Apocalyptic Number

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

Apocalyptic Number is a number such that 2^N contains “666” as the substring.  

Examples:

Input: N = 157 
Output: Yes 
Explanation:
2¹⁵⁷ = 182687704666362864775460604089535377456991567872 
which contains as the substring “666”.

Input: N = 10 
Output: No 

Approach: The idea is to calculate the value of 2^N and check if the resultant number contains “666” as a substring or not. If it has “666” as a substring then print “Yes” else print “No”.

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 a number
// N is Apocalyptic
bool isApocalyptic(int n)
{
    if (to_string(pow(2, n)).find("666") != string::npos)
        return true;
    return false;
}

// Driver Code
int main()
{
    // Given Number N
    int N = 157;

    // Function Call
    if (isApocalyptic(N))
        cout << "Yes";
    else
        cout << "No";
}

// This code is contributed by phasing17

Java

// Java program for the above approach
class GFG {

    // Function to check if a number
    // N is Apocalyptic
    static boolean isApocalyptic(int n)
    {
        if (String.valueOf((
                               Math.pow(2, n)))
                .contains("666"))
            return true;
        return false;
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Given Number N
        int N = 157;

        // Function Call
        if (isApocalyptic(N))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach

# Function to check if a number
# N is Apocalyptic
def isApocalyptic(n):
    if '666' in str(2**n):
        return True
    return False

# Driver Code

# Given Number N
N = 157

# Function Call
if(isApocalyptic(157)):
    print("Yes")
else:
    print("No")

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if a number
// N is Apocalyptic
static bool isApocalyptic(int n)
{
    if (Math.Pow(2, n).ToString().Contains("666"))
        return true;

    return false;
}

static public void Main()
{

    // Given Number N
    int N = 157;

    // Function Call
    if (isApocalyptic(N))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed by offbeat

Javascript

<script>
// Javascript implementation

// Function to check if N isApocalyptic
function isApocalyptic(n)
{
    var x = Math.pow(2,n);
    if(x.toString().indexOf('666'))
        return true
    return false
}

// Driver Code
// Given Number N
var N = 157;

// Function Call
if (isApocalyptic(N))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by shubhamsingh10
</script>

Output

Yes

Time Complexity: O(n)