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 that 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 that 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 which 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++

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// 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);
}

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Java

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

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Python3

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# 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

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C#

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// 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

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Output:

2333

Performance Analysis:

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

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