Skip to content
Related Articles

Related Articles

Kth largest N digit number divisible by M

View Discussion
Improve Article
Save Article
Like Article
  • Last Updated : 03 Feb, 2022

Given three positive integers N, K, and M. The task is to find Kth largest N digit number divisible by M

Note: K will be such an integer that Kth largest N digit number divisible by M always exists.

Examples

Input: N = 2, K = 2, M = 2
Output: 96
Explanation: The 2nd largest 2 digit number divisible by 2 is 96. 

Input: N = 9, K = 6, M = 4
Output: 999999976

 

Approach: The problem is maths-based. Given three numbers N, K, and M. It is required to find the Kth largest N digit number divisible by M. To get the largest N digit divisible by M, at first it is required to find the largest N digit number(say P), which is N times 9
Now the largest N digit number divisible by M is (P – (P%M)).
Therefore, subtract (K-1) times M from this value to get the Kth largest value of N digit number which is divisible by M.
Given below is the conditions and mathematical expression to get Kth largest N digit number divisible by M.

Let P be the largest N digit number
Then the largest N digit number divisible by M is: (P – (P % M))
Now the Kth largest N digit number divisible by M is: [(P – (P % M)) – ((K – 1) * M)]

Below is the code according to the above formula.

C++




// C++ program for above approach
#include <iostream>
using namespace std;
 
// Function to find Kth N
// digit number divisible by M
int findAnswer(int N, int K, int M)
{
    int i;
    long long int r = 0;
 
    // Loop to calculate the largest
    // N digit number.
    for (i = 1; i <= N; i++) {
        r = r * 10 + 9;
    }
 
    // Kth largest N digit number
    // divisible by M.
    long long int u = r - (r % M)
        - M * (K - 1);
 
    return u;
}
 
// Driver Code
int main()
{
    int N = 9;
    int K = 6;
    int M = 4;
 
    cout << findAnswer(N, K, M);
    return 0;
}


Java




// Java program for above approach
import java.util.*;
 
class GFG{
 
  // Function to find Kth N
  // digit number divisible by M
  static int findAnswer(int N, int K, int M)
  {
    int i;
    int r = 0;
 
    // Loop to calculate the largest
    // N digit number.
    for (i = 1; i <= N; i++) {
      r = r * 10 + 9;
    }
 
    // Kth largest N digit number
    // divisible by M.
    int u = r - (r % M)
      - M * (K - 1);
 
    return u;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 9;
    int K = 6;
    int M = 4;
 
    System.out.print(findAnswer(N, K, M));
  }
}
 
// This code is contributed by 29AjayKumar


Python3




# Python code for the above approach
 
# Function to find Kth N
# digit number divisible by M
def findAnswer(N, K, M):
    i = None
    r = 0;
 
    # Loop to calculate the largest
    # N digit number.
    for i in range(1, N + 1):
        r = r * 10 + 9;
 
    # Kth largest N digit number
    # divisible by M.
    u = r - (r % M) - M * (K - 1);
 
    return u;
 
# Driver Code
N = 9;
K = 6;
M = 4;
 
print(findAnswer(N, K, M));
 
# This code is contributed by Saurabh Jaiswal


C#




// C# program for above approach
using System;
class GFG
{
 
  // Function to find Kth N
  // digit number divisible by M
  static int findAnswer(int N, int K, int M)
  {
    long r = 0;
 
    // Loop to calculate the largest
    // N digit number.
    for (int i = 1; i <= N; i++) {
      r = r * 10 + 9;
    }
 
    // Kth largest N digit number
    // divisible by M.
    long u = r - (r % M)
      - M * (K - 1);
 
    return (int)u;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 9;
    int K = 6;
    int M = 4;
 
    Console.Write(findAnswer(N, K, M));
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
    // JavaScript code for the above approach
 
    // Function to find Kth N
    // digit number divisible by M
    function findAnswer(N, K, M) {
        let i;
        let r = 0;
 
        // Loop to calculate the largest
        // N digit number.
        for (i = 1; i <= N; i++) {
            r = r * 10 + 9;
        }
 
        // Kth largest N digit number
        // divisible by M.
        let u = r - (r % M)
            - M * (K - 1);
 
        return u;
    }
 
    // Driver Code
    let N = 9;
    let K = 6;
    let M = 4;
 
    document.write(findAnswer(N, K, M));
 
     // This code is contributed by Potta Lokesh
</script>


 
 

Output

999999976

 

Time Complexity: O(MaxDigit), Where maxDigit is the largest N digit number.
Auxiliary Space: O(1)

 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!