Minimum positive integer divisible by C and is not in range [A, B]

Given three positive integers **A**, **B** and **C**. The task is to find the minimum integer **X > 0** such that:

**X % C = 0**and**X**must not belong to the range**[A, B]**

**Examples:**

Input:A = 2, B = 4, C = 2

Output:6

Input:A = 5, B = 10, C = 4

Output:4

**Approach:**

- If
**C**doesn’t belong to**[A, B]**i.e.**C < A**or**C > B**then**C**is the required number. - Else get the first multiple of
**C**greater than**B**which is the required answer.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `// Function to return the required number ` `int` `getMinNum(` `int` `a, ` `int` `b, ` `int` `c) ` `{ ` ` ` ` ` `// If doesn't belong to the range ` ` ` `// then c is the required number ` ` ` `if` `(c < a || c > b) ` ` ` `return` `c; ` ` ` ` ` `// Else get the next multiple of c ` ` ` `// starting from b + 1 ` ` ` `int` `x = ((b / c) * c) + c; ` ` ` ` ` `return` `x; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `a = 2, b = 4, c = 4; ` ` ` `cout << getMinNum(a, b, c); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `import` `java.io.*; ` `import` `java.math.*; ` `public` `class` `GFG ` `{ ` ` ` `// Function to return the required number ` ` ` `int` `getMinNum(` `int` `a, ` `int` `b, ` `int` `c) ` ` ` `{ ` ` ` ` ` `// If doesn't belong to the range ` ` ` `// then c is the required number ` ` ` `if` `(c < a || c > b) ` ` ` `{ ` ` ` `return` `c; ` ` ` `} ` ` ` ` ` `// Else get the next multiple of c ` ` ` `// starting from b + 1 ` ` ` `int` `x = ((b / c) * c) + c; ` ` ` ` ` `return` `x; ` ` ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` ` ` `int` `a = ` `2` `; ` ` ` `int` `b = ` `4` `; ` ` ` `int` `c = ` `4` `; ` ` ` `GFG g = ` `new` `GFG(); ` ` ` `System.out.println(g.getMinNum(a, b, c)); ` `} ` `} ` ` ` `// This code is contributed by Shivi_Aggarwal ` |

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## Python3

`# Python3 implementation of the approach ` `# Function to return the required number ` `def` `getMinNum(a, b, c): ` ` ` ` ` `# If doesn't belong to the range ` ` ` `# then c is the required number ` ` ` `if` `(c < a ` `or` `c > b): ` ` ` `return` `c ` ` ` ` ` `# Else get the next multiple of c ` ` ` `# starting from b + 1 ` ` ` `x ` `=` `((b ` `/` `/` `c) ` `*` `c) ` `+` `c ` ` ` ` ` `return` `x ` ` ` `# Driver code ` `a, b, c ` `=` `2` `, ` `4` `, ` `4` `print` `(getMinNum(a, b, c)) ` ` ` `# This code is contributed by ` `# Mohit kumar 29 ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to return the required number ` ` ` `static` `int` `getMinNum(` `int` `a, ` `int` `b, ` `int` `c) ` ` ` `{ ` ` ` ` ` `// If doesn't belong to the range ` ` ` `// then c is the required number ` ` ` `if` `(c < a || c > b) ` ` ` `{ ` ` ` `return` `c; ` ` ` `} ` ` ` ` ` `// Else get the next multiple of c ` ` ` `// starting from b + 1 ` ` ` `int` `x = ((b / c) * c) + c; ` ` ` ` ` `return` `x; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `static` `public` `void` `Main () ` ` ` `{ ` ` ` `int` `a = 2, b = 4, c = 4; ` ` ` `Console.WriteLine( getMinNum(a, b, c)); ` ` ` `} ` `} ` ` ` `// This Code is contributed by ajit.. ` |

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## PHP

`<?php ` `// PHP implementation of the above approach ` ` ` `// Function to return the required number ` `function` `getMinNum(` `$a` `, ` `$b` `, ` `$c` `) ` `{ ` ` ` ` ` `// If doesn't belong to the range ` ` ` `// then c is the required number ` ` ` `if` `(` `$c` `< ` `$a` `|| ` `$c` `> ` `$b` `) ` ` ` `return` `$c` `; ` ` ` ` ` `// Else get the next multiple of c ` ` ` `// starting from b + 1 ` ` ` `$x` `= (` `floor` `((` `$b` `/ ` `$c` `)) * ` `$c` `) + ` `$c` `; ` ` ` ` ` `return` `$x` `; ` `} ` ` ` `// Driver code ` `$a` `= 2; ` `$b` `= 4; ` `$c` `= 4; ` ` ` `echo` `getMinNum(` `$a` `, ` `$b` `, ` `$c` `); ` ` ` `// This code is contributed by Ryuga ` `?> ` |

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**Output:**

8

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