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.

This article is contributed by Abhay. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.