# Minimum value that divides one number and divisible by other

Given two integer p and q, the task is to find the minimum possible number x such that q % x = 0 and x % p = 0. If the conditions aren’t true for any number then print -1.
Examples:

Input: p = 3, q = 99
Output:
99 % 3 = 0
3 % 3 = 0
Input: p = 2, q = 7
Output: -1

Approach: If a number x satisfies the given condition then it’s obvious that q will be divided by p i.e. q % p = 0 because x is a multiple of p and q is a multiple of x
So the minimum possible value of x will be the GCD of p and q and when q is not divisible by p then no number will satisfy the given condition.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the minimum valid number` `// that satisfies the given conditions` `int` `minValidNumber(``int` `p, ``int` `q)` `{` `    ``// If possible` `    ``if` `(q % p == 0)` `        ``return` `__gcd(p, q);` `    ``else` `        ``return` `-1;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `p = 2, q = 6;` `    ``cout << minValidNumber(p, q);` `    ``return` `0;` `}`

## Java

 `//Java  implementation of the approach`   `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Function to calculate gcd` `    ``static` `int` `__gcd(``int` `a, ``int` `b)` `    ``{` `        `  `        ``// Everything divides 0 ` `        ``if` `(a == ``0` `|| b == ``0``)` `            ``return` `0``;` `    `  `        ``// base case` `        ``if` `(a == b)` `            ``return` `a;` `    `  `        ``// a is greater` `        ``if` `(a > b)` `            ``return` `__gcd(a - b, b);` `            `  `        ``return` `__gcd(a, b - a);` `    ``}`   `// Function to return the minimum valid number` `// that satisfies the given conditions` `static` `int` `minValidNumber(``int` `p, ``int` `q)` `{` `    ``// If possible` `    ``if` `(q % p == ``0``)` `        ``return` `__gcd(p, q);` `    ``else` `        ``return` `-``1``;` `}`   `// Driver Code` `    ``public` `static` `void` `main (String[] args) {` `        ``int` `p = ``2``, q = ``6``;` `        ``System.out.print(minValidNumber(p, q));`     `    ``// THis code is contributed by  Sachin.` `    ``}` `}`

## Python3

 `# Python3 implementation of the approach ` `from` `math ``import` `gcd`   `# Function to return the minimum ` `# valid number that satisfies the` `# given conditions ` `def` `minValidNumber(p, q) :`   `    ``# If possible ` `    ``if` `(q ``%` `p ``=``=` `0``) : ` `        ``return` `gcd(p, q) ` `    ``else` `:` `        ``return` `-``1`   `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `:`   `    ``p, q ``=` `2``, ``6``; ` `    ``print``(minValidNumber(p, q))`   `# This code is contributed by Ryuga`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG` `{` `    ``// Function to calculate gcd` `    ``static` `int` `__gcd(``int` `a, ``int` `b)` `    ``{` `         `  `        ``// Everything divides 0 ` `        ``if` `(a == 0 || b == 0)` `            ``return` `0;` `     `  `        ``// base case` `        ``if` `(a == b)` `            ``return` `a;` `     `  `        ``// a is greater` `        ``if` `(a > b)` `            ``return` `__gcd(a - b, b);` `             `  `        ``return` `__gcd(a, b - a);` `    ``}`   `// Function to return the minimum valid number` `// that satisfies the given conditions` `static` `int` `minValidNumber(``int` `p, ``int` `q)` `{` `    ``// If possible` `    ``if` `(q % p == 0)` `        ``return` `__gcd(p, q);` `    ``else` `        ``return` `-1;` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``int` `p = 2, q = 6;` `    ``Console.Write(minValidNumber(p, q));` `}` `}`   `// THis code is contributed` `// by Mukul Singh`

## PHP

 ``

## Javascript

 ``

Output:

`2`

Time Complexity: O(logn)

Auxiliary Space: O(1)

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