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

08 Jun, 2022
We are given an integer N. We need to write a program to find the least positive integer X made up of only digits 9’s and 0’s, such that, X is a multiple of N.
Note: It is assumed that the value of X will not exceed 106.
Examples: 
Input : N = 5
Output : X = 90
Explanation: 90 is the smallest number made up 
of 9's and 0's which is divisible by 5.

Input : N = 7
Output : X = 9009
Explanation: 9009 is smallest number made up 
of 9's and 0's which is divisible by 7.
The idea to solve this problem is to generate and store all of the numbers which can be formed using digits 0 & 9. Then find the smallest number among these generated number which is divisible by N.We will use the method of generating binary numbers to generate all numbers which can be formed by using digits 0 & 9.
Below is the implementation of above idea:  
C++
Java
Python3
C#
// CPP program to find smallest multiple of a// given number made of digits 0 and 9 only#include <bits/stdc++.h>using namespace std; // Maximum number of numbers made of 0 and 9#define MAX_COUNT 10000 // vector to store all numbers that can be formed// using digits 0 and 9 and are less than 10^5vector<string> vec; /* Preprocessing function to generate all possible   numbers formed by 0 and 9 */void generateNumbersUtil(){      // Create an empty queue of strings    queue<string> q;             // enqueue the first number    q.push("9");         // This loops is like BFS of a tree with 9 as root    // 0 as left child and 9 as right child and so on    for (int count = MAX_COUNT; count > 0; count--)    {        string s1 = q.front();        q.pop();                 // storing the front of queue in the vector        vec.push_back(s1);                 string s2 = s1;                 // Append "0" to s1 and enqueue it        q.push(s1.append("0"));                 // Append "9" to s2 and enqueue it. Note that        // s2 contains the previous front        q.push(s2.append("9"));    }} // function to find smallest number made up of only// digits 9’s and 0’s, which is a multiple of n.string findSmallestMultiple(int n){      // traverse the vector to find the smallest    // multiple of n    for (int i = 0; i < vec.size(); i++)         // stoi() is used for string to int conversion        if (stoi(vec[i])%n == 0)            return vec[i];       } // Driver Codeint main(){    generateNumbersUtil();       int n = 7;       cout << findSmallestMultiple(n);       return 0;}
// Java program to find smallest// multiple of a given number// made of digits 0 and 9 onlyimport java.util.*; class GFG{     // Maximum number of    // numbers made of 0 and 9    static int MAX_COUNT = 10000;     // vector to store all numbers    // that can be formed using    // digits 0 and 9 and are    // less than 10^5    static List<String> vec = new LinkedList<String>();     /* Preprocessing function    to generate all possible    numbers formed by 0 and 9 */    static void generateNumbersUtil()    {        // Create an empty        // queue of Strings        Queue<String> q = new LinkedList<String>();         // enqueue the        // first number        q.add("9");         // This loops is like BFS of        // a tree with 9 as root        // 0 as left child and 9 as        // right child and so on        for (int count = MAX_COUNT;                count > 0; count--)        {            String s1 = q.peek();            q.remove();             // storing the Peek of            // queue in the vector            vec.add(s1);             String s2 = s1;             // Append "0" to s1            // and enqueue it            q.add(s1 + "0");             // Append "9" to s2 and            // enqueue it. Note that            // s2 contains the previous Peek            q.add(s2 + "9");        }    }     // function to find smallest    // number made up of only    // digits 9's and 0's, which    // is a multiple of n.    static String findSmallestMultiple(int n)    {        // traverse the vector        // to find the smallest        // multiple of n        for (int i = 0; i < vec.size(); i++) // stoi() is used for        // String to int conversion        {            if (Integer.parseInt(vec.get(i)) % n == 0)            {                return vec.get(i);            }        }        return "";    }     // Driver Code    public static void main(String[] args)    {        generateNumbersUtil();        int n = 7;        System.out.println(findSmallestMultiple(n));    }} // This code is contributed by Rajput-Ji
# Python3 program to find smallest multiple of# a given number made of digits 0 and 9 onlyfrom queue import Queue # Preprocessing function to generate# all possible numbers formed by 0 and 9def generateNumbersUtil():    global vec         # Create an empty queue of strings    q = Queue()             # enqueue the first number    q.put("9")         # This loops is like BFS of a tree    # with 9 as root, 0 as left child    # and 9 as right child and so on    for count in range(MAX_COUNT, -1, -1):        s1 = q.queue[0]        q.get()                 # storing the front of queue        # in the vector        vec.append(s1)                 s2 = s1                 # Append "0" to s1 and enqueue it        s1 += "0"        q.put(s1)                 # Append "9" to s2 and enqueue it. Note        # that s2 contains the previous front        s2 += "9"        q.put(s2) # function to find smallest number made# up of only digits 9’s and 0’s, which# is a multiple of n.def findSmallestMultiple(n):    global vec         # traverse the vector to find    # the smallest multiple of n    for i in range(len(vec)):         # int is used for string to        # conversion        if (int(vec[i]) % n == 0):            return vec[i] # Driver Code # Maximum number of numbers# made of 0 and 9MAX_COUNT = 10000 # stack to store all numbers that# can be formed using digits 0 and# 9 and are less than 10^5vec = []generateNumbersUtil()    n = 7   print(findSmallestMultiple(n)) # This code is contributed by PranchalK
// C# program to find smallest// multiple of a given number// made of digits 0 and 9 onlyusing System;using System.Collections.Generic; class GFG{    // Maximum number of    // numbers made of 0 and 9    static int MAX_COUNT = 10000;         // vector to store all numbers    // that can be formed using    // digits 0 and 9 and are    // less than 10^5    static List<string> vec = new List<string>();         /* Preprocessing function    to generate all possible    numbers formed by 0 and 9 */    static void generateNumbersUtil()    {        // Create an empty        // queue of strings        Queue<string> q = new Queue<string>();                     // enqueue the        // first number        q.Enqueue("9");                 // This loops is like BFS of        // a tree with 9 as root        // 0 as left child and 9 as        // right child and so on        for (int count = MAX_COUNT;                 count > 0; count--)        {            string s1 = q.Peek();            q.Dequeue();                         // storing the Peek of            // queue in the vector            vec.Add(s1);                         string s2 = s1;                         // Append "0" to s1            // and enqueue it            q.Enqueue(s1 + "0");                         // Append "9" to s2 and            // enqueue it. Note that            // s2 contains the previous Peek            q.Enqueue(s2 + "9");        }    }         // function to find smallest    // number made up of only    // digits 9’s and 0’s, which    // is a multiple of n.    static string findSmallestMultiple(int n)    {        // traverse the vector        // to find the smallest        // multiple of n        for (int i = 0; i < vec.Count; i++)                 // stoi() is used for            // string to int conversion            if (int.Parse(vec[i]) % n == 0)                return vec[i];            return "";    }         // Driver Code    static void Main()    {        generateNumbersUtil();        int n = 7;        Console.Write(findSmallestMultiple(n));    }} // This code is contributed by// Manish Shaw(manishshaw1)
Output: 
9009
Time Complexity: O(n), where n represents the given integer.Auxiliary Space: O(10000), no extra space is required, so it is a constant.
manishshaw1
PranchalKatiyar
Rajput-Ji
simmytarika5
sumitgumber28
samim2000
BFS
cpp-queue
number-digits
Queue
Queue
BFS
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