Related Articles

# Find smallest values of x and y such that ax – by = 0

• Difficulty Level : Easy
• Last Updated : 16 Apr, 2021

Given two values ‘a’ and ‘b’ that represent coefficients in “ax – by = 0”, find the smallest values of x and y that satisfy the equation. It may also be assumed that x > 0, y > 0, a > 0 and b > 0.

```Input: a = 25, b = 35
Output: x = 7, y = 5```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

A Simple Solution is to try every possible value of x and y starting from 1, 1 and stop when the equation is satisfied.
A Direct Solution is to use Least Common Multiple (LCM). LCM of ‘a’ and ‘b’ represents the smallest value that can make both sides equal. We can find LCM using below formula.

`   LCM(a, b) = (a * b) / GCD(a, b) `

Greatest Common Divisor (GCD) can be computed using Euclid’s algorithm.

## C++

 `// C++ program to find the smallest values of x and y that``// satisfy "ax - by = 0"``#include ``using` `namespace` `std;` `// To find GCD using Eculcid's algorithm``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(b == 0)``        ``return` `a;``    ``return` `(gcd(b, a % b));``}` `// Prints smallest values of x and y that``// satisfy "ax - by = 0"``void` `findSmallest(``int` `a, ``int` `b)``{``    ``// Find LCM``    ``int` `lcm = (a * b) / gcd(a, b);` `    ``cout << ``"x = "` `<< lcm / a``         ``<< ``"\ny = "` `<< lcm / b;``}` `// Driver program``int` `main()``{``    ``int` `a = 25, b = 35;``    ``findSmallest(a, b);``    ``return` `0;``}`

## Java

 `// Java program to find the smallest values of``// x and y that satisfy "ax - by = 0"``class` `GFG {` `    ``// To find GCD using Eculcid's algorithm``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{` `        ``if` `(b == ``0``)``            ``return` `a;``        ``return` `(gcd(b, a % b));``    ``}` `    ``// Prints smallest values of x and y that``    ``// satisfy "ax - by = 0"``    ``static` `void` `findSmallest(``int` `a, ``int` `b)``    ``{` `        ``// Find LCM``        ``int` `lcm = (a * b) / gcd(a, b);` `        ``System.out.print(``"x = "` `+ lcm / a``                         ``+ ``"\ny = "` `+ lcm / b);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a = ``25``, b = ``35``;``        ``findSmallest(a, b);``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python program to find the``# smallest values of x and y that``# satisfy "ax - by = 0"` `# To find GCD using Eculcid's algorithm``def` `gcd(a, b):``    ``if` `(b ``=``=` `0``):``        ``return` `a``    ``return``(gcd(b, a ``%` `b))` `# Prints smallest values of x and y that``# satisfy "ax - by = 0"``def` `findSmallest(a, b):` `    ``# Find LCM``    ``lcm ``=` `(a ``*` `b)``/``gcd(a, b)``    ``print``(``"x ="``, lcm ``/` `a, ``"\ny = "``, lcm ``/` `b)` `# Driver code``a ``=` `25``b ``=` `35``findSmallest(a, b)` `# This code is contributed``# by Anant Agarwal.`

## C#

 `// C# program to find the smallest``// values of x and y that``// satisfy "ax - by = 0"``using` `System;` `class` `GFG {` `    ``// To find GCD using``    ``// Eculcid's algorithm``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{` `        ``if` `(b == 0)``            ``return` `a;``        ``return` `(gcd(b, a % b));``    ``}` `    ``// Prints smallest values of x and``    ``// y that satisfy "ax - by = 0"``    ``static` `void` `findSmallest(``int` `a, ``int` `b)``    ``{` `        ``// Find LCM``        ``int` `lcm = (a * b) / gcd(a, b);` `        ``Console.Write(``"x = "` `+ lcm / a + ``"\ny = "` `+ lcm / b);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `a = 25, b = 35;``        ``findSmallest(a, b);``    ``}``}` `// This code is contributed by Sam007.`

## PHP

 ``

## Javascript

 ``

Output:

```x = 7
y = 5```

The above code for findSmallest() can be reduced:

```Since ax - by = 0,
ax = by, which means x/y = b/a
So we can calculate gcd and directly do as -

Value of x = b / gcd;
Value of y = a / gcd; ```

## C

 `// Prints smallest values of x and y that``// satisfy "ax - by = 0"``void` `findSmallest(``int` `a, ``int` `b)``{``    ``// Find GCD``    ``int` `g = gcd(a, b);` `    ``cout << ``"x = "` `<< b / g``         ``<< ``"\ny = "` `<< a / g;``}`

This article is contributed by Aakash Sachdeva. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.