# 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++ program for Lucky Numbers ` `#include ` `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 ` `#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 `

Output:

```5 is not a lucky no.
```

## 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

 ` ``\$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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