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 given integer n.

Below is the implementation based on this idea.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

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

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java program to find nth 


// magic numebr 


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 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python program to find nth magic numebr 


  


# Function to find nth magic numebr 


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 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# program to find nth 


// magic numebr 


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 



















                        chevron_right







                        filter_none









PHP














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















<?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. 


?> 



















                        chevron_right







                        filter_none










Output : 


 nth magic number is 130 


Thanks to manrajsingh for suggesting 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.


GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details 





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:

Improved By :  Sam007, jit_t