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

Recommended: Please solve it on “PRACTICE” 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 using namespace std; // To find GCD using Eculcid's algorithmint 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 programint 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 algorithmdef 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 codea = 25b = 35findSmallest(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.



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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.