# Lucky Numbers

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
1,2,3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,……

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)
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Algorithm:
Before every iteration, if we calculate position of the given no, then in a given iteration, we can determine if the no will be deleted. Suppose calculated position for the given no. is P before some iteration, and each Ith no. is going to be removed in this iteration, if P < I then input no is lucky, if P is such that P%I == 0 (I is a divisor of P), then input no is not lucky.   Recursive Way:

## C/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

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
```

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.

# GATE CS Corner    Company Wise Coding Practice

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