

## Java

 // Java implementation to check that  // a number is icositetragonal number or not  class GFG{     // Function to check that the  // number is a icositetragonal number  static boolean isicositetragonal(int N)  {      float n = (float)((10 + Math.sqrt(44 * N +                                         100)) / 22);         // Condition to check if the      // number is a icositetragonal number      return (n - (int)n) == 0;  }     // Driver Code  public static void main(String[] args)  {      int i = 24;         // Function call      if (isicositetragonal(i))      {          System.out.print("Yes");      }      else     {          System.out.print("No");      }  }  }     // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation to check that   # a number is icositetragonal number   # or not   import math     # Function to check that the number  # is a icositetragonal number   def isicositetragonal(N):          n = (10 + math.sqrt(44 * N + 100)) / 22        # Condition to check if the number       # is a icositetragonal number       return (n - int(n)) == 0    # Driver Code   i = 24    # Function call   if (isicositetragonal(i)):       print("Yes")   else:       print("No")      # This code is contributed by divyamohan123

## C#

 // C# implementation to check that  // a number is icositetragonal number or not  using System;  class GFG{     // Function to check that the  // number is a icositetragonal number  static bool isicositetragonal(int N)  {      float n = (float)((10 + Math.Sqrt(44 * N +                                         100)) / 22);         // Condition to check if the      // number is a icositetragonal number      return (n - (int)n) == 0;  }     // Driver Code  public static void Main()  {      int i = 24;         // Function call      if (isicositetragonal(i))      {          Console.Write("Yes");      }      else     {          Console.Write("No");      }  }  }     // This code is contributed by Akanksha_Rai

Output:

Yes


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