# Minimum positive integer value possible of X for given A and B in X = P*A + Q*B

Given values of A and B, find the minimum positive integer value of X that can be achieved in the equation X = P*A + P*B, Here P and Q can be zero or any positive or negative integer.

Examples:

```Input : A = 3
B = 2
Output : 1

Input  : A = 2
B = 4
Output : 2
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Basically we need to find P and Q such that P*A > P*B and P*A – P*B is minimum positive integer. This problem can be easily solved by calculating GCD of both numbers.
For example:

```For A = 2
And B = 4
Let P  = 1
And Q = 0
X = P*A + Q*B
= 1*2 + 0*4
= 2 + 0
= 2 (i. e GCD of 2 and 4)

For A = 3
and B = 2
let P = -1
And Q = 2
X = P*A + Q*B
= -1*3 + 2*2
= -3 + 4
= 1 ( i.e GCD of 2 and 3 )
```

Below is the implementation of above idea .

## CPP

 `// CPP Program to find ` `// minimum value of X ` `// in equation X = P*A + Q*B ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Utility function to calculate GCD ` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(a == 0) ` `        ``return` `b; ` `    ``return` `(b % a, a); ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `a = 2; ` `    ``int` `b = 4; ` `    ``cout << gcd(a, b); ` `    ``return` `0; ` `} `

## Java

 `// Java Program to find ` `// minimum value of X ` `// in equation X = P*A + Q*B ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG { ` `    ``// utility function to calculate gcd ` ` `  `    ``public` `static` `int` `gcd(``int` `a, ``int` `b) ` `    ``{ ` `        ``if` `(a == ``0``) ` `            ``return` `b; ` ` `  `        ``return` `gcd(b % a, a); ` `    ``} ` ` `  `    ``// Driver Program ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `a = ``2``; ` `        ``int` `b = ``4``; ` ` `  `        ``System.out.println(gcd(a, b)); ` `    ``} ` `} `

## Python3

 `# Python3 Program to find  ` `# minimum value of X ` `# in equation X = P * A + Q * B  ` ` `  `# Function to return gcd of a and b ` `def` `gcd(a, b):  ` `    ``if` `a ``=``=` `0` `: ` `        ``return` `b       ` `    ``return` `gcd(b ``%` `a, a) ` `  `  `a ``=` `2` `b ``=` `4` `print``(gcd(a, b)) `

## C#

 `// CSHARP Program to find ` `// minimum value of X ` `// in equation X = P*A + Q*B ` `using` `System; ` ` `  `class` `GFG { ` `    ``// function to get gcd of a and b ` `    ``public` `static` `int` `gcd(``int` `a, ``int` `b) ` `    ``{ ` `        ``if` `(a == 0) ` `            ``return` `b; ` ` `  `        ``return` `gcd(b % a, a); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``static` `public` `void` `Main() ` `    ``{ ` `        ``int` `a = 2; ` `        ``int` `b = 4; ` `        ``Console.WriteLine(gcd(a, b)); ` `    ``} ` `} `

## PHP

 `// PHP Program to find  ` `// minimum value of X ` `// in equation X = P*A + Q*B  ` ` `  ` `

Output:

```2
```

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