Given a number N and a number K, the task is to find the smallest number greater than or equal to N which is divisible by K.

**Examples:**

Input:N = 45, K = 6Output:48 48 is the smallest number greater than or equal to 45 which is divisible by 6.Input:N = 11, K = 3Output:12

**Approach:** The idea is to divide the N+K by K. If the remainder is 0 then print **N** else print **N + K – remainder**.

Below is the implementation of the above approach :

## C++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the smallest number ` `// greater than or equal to N ` `// that is divisible by k ` `int` `findNum(` `int` `N, ` `int` `K) ` `{ ` ` ` `int` `rem = (N + K) % K; ` ` ` ` ` `if` `(rem == 0) ` ` ` `return` `N; ` ` ` `else` ` ` `return` `N + K - rem; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `N = 45, K = 6; ` ` ` ` ` `cout << ` `"Smallest number greater than or equal to "` `<< N ` ` ` `<< ` `"\nthat is divisible by "` `<< K << ` `" is "` `<< findNum(N, K); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the above approach ` ` ` `public` `class` `GFG{ ` ` ` ` ` `// Function to find the smallest number ` ` ` `// greater than or equal to N ` ` ` `// that is divisible by k ` ` ` `static` `int` `findNum(` `int` `N, ` `int` `K) ` ` ` `{ ` ` ` `int` `rem = (N + K) % K; ` ` ` ` ` `if` `(rem == ` `0` `) ` ` ` `return` `N; ` ` ` `else` ` ` `return` `N + K - rem; ` ` ` `} ` ` ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main(String []args){ ` ` ` ` ` `int` `N = ` `45` `, K = ` `6` `; ` ` ` ` ` `System.out.println(` `"Smallest number greater than or equal to "` `+ N ` ` ` `+` `"\nthat is divisible by "` `+ K + ` `" is "` `+ findNum(N, K)); ` ` ` ` ` `} ` ` ` `// This code is contributed by ANKITRAI1 ` `} ` |

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## Python

`# Python 3 implementation of the ` `# above approach ` ` ` `# Function to find the smallest number ` `# greater than or equal to N ` `# that is divisible by k ` `def` `findNum(N, K): ` ` ` `rem ` `=` `(N ` `+` `K) ` `%` `K; ` ` ` ` ` `if` `(rem ` `=` `=` `0` `): ` ` ` `return` `N ` ` ` `else` `: ` ` ` `return` `(N ` `+` `K ` `-` `rem) ` ` ` `# Driver Code ` `N ` `=` `45` `K ` `=` `6` `print` `(` `'Smallest number greater than'` `, ` ` ` `'or equal to'` `, N, ` ` ` `'that is divisible by'` `, K, ` ` ` `'is'` `, findNum(` `45` `, ` `6` `)) ` ` ` `# This code is contributed by Arnab Kundu ` |

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## C#

`// C# implementation of the above approach ` ` ` `public` `class` `GFG{ ` ` ` ` ` `// Function to find the smallest number ` ` ` `// greater than or equal to N ` ` ` `// that is divisible by k ` ` ` `static` `int` `findNum(` `int` `N, ` `int` `K) ` ` ` `{ ` ` ` `int` `rem = (N + K) % K; ` ` ` ` ` `if` `(rem == 0) ` ` ` `return` `N; ` ` ` `else` ` ` `return` `N + K - rem; ` ` ` `} ` ` ` ` ` ` ` `// Driver Code ` ` ` `static` `void` `Main(){ ` ` ` ` ` `int` `N = 45, K = 6; ` ` ` ` ` `System.Console.WriteLine(` `"Smallest number greater than or equal to "` `+ N ` ` ` `+` `"\nthat is divisible by "` `+ K + ` `" is "` `+ findNum(N, K)); ` ` ` ` ` `} ` ` ` `// This code is contributed by mits ` `} ` |

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## PHP

`<?php ` `// PHP implementation of the above approach ` ` ` `// Function to find the smallest number ` `// greater than or equal to N that is ` `// divisible by k ` `function` `findNum(` `$N` `, ` `$K` `) ` `{ ` ` ` `$rem` `= (` `$N` `+ ` `$K` `) % ` `$K` `; ` ` ` ` ` `if` `(` `$rem` `== 0) ` ` ` `return` `$N` `; ` ` ` `else` ` ` `return` `$N` `+ ` `$K` `- ` `$rem` `; ` `} ` ` ` `// Driver code ` `$N` `= 45; ` `$K` `= 6; ` ` ` `echo` `"Smallest number greater than "` `. ` ` ` `"or equal to "` `, ` `$N` `; ` `echo` `"\nthat is divisible by "` `, ` `$K` `, ` ` ` `" is "` `, findNum(` `$N` `, ` `$K` `); ` ` ` `// This code is contributed by anuj_67 ` `?> ` |

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**Output:**

Smallest number greater than or equal to 45 that is divisible by 6 is 48

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