# 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 = 2Output:6Input:A = 5, B = 10, C = 4Output: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;` `}` |

## 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` |

## 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` |

## 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..` |

## 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` `?>` |

## Javascript

`<script>` `// Javascript implementation of the 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` ` ` `let x = (parseInt(b / c) * c) + c;` ` ` `return` `x;` `}` `// Driver code` ` ` `let a = 2, b = 4, c = 4;` ` ` `document.write(getMinNum(a, b, c));` `// This code is contributed by souravmahato348` `</script>` |

**Output:**

8

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