# Find a N-digit number such that it is not divisible by any of its digits

Given an integer N, the task is to find any N-digit positive number (except for zeros) such that it is not divisible by any of its digits. If it is not possible to find any such number then print -1.

Note: There can be more than one such number for the same N-digit.

Examples:

```Input: N = 2
Output: 23
23 is not divisible by 2 or 3

Input: N = 3
Output: 239
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The easiest solution to this problem can be thought of with the help of digits ‘4’ and ‘5’.

1. Since, in order for a number to be divisible by 5, the number must end with 0 or 5; and in order for it to be divisible by 4, the last two digits if the number must be divisible by 4.
2. Therefore, a shortcut method can be applied to prevent both of the divisibility criteria of 4 and as well as of 5, as:
• To prevent a number from being divisible by 5, the number can contain 5 for every other digit except for last digit.
```Therefore for N digit number,
(N - 1) digits must be 5 = 5555...(N-1 times)d
where d is the Nth digit
```
• To prevent a number from being divisible by 4, the number can contain 5 at the second last digit and 4 at the last digit.
```Therefore for N digit number,
Last digit must be 4 = 5555...(N-1 times)4
```

Below is the implementation of the above approach:

## CPP

 `// CPP program to find N digit number such ` `// that it is not divisible by any of its digits ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function that print the answer ` `void` `findTheNumber(``int` `n) ` `{ ` `    ``// if n == 1 then it is ` `    ``// not possible ` `    ``if` `(n == 1) { ` `        ``cout << ``"Impossible"` `<< endl; ` `        ``return``; ` `    ``} ` ` `  `    ``// loop to n-1 times ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``cout << ``"5"``; ` `    ``} ` ` `  `    ``// print 4 as last digit of ` `    ``// the number ` `    ``cout << ``"4"``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 12; ` ` `  `    ``// Function call ` `    ``findTheNumber(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// JAVA program to find N digit number such ` `// that it is not divisible by any of its digits ` `class` `GFG{ ` `  `  `// Function that print the answer ` `static` `void` `findTheNumber(``int` `n) ` `{ ` `    ``// if n == 1 then it is ` `    ``// not possible ` `    ``if` `(n == ``1``) { ` `        ``System.out.print(``"Impossible"` `+``"\n"``); ` `        ``return``; ` `    ``} ` `  `  `    ``// loop to n-1 times ` `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) { ` `        ``System.out.print(``"5"``); ` `    ``} ` `  `  `    ``// print 4 as last digit of ` `    ``// the number ` `    ``System.out.print(``"4"``); ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``12``; ` `  `  `    ``// Function call ` `    ``findTheNumber(n); ` `  `  `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program to find N digit number such ` `# that it is not divisible by any of its digits ` `  `  `# Function that prthe answer ` `def` `findTheNumber(n): ` `    ``# if n == 1 then it is ` `    ``# not possible ` `    ``if` `(n ``=``=` `1``): ` `        ``print``(``"Impossible"``) ` `        ``return` `  `  `    ``# loop to n-1 times ` `    ``for` `i ``in` `range``(n``-``1``): ` `        ``print``(``"5"``,end``=``"") ` `  `  `    ``# print as last digit of ` `    ``# the number ` `    ``print``(``"4"``) ` `  `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `12` `  `  `    ``#Function call ` `    ``findTheNumber(n) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# program to find N digit number such ` `// that it is not divisible by any of its digits ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function that print the answer ` `static` `void` `findTheNumber(``int` `n) ` `{ ` `    ``// if n == 1 then it is ` `    ``// not possible ` `    ``if` `(n == 1) { ` `        ``Console.Write(``"Impossible"` `+``"\n"``); ` `        ``return``; ` `    ``} ` ` `  `    ``// loop to n-1 times ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``Console.Write(``"5"``); ` `    ``} ` ` `  `    ``// print 4 as last digit of ` `    ``// the number ` `    ``Console.Write(``"4"``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `n = 12; ` ` `  `    ``// Function call ` `    ``findTheNumber(n); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```555555555554
```

Time complexity: 0(N)

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