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++ 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 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 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# 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 |
<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(log N), it is time taken by inbuilt sqrt() function
Auxiliary Space: O(1)