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 an N-digit number such that it is not divisible by any of its digits.
Note: There can be multiple answers for each value of N. 

Examples:  

Input: N = 4 
Output: 6789 
Explanation: 
As the number 6789 is not divisible by any of its digits, it is 6, 7, 8 and 9 and it is also a four-digit number. Hence, it can be the desired number. 

Input: N = 2 
Output: 57 
Explanation: 
As the number 57 is not divisible by any of its digits, it is 5 and 7 and it is also a 2-digit number. Hence, it can be the desired number.   

Approach: The key observation in the problem is that 2 and 3 are those numbers that don't divide each other. Also, the numbers "23, 233, 2333, ..." are not divisible by neither 2 nor 3. Hence, for any N-digit number, the most significant digit will be 2 and the rest of the digits will be 3 to get the desired number.

Algorithm:  

• Check if the value of the N is equal to 1, then there is no such number is possible, hence return -1.
• Otherwise, initialize a variable num, to store the number by 2.
• Run a loop from 1 to N and then, for each iteration, multiply the number by 10 and add 3 to it. 

num = (num * 10) + 3 

Below is the implementation of the above approach: 

// C++ implementation to find a
// N-digit number such that the number
// it is not divisible by its digits

#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;

// Function to find the number
// such that it is not divisible
// by its digits
void solve(ll n)
{
    // Base Cases
    if (n == 1)
    {
        cout << -1;
    }
    else {
        
        // First Digit of the
        // number will be 2
        int num = 2;
        
        // Next digits of the numbers
        for (ll i = 0; i < n - 1; i++) {
            num = (num * 10) + 3;
        }
        cout << num;
    }
}

// Driver Code
int main()
{
    ll n = 4;
    
    // Function Call
    solve(n);
}

// Java implementation to find a
// N-digit number such that the number
// it is not divisible by its digits
import java.io.*;
public class GFG {

    long ll;
    
    // Function to find the number
    // such that it is not divisible
    // by its digits
    static void solve(long n)
    {
        // Base Cases
        if (n == 1)
        {
            System.out.println(-1);
        }
        else {
            
            // First Digit of the
            // number will be 2
            int num = 2;
            
            // Next digits of the numbers
            for (long i = 0; i < n - 1; i++) {
                num = (num * 10) + 3;
            }
            System.out.println(num);
        }
    }
    
    // Driver Code
    public static void main (String[] args)
    {
        long n = 4;
        
            // Function Call
            solve(n);
    }
}

// This code is contributed by AnkitRai01

# Python3 implementation to find a 
# N-digit number such that the number 
# it is not divisible by its digits 

# Function to find the number 
# such that it is not divisible 
# by its digits 
def solve(n) : 

    # Base Cases 
    if (n == 1) :

        print(-1); 
    
    else :
        
        # First Digit of the 
        # number will be 2 
        num = 2; 
        
        # Next digits of the numbers 
        for i in range(n - 1) : 
            num = (num * 10) + 3; 
         
        print(num); 

# Driver Code 
if __name__ == "__main__" : 

    n = 4; 
    
    # Function Call 
    solve(n); 
    
# This code is contributed by AnkitRai01

// C# implementation to find a
// N-digit number such that the number
// it is not divisible by its digits
using System;

class GFG {
 
    long ll;
     
    // Function to find the number
    // such that it is not divisible
    // by its digits
    static void solve(long n)
    {
        // Base Cases
        if (n == 1)
        {
            Console.WriteLine(-1);
        }
        else {
             
            // First Digit of the
            // number will be 2
            int num = 2;
             
            // Next digits of the numbers
            for (long i = 0; i < n - 1; i++) {
                num = (num * 10) + 3;
            }
            Console.WriteLine(num);
        }
    }
     
    // Driver Code
    public static void Main(String[] args)
    {
        long n = 4;
         
            // Function Call
            solve(n);
    }
}

// This code is contributed by sapnasingh4991

<script>
//Javascript  implementation to find a
// N-digit number such that the number
// it is not divisible by its digits


// Function to find the number
// such that it is not divisible
// by its digits
function solve(n)
{
    // Base Cases
    if (n == 1)
    {
        document.write(  -1);
    }
    else {
        
        // First Digit of the
        // number will be 2
        var num = 2;
        
        // Next digits of the numbers
        for (var i = 0; i < n - 1; i++) {
            num = (num * 10) + 3;
        }
        document.write(  num);
    }
}


// Given N 
var n = 4;
// Function Call
solve(n);

// This code is contributed by SoumikMondal
</script>

2333

Performance Analysis: 

• Time Complexity: O(N).
• Auxiliary Space: O(1).