Lucky Numbers

Difficulty Level : Medium
Last Updated : 16 Oct, 2020

Lucky numbers are subset of integers. Rather than going into much theory, let us see the process of arriving at lucky numbers,
Take the set of integers
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,……
First, delete every second number, we get following reduced set.
1,3,5,7,9,11,13,15,17,19,…………
Now, delete every third number, we get
1, 3, 7, 9, 13, 15, 19,….….
Continue this process indefinitely……
Any number that does NOT get deleted due to above process is called “lucky”.
Therefore, set of lucky numbers is 1, 3, 7, 13,………
Now, given an integer ‘n’, write a function to say whether this number is lucky or not.

    bool isLucky(int n)

Algorithm:
Before every iteration, if we calculate position of the given number, then in a given iteration, we can determine if the number will be deleted. Suppose calculated position for the given number is P before some iteration, and each Ith number is going to be removed in this iteration, i

f P < I then input number is lucky,

if P is such that P%I == 0 (I is a divisor of P), then input no is not lucky.

How to calculate Next position of the number:

We know that initially the position of the number is nth itself. Now any next position will be equal to the previous position minus the number of elements (or say items) removed.

That is, next_position = current_position – count of numbers removed

For example, take the case of n=13.

We have: Initial position: n, ie. 13 itself.

1,2,3,4,5,6,7,8,9,10,11,12,13

Now after removing every second elements , we actually removed n/2 elements. So now the position of 13 will be : n-n/2=13-6=7 (n=13), i=2

1,3,5,7,9,11,13.

After that, we remove n/3 items. Note that n now is n=7. So position of 13 : n-n/3 = 7-7/3 = 7-2 = 5 (n=7), i=3

1,3,7,9,13

So next it will be : n-n/4 = 5-5/4 = 4 (n=5), i=4

1,3,7,13

So now i=5, but since position of 13 is 4 only, so it will be saved. Hence a lucky number! n=4, i=5

Recursive Way:

C++

// C++ program for Lucky Numbers
#include <bits/stdc++.h>
using namespace std;
#define bool int 
 
/* Returns 1 if n is a lucky no. 
otherwise returns 0*/
bool isLucky(int n) 
{ 
    static int counter = 2; 
     
    /*variable next_position is just for 
    readability of the program we can
    remove it and use n only */
    int next_position = n; 
    if(counter > n) 
        return 1; 
    if(n % counter == 0) 
        return 0; 
     
    /*calculate next position of input no*/
    next_position -= next_position / counter; 
     
    counter++; 
    return isLucky(next_position); 
} 
 
// Driver Code
int main() 
{ 
    int x = 5; 
    if( isLucky(x) ) 
        cout << x << " is a lucky no."; 
    else
        cout << x << " is not a lucky no."; 
} 
 
// This code is contributed 
// by rathbhupendra

C

#include <stdio.h>
#define bool int
 
/* Returns 1 if n is a lucky no. ohterwise returns 0*/
bool isLucky(int n)
{
    static int counter = 2;
     
    /*variable next_position is just for readability of
        the program we can remove it and use n only */
    int next_position = n;
    if(counter > n)
        return 1;
    if(n%counter == 0)
        return 0;     
     
    /*calculate next position of input no*/
    next_position -= next_position/counter;
     
    counter++;
    return isLucky(next_position);
}
 
/*Driver function to test above function*/
int main()
{
    int x = 5;
    if( isLucky(x) )
        printf("%d is a lucky no.", x);
    else
        printf("%d is not a lucky no.", x);
    getchar();
}

Java

// Java program to check Lucky Number
import java.io.*;
 
class GFG 
{
    public static int counter = 2;    
 
    // Returns 1 if n is a lucky no. 
    // ohterwise returns 0
    static boolean isLucky(int n)
    {
        // variable next_position is just 
        // for readability of
        // the program we can remove it 
        // and use n only 
        int next_position = n;
        if(counter > n)
            return true;
        if(n%counter == 0)
            return false;      
  
        // calculate next position of input no
        next_position -= next_position/counter;
    
        counter++;
        return isLucky(next_position);
    }
     
    // driver program
    public static void main (String[] args) 
    {
        int x = 5;
        if( isLucky(x) )
            System.out.println(x+" is a lucky no.");
        else
            System.out.println(x+" is not a lucky no.");
    }
}
 
// Contributed by Pramod Kumar

Python

# Python program to check for lucky number
 
# Returns 1 if n is a lucky number 
# otherwise returns 0
def isLucky(n):
     
    # Function attribute will act 
    # as static variable   
    # just for readability, can be 
    # removed and used n instead
    next_position = n
     
    if isLucky.counter > n:
        return 1
    if n % isLucky.counter == 0:
        return 0
     
    # Calculate next position of input number
    next_position = next_position - next_position /
                                   isLucky.counter
     
    isLucky.counter = isLucky.counter + 1
     
    return isLucky(next_position)
     
     
# Driver Code
 
isLucky.counter = 2 # Acts as static variable
x = 5
if isLucky(x):
    print x,"is a Lucky number"
else:
    print x,"is not a Lucky number"
     
# Contributed by Harshit Agrawal

C#

// C# program to check Lucky Number
using System;
 
class GFG {
     
    public static int counter = 2; 
 
    // Returns 1 if n is a lucky no.
    // ohterwise returns 0
    static bool isLucky(int n)
    {
         
        // variable next_position is 
        // just for readability of
        // the program we can remove
        // it and use n only 
        int next_position = n;
         
        if(counter > n)
            return true;
        if(n % counter == 0)
            return false;     
 
        // calculate next position of
        // input no
        next_position -= next_position 
                            / counter;
     
        counter++;
         
        return isLucky(next_position);
    }
     
    // driver program
    public static void Main () 
    {
        int x = 5;
         
        if( isLucky(x) )
            Console.Write(x + " is a "
                         + "lucky no.");
        else
            Console.Write(x + " is not"
                      + " a lucky no.");
    }
}
 
// This code is contributed by
// nitin mittal.

PHP

<?php
// PHP program for Lucky Numbers
 
/* Returns 1 if n is a lucky 
   no. ohterwise returns 0 */
function isLucky($n)
{
    $counter = 2;
     
    /* variable next_position is 
       just for readability of
       the program we can remove 
       it and use n only */
    $next_position = $n;
     
    if($counter > $n)
        return 1;
    if($n % $counter == 0)
        return 0; 
     
    /* calculate next position
       of input no */
    $next_position -= $next_position / $counter;
     
    $counter++;
    return isLucky($next_position);
}
 
    // Driver Code
    $x = 5;
    if(isLucky($x) )
        echo $x ," is a lucky no.";
    else
        echo $x ," is not a lucky no.";
         
// This code is contributed by anuj_67.
?>

Output:

5 is not a lucky no.

Example:
Let’s us take an example of 19
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,15,17,18,19,20,21,……
1,3,5,7,9,11,13,15,17,19,…..
1,3,7,9,13,15,19,……….
1,3,7,13,15,19,………
1,3,7,13,19,………
In next step every 6th no .in sequence will be deleted. 19 will not be deleted after this step because position of 19 is 5th after this step. Therefore, 19 is lucky. Let’s see how above C code finds out:

Current function call Position after this call Counter for next call Next Call isLucky(19 ) 10 3 isLucky(10) isLucky(10) 7 4 isLucky(7) isLucky(7) 6 5 isLucky(6) isLucky(6) 5 6 isLucky(5)

When isLucky(6) is called, it returns 1 (because counter > n).
Please write comments if you find any bug in the given programs or other ways to solve the same problem.

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