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Program to check if N is a Icositetragonal number

  • Last Updated : 23 Jun, 2021

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 
 

 



Approach: 
 

  1. The Kth term of the icositetragonal number is given as
    K^{th} Term = \frac{22*K^{2} - 20*K}{2}
     
  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 – 
     

=> N = \frac{22*K^{2} - 20*K}{2}
=> K = \frac{10 + \sqrt{44*N + 100}}{22}
 

  1.  
  2. 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 <bits/stdc++.h>
 
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

Javascript




<script>
 
// JavaScript implementation to check that
// a number is icositetragonal number or not
 
// Function to check that the
// number is a icositetragonal number
function isicositetragonal(N)
{
    var n = (10 + Math.sqrt(44 * N + 100))
          / 22;
 
    // Condition to check if the
    // number is a icositetragonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
var i = 24;
// Function call
if (isicositetragonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}
     
</script>
Output: 
Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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