Curzon Numbers

Given an integer N, check whether the given number is a Curzon Number or not.

A number N is said to be a Curzon Number if 2^N + 1 is divisible by 2*N + 1.

Example:

Input: N = 5 
Output: Yes 
Explanation:
2^5 + 1 = 33 and 2*5 + 1 = 11 
Since 11 divides 33, so 5 is a curzon number.

Input: N = 10 
Output: No 
Explanation:
2^10 + 1 = 1025 and 2*10 + 1 = 21 
1025 is not divisible by 21, so 10 is not a curzon number. 

Approach: The approach is to compute and check if 2^N + 1 is divisible by 2*N + 1 or not.  

• First find the value of 2*N + 1
• Then find the value of 2^N + 1
• Check if the second value is divisible by the first value, then it is a Curzon Number, else not.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if a number
// is a Curzon number or not
void checkIfCurzonNumber(int N)
{

    long int powerTerm, productTerm;

    // Find 2^N + 1
    powerTerm = pow(2, N) + 1;

    // Find 2*N + 1
    productTerm = 2 * N + 1;

    // Check for divisibility
    if (powerTerm % productTerm == 0)
        cout << "Yes\n";
    else
        cout << "No\n";
}

// Driver code
int main()
{
    long int N = 5;
    checkIfCurzonNumber(N);

    N = 10;
    checkIfCurzonNumber(N);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
import java.util.*;

class GFG {

// Function to check if a number
// is a Curzon number or not
static void checkIfCurzonNumber(long N)
{
    double powerTerm, productTerm;

    // Find 2^N + 1
    powerTerm = Math.pow(2, N) + 1;

    // Find 2*N + 1
    productTerm = 2 * N + 1;

    // Check for divisibility
    if (powerTerm % productTerm == 0)
        System.out.println("Yes");
    else
        System.out.println("No");
}

// Driver code
public static void main(String[] args)
{
    long N = 5;
    checkIfCurzonNumber(N);

    N = 10;
    checkIfCurzonNumber(N);
}
}

// This code is contributed by coder001

Python3

# Python3 implementation of the approach

# Function to check if a number
# is a Curzon number or not
def checkIfCurzonNumber(N):

    powerTerm, productTerm = 0, 0

    # Find 2^N + 1
    powerTerm = pow(2, N) + 1

    # Find 2*N + 1
    productTerm = 2 * N + 1

    # Check for divisibility
    if (powerTerm % productTerm == 0):
        print("Yes")
    else:
        print("No")

# Driver code
if __name__ == '__main__':

    N = 5
    checkIfCurzonNumber(N)

    N = 10
    checkIfCurzonNumber(N)

# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;

class GFG{

// Function to check if a number
// is a curzon number or not
static void checkIfCurzonNumber(long N)
{
    double powerTerm, productTerm;

    // Find 2^N + 1
    powerTerm = Math.Pow(2, N) + 1;

    // Find 2*N + 1
    productTerm = 2 * N + 1;

    // Check for divisibility
    if (powerTerm % productTerm == 0)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}

// Driver code
static public void Main ()
{
    long N = 5;
    checkIfCurzonNumber(N);

    N = 10;
    checkIfCurzonNumber(N);
}
}

// This code is contributed by shubhamsingh10

Javascript

<script>

// Javascript implementation of the approach

// Function to check if a number
// is a Curzon number or not
function checkIfCurzonNumber(N)
{
    var powerTerm, productTerm;

    // Find 2^N + 1
    powerTerm = Math.pow(2, N) + 1;

    // Find 2*N + 1
    productTerm = 2 * N + 1;

    // Check for divisibility
    if (powerTerm % productTerm == 0)
    {
        document.write("Yes" + "</br>");
    }
    else
    {
        document.write("No");
    }
}

// Driver code
var N = 5;
checkIfCurzonNumber(N);

N = 10;
checkIfCurzonNumber(N);

// This code is contributed by Ankita saini

</script>

Output:

Yes
No

Time complexity: O(log N)

Auxiliary Space: O(1)