Program to check if N is a Myriagon Number

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

Myriagon Number is a polygon with 10000 sides. The first few Myriagon numbers are 1, 10000, 29997, 59992, 99985, 149976 …

Examples:

Input: N = 10000
Output: Yes
Explanation:
Second Myriagon number is 10000.

Input: N = 300
Output: No



Approach:

  1. The Kth term of the Myriagon number is given as:
    K^{th} Term = \frac{9998*K^{2} - 9996*K}{2}
  2. As we have to check that the given number can be expressed as a Myriagon Number or not. This can be checked as:

    => N =  \frac{9998*K^{2} - 9996*K}{2}
    => K = \frac{9996 + \sqrt{79984*N + 99920016}}{19996}

  3. If the value of K calculated using the above formula is an integer, then N is a Myriagon Number.
  4. Else N is not a Myriagon Number.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if N is a
// Myriagon Number
bool isMyriagon(int N)
{
    float n
        = (9996 + sqrt(79984 * N + 99920016))
          / 19996;
  
    // Condition to check if the
    // number is a Myriagon number
    return (n - (int)n) == 0;
}
  
// Driver Code
int main()
{
    // Given Number
    int N = 10000;
  
    // Function call
    if (isMyriagon(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

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Java

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// Java program for the above approach
import java.io.*;
  
class GFG {
      
// Function to check if N 
// is a myriagon number
static boolean isMyriagon(int N)
{
    double n = (9996 + Math.sqrt(79984 * N + 
                                 99920016)) / 19996;
          
    // Condition to check if the
    // number is a myriagon number
    return (n - (int)n) == 0;
}
          
// Driver Code
public static void main (String[] args)
{
          
    // Given Number
    int N = 10000;
          
    // Function call
    if (isMyriagon(N))
    {
        System.out.println("Yes" );
    }
    else
    {
        System.out.println("No" );
    }
}
}
  
// This code is contributed by ShubhamCoder

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Python3

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# Python3 implementation to check that
# a number is a myriagon number or not
import math
  
# Function to check that the
# number is a myriagon number
def isMyriagon(N):
      
    n = (9996 + math.sqrt(79984 * N + 
                          99920016)) / 19996
      
    # Condition to check if the
    # number is a myriagon number
    return (n - int(n)) == 0
      
  
# Driver Code
n = 10000
  
# Function call
if (isMyriagon(n)):
    print("Yes")
else:
    print("No")
      
# This code is contributed by ShubhamCoder

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C#

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// C# program for the above approach
using System;
  
class GFG{
      
// Function to check if N 
// is a myriagon number
static bool isMyriagon(int N)
{
    double n = (9996 + Math.Sqrt(79984 * N + 
                                 99920016)) / 19996;
      
    // Condition to check if the
    // number is a myriagon number
    return (n - (int)n) == 0;
}
      
// Driver Code
static public void Main ()
{
      
    // Given Number
    int N = 10000;
      
    // Function call
    if (isMyriagon(N))
    {
        Console.Write( "Yes" );
    }
    else 
    {
        Console.Write( "No" );
    }
}
}
  
// This code is contributed by ShubhamCoder

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Output:

Yes

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