Maximum multiple of a number with all digits same

Last Updated : 07 Mar, 2024

Given two positive integers X and Y (1 ? Y ? 10⁵), find the maximum positive multiple (say M) of X such that M is less than 10^Y and all the digits in the decimal representation of M are equal.

Note: If no positive multiple exists that satisfies the conditions, return -1.

Examples:

Input: X = 1, Y = 6
Output: 999999
Explanation: 999999 is the largest integer which is a multiple of 1, has all the same digits and less than 10⁶.

Input: X = 12, Y = 7
Output: 888888
Explanation: 888888 is the largest multiple of 12 which have all digits same and is less than 10⁷.

Input: X = 25, Y = 10
Output: -1
Explanation: No positive integers M satisfy the conditions.

Approach:

To solve the problem follow the below idea:

The main idea is to consider all the numbers less than 10^Y which have the same digits in decimal representation and check if the number is a multiple of X, from all such multiples, the maximum one as the result.

Step-by-step approach:

Consider a function f(n, d) which denotes the remainder of the number with length n and all digits d when divided by X.
Calculate f(n, d) for every {n, d}.
Notice that f(n ? 1, d) is a prefix of f(n, d). So: f(n, d) = (f(n?1, d)*10 + d)%X
Find the maximum value of n for which f(n, d) = 0.
- Choose the one with the maximum d among the ones with the maximum n.

Below is the implementation of the above approach.

C++

// C++ program for finding maximum multiple
// of a number with all the digits same

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

// Function for calculating maximum
// multiple of X less than 10^Y
string maxMultiple(int X, int Y)
{
    // f(n, d) -> d digit is repeated
    // n times (ddd....d)
    pair<int, int> res = { 0, 0 };
    int number = 0;
    for (int d = 1; d <= 9; ++d) {
        for (int n = 1; n < Y; ++n) {
            number = (10 * number + d) % X;
            if (number == 0)
                res = max(res, { n, d });
        }
    }

    // No such multiple exits so return -1
    if (res.first == 0)
        return "-1";

    // Generating the multiple
    // from pair res
    string ans = "";
    for (int i = 1; i <= res.first; i++)
        ans += (res.second + '0');

    return ans;
}

// Driver function
int main()
{
    int X = 12, Y = 7;

    // Function Call
    cout << maxMultiple(X, Y);

    return 0;
}

Java

// Java program for finding maximum 
// multiple of a number with all the digits same
class Main
{
  
  // Function for calculating maximum multiple of X less than 10^Y
  public static String maxMultiple(int X, int Y)
  {
    
    // f(n, d) -> d digit is repeated
    // n times (ddd....d)
    int res[] = {0, 0};
    int number = 0;
    for (int d = 1; d <= 9; ++d) {
      for (int n = 1; n < Y; ++n) {
        number = (10 * number + d) % X;
        if (number == 0) {
          // res = max(res, {n, d});
          // doing max of res & {n,d}
          if (res[0] > n) {
            // no change
          } else if (res[0] == n) {
            if (res[1] >= d) {
              // no change
            } else {
              res[0] = n;
              res[1] = d;
            }
          } else {
            res[0] = n;
            res[1] = d;
          }
        }
      }
    }

    // No such multiple exits so return -1
    if (res[0] == 0)
      return "-1";

    // Generating the multiple from pair res
    String ans = "";
    char c = '0';
    while (res[1] > 0) {
      c++;
      res[1]--;
    }
    for (int i = 1; i <= res[0]; i++) {
      ans += c;
    }

    return ans;
  }

  // Driver function
  public static void main(String[] args) {
    int X = 12, Y = 7;

    // Function Call
    System.out.println(maxMultiple(X, Y));
  }
}

// This code is contributed by ajaymakavana.

// C# program for finding maximum 
// multiple of a number with all the digits same
using System;

class GFG
{

  // Function for calculating maximum multiple of X less than 10^Y
  static string maxMultiple(int X, int Y)
  {

    // f(n, d) -> d digit is repeated
    // n times (ddd....d)
    int[] res = {0, 0};
    int number = 0;
    for (int d = 1; d <= 9; ++d) {
      for (int n = 1; n < Y; ++n) {
        number = (10 * number + d) % X;
        if (number == 0) {
          // res = max(res, {n, d});
          // doing max of res & {n,d}
          if (res[0] > n) {
            // no change
          } else if (res[0] == n) {
            if (res[1] >= d) {
              // no change
            } else {
              res[0] = n;
              res[1] = d;
            }
          } else {
            res[0] = n;
            res[1] = d;
          }
        }
      }
    }

    // No such multiple exits so return -1
    if (res[0] == 0)
      return "-1";

    // Generating the multiple from pair res
    string ans = "";
    char c = '0';
    while (res[1] > 0) {
      c++;
      res[1]--;
    }
    for (int i = 1; i <= res[0]; i++) {
      ans += c;
    }

    return ans;
  }

  // Driver function
  public static void Main(String[] args) {
    int X = 12, Y = 7;

    // Function Call

    Console.Write(maxMultiple(X, Y));
  }
}

// This code is contributed by Aman Kumar.

JavaScript

    // JavaScript program for finding maximum
    // multiple of a number with all the digits same

    // Function for calculating maximum 
    // multiple of X less than 10^Y
    function maxMultiple(X, Y){
        
        var res = [ 0, 0 ];
        var number = 0;
        for (let d = 1; d <= 9; ++d) {
            for (let n = 1; n < Y; ++n) {
                number = (10 * number + d) % X;
                if (number == 0) {
                    // res = max(res, {n, d});
                    // doing max of res & {n,d}
                    if (res[0] > n) {
                          // no change
                    } 
                    else if (res[0] == n) {
                          if (res[1] >= d) {
                            // no change
                          } 
                        else {
                            res[0] = n;
                            res[1] = d;
                        }
                    } 
                    else {
                          res[0] = n;
                          res[1] = d;
                    }
                }
            }
        }
        
        // No such multiple exits so return -1
        if (res[0] == 0)
              return "-1";

        // Generating the multiple from pair res
        var ans = "";
        var c = '0';
        while (res[1] > 0) {
            c++;
            res[1]--;
        }
        for (let i = 1; i <= res[0]; i++) {
              ans += c;
        }

        return ans;
    }

    var X = 12, Y = 7;
    
    // Function Call
    console.log(maxMultiple(X, Y));
    
    
    // This code is contributed by lokeshmvs21.

Python3

# Python3 program for finding maximum multiple
# of a number with all the digits same

# Function for calculating maximum
# multiple of X less than 10^Y
def maxMultiple(X, Y) :

    # f(n, d) -> d digit is repeated
    # n times (ddd....d)
    res = ( 0, 0 );
    number = 0;
    for d in range(1, 10) :
        for n in range(1, Y) :
            number = (10 * number + d) % X;
            if (number == 0) :
                res = max(res, ( n, d ));
    
    # No such multiple exits so return -1
    if (res[0] == 0) :
        return "-1";

    # Generating the multiple
    # from pair res
    ans = "";
    for i in range(1, res[0] + 1) :
        ans += str(res[1]);

    return ans;

# Driver function
if __name__ == "__main__" :

    X = 12; Y = 7;

    # Function Call
    print(maxMultiple(X, Y));

    # This code is contributed by AnkThon

Output

Time Complexity: O(Y)
Auxiliary Space: O(1)

Smallest multiple of a given number made of digits 0 and 9 only

codelohani

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