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
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 computed using this formula 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> |
Yes
Time Complexity: O(log N), it is time taken by inbuilt sqrt() function
Auxiliary Space: O(1)
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