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

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

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

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 <iostream> 
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

<?php 
// PHP program to find the  
// smallest values of x  
// and y that satisfy  
// "ax - by = 0" 
  
// To find GCD using  
// Eculcid's algorithm 
function 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" 
function findSmallest($a, $b) 
{ 
      
    // Find LCM 
    $lcm = ($a * $b) / gcd($a, $b); 
  
    echo "x = ", $lcm/$a, "\ny = ", $lcm/$b; 
} 
  
    // Driver Code 
    $a = 25; 
    $b = 35; 
    findSmallest($a, $b); 
      
// This code is contributed by ajit 
?> 

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;

// 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes

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

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

C++

Java

Python3

C#

PHP

Recommended Posts: