Append X digits to the end of N to make it divisible by M

Given three positive integers N, M, and X, the task is to generate a number by appending X digits on the right side of N such that the number is divisible by M. If multiple solutions exist, then print any of them. Otherwise, print -1.

Examples:

Input: N = 10, M = 5, X = 4 
Output: 105555 
Explanation: One of possible values of N (= 10) by appending X(= 4) digits on the right side of N is 105555, which is divisible by M (= 5).

Input: N = 4, M = 50, X = 2 
Output: 400 

Approach: The idea is to append X digits on the right side of N by trying out all possible digits from the range [0, 9] and after appending X digits on the right side of N, check if the number is divisible by M or not. If found to be true, then print the number. Otherwise, print -1. Following are the recurrence relation:

Follow the steps below to solve the problem:

• Use the above recurrence relation, check if the number N is divisible by M or not by appending X digits on the right side of N. If found to be true, then print the value of N by appending X digits.
• Otherwise, print -1.

Below is the implementation of the above approach :

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if the value of N by
// appending X digits on right side of N
// is divisible by M or not
bool isDiv(int N, int X, int M, int& res)
{

    // Base Case
    if (X == 0) {

        // If N is divisible
        // by M
        if (N % M == 0) {

            // Update res
            res = N;
            return true;
        }

        return false;
    }

    // Iterate over the range [0, 9]
    for (int i = 0; i <= 9; i++) {

        // If N is divisible by M by
        // appending X digits
        if (isDiv(N * 10 + i, X - 1, M, res)) {

            return true;
        }
    }
}

// Driver Code
int main()
{
    int N = 4, M = 50, X = 2;

    // Stores the number by appending
    // X digits on the right side of N
    int res = -1;

    isDiv(N, X, M, res);
    cout << res;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;

class GFG
{

  // Function to check if the value of N by
  // appending X digits on right side of N
  // is divisible by M or not
  static int isDiv(int N, int X, int M, int res)
  {

    // Base Case
    if (X == 0)
    {

      // If N is divisible
      // by M
      if (N % M == 0)
      {

        // Update res
        res = N;
        return res;
      }

      return res;
    }

    // Iterate over the range [0, 9]
    for (int i = 0; i < 9; i++)
    {

      // If N is divisible by M by
      // appending X digits
      int temp = isDiv(N * 10 + i, X - 1, M, res);
      if (temp != -1)
      {
        return temp;
      }
    }
    return res;
  }

  // Driver code
  public static void main(String[] args)
    throws java.lang.Exception
  {
    int N = 4, M = 50, X = 2;

    // Stores the number by appending
    // X digits on the right side of N
    int res = -1;

    res = isDiv(N, X, M, res);
    System.out.println(res);
  }
}

// This code is contributed by 18bhupenderyadav18.

Python3

# Python3 program to implement
# the above approach

# Function to check if the value of N by
# appending X digits on right side of N
# is divisible by M or not

# global variable to store result
res = -1


def isDiv(N, X, M):
    # Base case
    if(X == 0):
        # If N is divisible
        # by M
        if(N % M == 0):
            global res
            res = N
            return True

        return False

    # Iterate over the range [0, 9]
    for i in range(10):
        # if N is Divisible by M upon appending X digits
        if(isDiv(N*10+i, X-1, M)):
            return True


# Driver Code
if __name__ == "__main__":
    N, M, X = 4, 50, 2

    if(isDiv(N, X, M)):
        print(res)
    else:
        print("-1")

C#

// C# program to implement
// the above approach
using System;

class GFG
{

  // Function to check if the value of N by
  // appending X digits on right side of N
  // is divisible by M or not
  static int isDiv(int N, int X, int M, int res)
  {

    // Base Case
    if (X == 0)
    {

      // If N is divisible
      // by M
      if (N % M == 0)
      {

        // Update res
        res = N;
        return res;
      }

      return res;
    }

    // Iterate over the range [0, 9]
    for (int i = 0; i < 9; i++)
    {

      // If N is divisible by M by
      // appending X digits
      int temp = isDiv(N * 10 + i, X - 1, M, res);
      if (temp != -1)
      {
        return temp;
      }
    }
    return res;
  }

  // Driver code
  public static void Main()
  {
    int N = 4, M = 50, X = 2;

    // Stores the number by appending
    // X digits on the right side of N
    int res = -1;

    res = isDiv(N, X, M, res);
    Console.WriteLine(res);
  }
}

// This code is contributed by sanjoy_62

Javascript

<script>

// Javascript program to implement
// the above approach

var res = -1;

// Function to check if the value of N by
// appending X digits on right side of N
// is divisible by M or not
function isDiv(N, X, M)
{

    // Base Case
    if (X == 0) {

        // If N is divisible
        // by M
        if (N % M == 0) {

            // Update res
            res = N;
            return true;
        }

        return false;
    }

    // Iterate over the range [0, 9]
    for (var i = 0; i <= 9; i++) {

        // If N is divisible by M by
        // appending X digits
        if (isDiv(N * 10 + i, X - 1, M)) {

            return true;
        }
    }
}

// Driver Code

var N = 4, M = 50, X = 2;

// Stores the number by appending
// X digits on the right side of N
isDiv(N, X, M, res);

document.write(res);



</script>

Output

400

Time Complexity: O(10^X)
Auxiliary Space: O(1)