# Smallest K digit number divisible by X

Integers X and K are given. The task is to find the smallest K-digit number divisible by X.

Examples :

Input : X = 83, K = 5 Output : 10043 10040 is the smallest 5 digit number that is multiple of 83. Input : X = 5, K = 2 Output : 10

A **simple solution** is to try all numbers starting from the smallest K digit number

(which is 100…(K-1)times) and return the first number divisible by X.

An **efficient solution** would be :

Compute MIN : smallest K-digit number (1000...K-times) If, MIN % X is 0, ans = MIN else, ans = (MIN + X) - ((MIN + X) % X)) This is because there will be a number in range [MIN...MIN+X] divisible by X.

## CPP

`// CPP code to find smallest K-digit number ` `// divisible by X ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to compute the result ` `int` `answer(` `int` `X, ` `int` `K) ` `{ ` ` ` `// Computing MIN ` ` ` `int` `MIN = ` `pow` `(10, K - 1); ` ` ` ` ` `// MIN is the result ` ` ` `if` `(MIN % X == 0) ` ` ` `return` `MIN; ` ` ` ` ` `// returning ans ` ` ` `return` `((MIN + X) - ((MIN + X) % X)); ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `// Number whose divisible is to be found ` ` ` `int` `X = 83; ` ` ` ` ` `// Max K-digit divisible is to be found ` ` ` `int` `K = 5; ` ` ` ` ` `cout << answer(X, K); ` `} ` |

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## Java

`// Java code to find smallest K-digit ` `// number divisible by X ` ` ` `import` `java.io.*; ` `import` `java.lang.*; ` ` ` `class` `GFG { ` ` ` `public` `static` `double` `answer(` `double` `X, ` `double` `K) ` ` ` `{ ` ` ` `double` `i = ` `10` `; ` ` ` `// Computing MIN ` ` ` `double` `MIN = Math.pow(i, K - ` `1` `); ` ` ` ` ` `// returning ans ` ` ` `if` `(MIN % X == ` `0` `) ` ` ` `return` `(MIN); ` ` ` `else` ` ` `return` `((MIN + X) - ((MIN + X) % X)); ` ` ` `} ` ` ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` ` ` `// Number whose divisible is to be found ` ` ` `double` `X = ` `83` `; ` ` ` `double` `K = ` `5` `; ` ` ` ` ` `System.out.println((` `int` `)answer(X, K)); ` ` ` `} ` `} ` ` ` `// Code contributed by Mohit Gupta_OMG <(0_o)> ` |

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## Python3

`# Python code to find smallest K-digit ` `# number divisible by X ` ` ` `def` `answer(X, K): ` ` ` ` ` `# Computing MAX ` ` ` `MIN` `=` `pow` `(` `10` `, K` `-` `1` `) ` ` ` ` ` `if` `(` `MIN` `%` `X ` `=` `=` `0` `): ` ` ` `return` `(` `MIN` `) ` ` ` ` ` `else` `: ` ` ` `return` `((` `MIN` `+` `X) ` `-` `((` `MIN` `+` `X) ` `%` `X)) ` ` ` ` ` `X ` `=` `83` `; ` `K ` `=` `5` `; ` ` ` `print` `(answer(X, K)); ` ` ` `# Code contributed by Mohit Gupta_OMG <(0_o)> ` |

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## C#

`// C# code to find smallest K-digit ` `// number divisible by X ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// Function to compute the result ` ` ` `public` `static` `double` `answer(` `double` `X, ` `double` `K) ` ` ` `{ ` ` ` ` ` `double` `i = 10; ` ` ` ` ` `// Computing MIN ` ` ` `double` `MIN = Math.Pow(i, K - 1); ` ` ` ` ` `// returning ans ` ` ` `if` `(MIN % X == 0) ` ` ` `return` `MIN; ` ` ` `else` ` ` `return` `((MIN + X) - ((MIN + X) % X)); ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` ` ` `// Number whose divisible is to be found ` ` ` `double` `X = 83; ` ` ` `double` `K = 5; ` ` ` ` ` `Console.WriteLine((` `int` `)answer(X, K)); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

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## PHP

`<?php ` `// PHP code to find smallest ` `// K-digit number divisible by X ` ` ` `// Function to compute ` `// the result ` `function` `answer(` `$X` `, ` `$K` `) ` `{ ` ` ` `// Computing MIN ` ` ` `$MIN` `= pow(10, ` `$K` `- 1); ` ` ` ` ` `// MIN is the result ` ` ` `if` `(` `$MIN` `% ` `$X` `== 0) ` ` ` `return` `$MIN` `; ` ` ` ` ` `// returning ans ` ` ` `return` `((` `$MIN` `+ ` `$X` `) - ` ` ` `((` `$MIN` `+ ` `$X` `) % ` `$X` `)); ` `} ` ` ` `// Driver Code ` ` ` `// Number whose divisible ` `// is to be found ` `$X` `= 83; ` ` ` `// Max K-digit divisible ` `// is to be found ` `$K` `= 5; ` ` ` `echo` `answer(` `$X` `, ` `$K` `); ` ` ` `// This code is contributed by ajit ` `?> ` |

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**Output :**

10043

This article is contributed by **Rohit Thapliyal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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