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Program to check if N is a Pentadecagonal Number

  • Last Updated : 23 Jun, 2021

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

Pentadecagon Number is a 15-sided polygon..The first few Pentadecagon numbers are 1, 15, 42, 82, 135, 201, … 
 

Examples: 
 

Input: N = 15 
Output: Yes 
Explanation: 
Second Pentadecagon number is 15.
Input: N = 30 
Output: No 
 

 



Approach: 
 

  1. The Kth term of the Pentadecagon number is given as
    K^{th} Term = \frac{13*K^{2} - 11*K}{2}
     
  2. As we have to check that the given number can be expressed as a Pentadecagon Number or not. This can be checked as: 
     

=> N = \frac{13*K^{2} - 11*K}{2}
=> K = \frac{11 + \sqrt{104*N + 121}}{26}
 

  1.  
  2. If the value of K calculated using the above formula is an integer, then N is a Pentadecagon Number.
  3. Else N is not a Pentadecagon 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
// Pentadecagon number
bool isPentadecagon(int N)
{
    float n
        = (11 + sqrt(104 * N + 121))
          / 26;
 
    // Condition to check if the
    // number is a Pentadecagon number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 15;
 
    // Function call
    if (isPentadecagon(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
// Function to check if N is
// a pentadecagon number
public static boolean isPentadecagon(int N)
{
    double n = (11 + Math.sqrt(104 * N +
                               121)) / 26;
     
    // Condition to check if the number
    // is a pentadecagon number
    return (n - (int)n) == 0;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Number
    int N = 15;
     
    // Function call
    if (isPentadecagon(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by coder001

Python3




# Python3 program for the above approach
from math import sqrt
 
# Function to check if N is a
# pentadecagon number
def isPentadecagon(N):
 
    n = (11 + sqrt(104 * N + 121)) / 26;
     
    # Condition to check if the
    # number is a pentadecagon number
    return (n - int(n) == 0);
 
# Driver Code
if __name__ == "__main__":
 
    # Given number
    N = 15;
 
    # Function call
    if (isPentadecagon(N)):
        print("Yes");
     
    else :
        print("No");
     
# This code is contributed by AnkitRai01

C#




// C# program for the above approach
using System;
 
class GFG {
 
// Function to check if N is
// a pentadecagon number
public static bool isPentadecagon(int N)
{
    double n = (11 + Math.Sqrt(104 * N +
                               121)) / 26;
     
    // Condition to check if the number
    // is a pentadecagon number
    return (n - (int)n) == 0;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given Number
    int N = 15;
     
    // Function call
    if (isPentadecagon(N))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to check if N is a
// Pentadecagon number
function isPentadecagon(N)
{
    var n = (11 + Math.sqrt(104 * N + 121))
          / 26;
 
    // Condition to check if the
    // number is a Pentadecagon number
    return (n - parseInt(n)) == 0;
}
 
// Given Number
var N = 15;
// Function call
if (isPentadecagon(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
</script>
Output: 
Yes

 

Time Complexity: O(1)

Auxiliary Space: O(1)

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