# Find two numbers made up of a given digit such that their difference is divisible by N

Given two numbers, **N** and **M**, the task is to find two numbers made up of M as all its digits such that their difference is divisible by **N**. **Examples:**

Input:N = 8, M = 2Output:22 222Explanation:The difference between 222 and 22 (200) is divisible by 8Input:N = 17, M = 6Output:6 66666666666666666

**Approach:**

In this problem, we have to find numbers consisting of only one unique digit. Let’s say **M** is equal to 2, then we have to find A and B from numbers like 2, 22, 222, 2222 …so on. The difference between A and B should be divisible by **N**. For this condition to satisfy, we have to pick A and B such that the remainder of A and B when it is divided by **N**, is the same.

For a digit of length **N+1** length, consisting of only one unique digit **M**, we will have **N+1** numbers. If we divide these **N+1 **numbers by **N** we will have **N+1** remainders which will range from **[0, N]**. Since the numbers can exceed the range of integer values, we are storing remainder-length of the number as key-value pairings in a Map. Once a remainder occurs with a value already paired in the Map, the current length and the mapped lengths are the lengths of the desired numbers.

The below code is the implementation of the above approach:

## C++

`// C++ implementation` `// of the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to implement` `// the above approach` `void` `findNumbers(` `int` `N, ` `int` `M)` `{` ` ` `int` `m = M;` ` ` `// Hashmap to store` ` ` `// remainder-length of the` ` ` `// number as key-value pairs` ` ` `map<` `int` `, ` `int` `> remLen;` ` ` `int` `len, remainder;` ` ` `// Iterate till N + 1 length` ` ` `for` `(len = 1; len <= N + 1; ++len) {` ` ` `remainder = M % N;` ` ` `// Search remainder in the map` ` ` `if` `(remLen.find(remainder)` ` ` `== remLen.end())` ` ` `// If remainder is not` ` ` `// already present insert` ` ` `// the length for the` ` ` `// corresponding remainder` ` ` `remLen[remainder] = len;` ` ` `else` ` ` `break` `;` ` ` `// Keep increasing M` ` ` `M = M * 10 + m;` ` ` `// To keep M in range of integer` ` ` `M = M % N;` ` ` `}` ` ` `// Length of one number` ` ` `// is the current Length` ` ` `int` `LenA = len;` ` ` `// Length of the other number` ` ` `// is the length paired with` ` ` `// current remainder in map` ` ` `int` `LenB = remLen[remainder];` ` ` `for` `(` `int` `i = 0; i < LenB; ++i)` ` ` `cout << m;` ` ` `cout << ` `" "` `;` ` ` `for` `(` `int` `i = 0; i < LenA; ++i)` ` ` `cout << m;` ` ` `return` `;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 8, M = 2;` ` ` `findNumbers(N, M);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to implement` `// the above approach` `static` `void` `findNumbers(` `int` `N, ` `int` `M)` `{` ` ` `int` `m = M;` ` ` `// Hashmap to store` ` ` `// remainder-length of the` ` ` `// number as key-value pairs` ` ` `Map<Integer, Integer> remLen = ` `new` `HashMap<>();` ` ` `int` `len, remainder = ` `0` `;` ` ` ` ` `// Iterate till N + 1 length` ` ` `for` `(len = ` `1` `; len <= N + ` `1` `; ++len)` ` ` `{` ` ` `remainder = M % N;` ` ` ` ` `// Search remainder in the map` ` ` `if` `(!remLen.containsKey(remainder))` ` ` `{` ` ` ` ` `// If remainder is not` ` ` `// already present insert` ` ` `// the length for the` ` ` `// corresponding remainder` ` ` `remLen.put(remainder, len);` ` ` `}` ` ` `else` ` ` `{` ` ` `break` `;` ` ` `}` ` ` ` ` `// Keep increasing M` ` ` `M = M * ` `10` `+ m;` ` ` ` ` `// To keep M in range of integer` ` ` `M = M % N;` ` ` `}` ` ` ` ` `// Length of one number` ` ` `// is the current Length` ` ` `int` `LenA = len;` ` ` ` ` `// Length of the other number` ` ` `// is the length paired with` ` ` `// current remainder in map` ` ` `int` `LenB = remLen.getOrDefault(remainder, ` `0` `);` ` ` `for` `(` `int` `i = ` `0` `; i < LenB; ++i)` ` ` `System.out.print(m);` ` ` `System.out.print(` `" "` `);` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < LenA; ++i)` ` ` `System.out.print(m);` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `8` `, M = ` `2` `;` ` ` `findNumbers(N, M);` `}` `}` `// This code is contributed by offbeat` |

