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 2N + 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 2N + 1 is divisible by 2*N + 1 or not.
- First find the value of 2*N + 1
- Then find the value of 2N + 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 notvoid 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 codeint main(){ long int N = 5; checkIfCurzonNumber(N); N = 10; checkIfCurzonNumber(N); return 0;} |
Java
// Java implementation of the approachimport java.io.*; import java.util.*; class GFG { // Function to check if a number// is a Curzon number or notstatic 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 notdef 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 codeif __name__ == '__main__': N = 5 checkIfCurzonNumber(N) N = 10 checkIfCurzonNumber(N)# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing System;class GFG{ // Function to check if a number// is a curzon number or notstatic 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 notfunction 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 codevar 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)
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