Find nth Magic Number

A magic number is defined as a number which can be expressed as a power of 5 or sum of unique powers of 5. First few magic numbers are 5, 25, 30(5 + 25), 125, 130(125 + 5), ….
Write a function to find the nth Magic number.
Example: 

Input: n = 2
Output: 25

Input: n = 5
Output: 130 

If we notice carefully the magic numbers can be represented as 001, 010, 011, 100, 101, 110 etc, where 001 is 0*pow(5,3) + 0*pow(5,2) + 1*pow(5,1). So basically we need to add powers of 5 for each bit set in a given integer n. 
Below is the implementation based on this idea. 

Approach : 

Step 1 : declare and assign a number for which you want to find the magic number.

Step 2 : assign a pow = 1, and ans = 0

Step 3 : use while loop to iterate each bit until ends (while n > 0)

Step 4 : inside loop, find last bit using & operation and  keep updating answer and  power as well

Step 5 : Once loop exit return answer

C++

// C++ program to find nth magic number
#include <bits/stdc++.h>
using namespace std;

// Function to find nth magic number
int nthMagicNo(int n)
{
    int pow = 1, answer = 0;

    // Go through every bit of n
    while (n)
    {
       pow = pow*5;

       // If last bit of n is set
       if (n & 1)
         answer += pow;

       // proceed to next bit
       n >>= 1;  // or n = n/2
    }
    return answer;
}

// Driver program to test above function
int main()
{
    int n = 5;
    cout << "nth magic number is " << nthMagicNo(n) << endl;
    return 0;
}

Java

// Java program to find nth
// magic number
import java.io.*;

class GFG
{
  // Function to find nth magic number
  static int nthMagicNo(int n)
  {
     int pow = 1, answer = 0;

     // Go through every bit of n
     while (n != 0)
     {
       pow = pow*5;

       // If last bit of n is set
       if ((int)(n & 1) == 1)
         answer += pow;

       // proceed to next bit
       // or n = n/2
       n >>= 1;
     }
     return answer;
  }

  // Driver program to test
  // above function
  public static void main(String[] args)
  {
    int n = 5;
    System.out.println("nth magic" +
    " number is " + nthMagicNo(n));
  }
}


// This code is contributed by
// prerna saini

Python3

# Python program to find nth magic number

# Function to find nth magic number
def nthMagicNo(n):

    pow = 1
    answer = 0

    # Go through every bit of n
    while (n):

        pow = pow*5

        # If last bit of n is set
        if (n & 1):
            answer += pow

        # proceed to next bit
        n >>= 1 # or n = n/2

    return answer


# Driver program to test above function
n = 5
print("nth magic number is", nthMagicNo(n))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to find nth
// magic number
using System;

public class GFG
{

// Function to find nth magic number
static int nthMagicNo(int n)
{
    int pow = 1, answer = 0;

    // Go through every bit of n
    while (n != 0)
    {
        pow = pow * 5;

        // If last bit of n is set
        if ((int)(n & 1) == 1)
            answer += pow;

        // proceed to next bit
        // or n = n/2
        n >>= 1;
    }
    return answer;
}

// Driver Code
public static void Main()
{
    int n = 5;
    Console.WriteLine("nth magic" +    " number is "
                       + nthMagicNo(n));

}
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to find nth
// magic number

// Function to find nth
// magic number
function nthMagicNo($n)
{
    $pow = 1;
    $answer = 0;

    // Go through every bit of n
    while ($n)
    {
    $pow = $pow * 5;

    // If last bit of n is set
    if ($n & 1)
        $answer += $pow;

    // proceed to next bit
    $n >>= 1; // or $n = $n/2
    }
    return $answer;
}

// Driver Code
$n = 5;
echo "nth magic number is ",
       nthMagicNo($n), "\n";

// This code is contributed by Ajit.
?>

Javascript

<script>

    // Javascript program to find nth
    // magic number

    // Function to find nth magic number
    function nthMagicNo(n)
    {
        let pow = 1, answer = 0;

        // Go through every bit of n
        while (n != 0)
        {
            pow = pow * 5;

            // If last bit of n is set
            if ((n & 1) == 1)
                answer += pow;

            // proceed to next bit
            // or n = n/2
            n >>= 1;
        }
        return answer;
    }

    let n = 5;
    document.write("nth magic" + " number is " + nthMagicNo(n));

</script>

Output :

 nth magic number is 130 

Complexity : 

Time complexity : O(logN)

Auxiliary Space : O(1)

Thanks to manrajsingh for suggesting the above solution.