## Python3

`# Python3 implementation` `# of the above approach` `# Function to implement` `# the above approach` `def` `findNumbers(N, M):` ` ` `m ` `=` `M` ` ` `# Hashmap to store` ` ` `# remainder-length of the` ` ` `# number as key-value pairs` ` ` `remLen ` `=` `{}` ` ` `# Iterate till N + 1 length` ` ` `for` `len1 ` `in` `range` `(` `1` `, N ` `+` `1` `, ` `1` `):` ` ` `remainder ` `=` `M ` `%` `N` ` ` ` ` `# Search remainder in the map` ` ` `if` `(remLen.get(remainder) ` `=` `=` `None` `):` ` ` ` ` `# If remainder is not` ` ` `# already present insert` ` ` `# the length for the` ` ` `# corresponding remainder` ` ` `remLen[remainder] ` `=` `len1` ` ` `else` `:` ` ` `break` ` ` `# Keep increasing M` ` ` `M ` `=` `M ` `*` `10` `+` `m` ` ` ` ` `# To keep M in range of integer` ` ` `M ` `=` `M ` `%` `N` ` ` ` ` `# Length of one number` ` ` `# is the current Length` ` ` `LenA ` `=` `len1` ` ` ` ` `# Length of the other number` ` ` `# is the length paired with` ` ` `# current remainder in map` ` ` `LenB ` `=` `remLen[remainder]` ` ` `for` `i ` `in` `range` `(LenB):` ` ` `print` `(m, end ` `=` `"")` ` ` `print` `(` `" "` `, end ` `=` `"")` ` ` ` ` `for` `i ` `in` `range` `(LenA):` ` ` `print` `(m, end ` `=` `"")` ` ` `return` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `N ` `=` `8` ` ` `M ` `=` `2` ` ` `findNumbers(N, M)` `# This code is contributed by Bhupendra_Singh` |

## C#

`// C# implementation of the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` `// Function to implement` `// the above approach` `static` `void` `findNumbers(` `int` `N, ` `int` `M)` `{` ` ` `int` `m = M;` ` ` `// To store remainder-length of` ` ` `// the number as key-value pairs` ` ` `Dictionary<` `int` `,` ` ` `int` `> remLen = ` `new` `Dictionary<` `int` `,` ` ` `int` `>();` ` ` `int` `len, remainder = 0;` ` ` ` ` `// Iterate till N + 1 length` ` ` `for` `(len = 1; len <= N + 1; ++len)` ` ` `{` ` ` `remainder = M % N;` ` ` ` ` `// Search remainder in the map` ` ` `if` `(!remLen.ContainsKey(remainder))` ` ` `{` ` ` ` ` `// If remainder is not` ` ` `// already present insert` ` ` `// the length for the` ` ` `// corresponding remainder` ` ` `remLen.Add(remainder, len);` ` ` `}` ` ` `else` ` ` `{` ` ` `break` `;` ` ` `}` ` ` ` ` `// Keep increasing M` ` ` `M = M * 10 + m;` ` ` ` ` `// To keep M in range of integer` ` ` `M = M % N;` ` ` `}` ` ` ` ` `// Length of one number` ` ` `// is the current Length` ` ` `int` `LenA = len;` ` ` ` ` `// Length of the other number` ` ` `// is the length paired with` ` ` `// current remainder in map` ` ` `int` `LenB = remLen[remainder];` ` ` `for` `(` `int` `i = 0; i < LenB; ++i)` ` ` `Console.Write(m);` ` ` ` ` `Console.Write(` `" "` `);` ` ` ` ` `for` `(` `int` `i = 0; i < LenA; ++i)` ` ` `Console.Write(m);` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 8, M = 2;` ` ` `findNumbers(N, M);` `}` `}` `// This code is contributed by Amit Katiyar` |

## Javascript

`<script>` ` ` `// Javascript implementation of the above approach` ` ` ` ` `// Function to implement` ` ` `// the above approach` ` ` `function` `findNumbers(N, M)` ` ` `{` ` ` `let m = M;` ` ` `// To store remainder-length of` ` ` `// the number as key-value pairs` ` ` `let remLen = ` `new` `Map();` ` ` `let len, remainder = 0;` ` ` `// Iterate till N + 1 length` ` ` `for` `(len = 1; len <= N + 1; ++len)` ` ` `{` ` ` `remainder = M % N;` ` ` `// Search remainder in the map` ` ` `if` `(!remLen.has(remainder))` ` ` `{` ` ` `// If remainder is not` ` ` `// already present insert` ` ` `// the length for the` ` ` `// corresponding remainder` ` ` `remLen.set(remainder, len);` ` ` `}` ` ` `else` ` ` `{` ` ` `break` `;` ` ` `}` ` ` `// Keep increasing M` ` ` `M = M * 10 + m;` ` ` `// To keep M in range of integer` ` ` `M = M % N;` ` ` `}` ` ` `// Length of one number` ` ` `// is the current Length` ` ` `let LenA = len;` ` ` `// Length of the other number` ` ` `// is the length paired with` ` ` `// current remainder in map` ` ` `let LenB = remLen.get(remainder);` ` ` `for` `(let i = 0; i < LenB; ++i)` ` ` `document.write(m);` ` ` `document.write(` `" "` `);` ` ` `for` `(let i = 0; i < LenA; ++i)` ` ` `document.write(m);` ` ` `}` ` ` ` ` `let N = 8, M = 2;` ` ` ` ` `findNumbers(N, M);` `// This code is contributed by divyeshrabadiya07.` `</script>` |

**Output:**

22 222

**Time Complexity:** *O(N)* **Auxiliary Space:** *O(1)*