Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

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

 

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
?>
Javascript



<script>
 
    // Javascript 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
        let lcm = parseInt((a * b) / gcd(a, b), 10);
   
        document.write("x = " + parseInt(lcm / a, 10) +
        "</br>y = " + parseInt(lcm / b, 10));
    }
     
    let a = 25, b = 35;
      findSmallest(a, b);
     
</script>
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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

My Personal Notes
arrow_drop_up
Previous
Top 5 Topics for Each Section of GATE CS Syllabus
Next

Redirecting System.out.println() output to a file in Java
Recommended Articles
Page :
Find range of values for S in given Array with values satisfying [ arr[i] = floor((i*S)/K) ]
01, Sep 21
Find smallest positive number Y such that Bitwise AND of X and Y is Zero
23, Jul 19
Find N values of X1, X2, ... Xn such that X1 < X2 < ... < XN and sin(X1) < sin(X2) < ... < sin(XN)
05, May 20
Find smallest number K such that K % p = 0 and q % K = 0
28, Feb 19
Find smallest value of K such that bitwise AND of numbers in range [N, N-K] is 0
28, Jun 21
All possible values of floor(N/K) for all values of K
17, Jun 20
Find all possible values of K such that the sum of first N numbers starting from K is G
22, Jul 21
Find the smallest number X such that X! contains at least Y trailing zeros.
14, Mar 19
Find Kth smallest value for b such that a + b = a | b
17, Jan 20
Find smallest perfect square number A such that N + A is also a perfect square number
02, May 20
Find the smallest value of N such that sum of first N natural numbers is ≥ X
29, Jan 21
Find K'th smallest number such that A + B = A | B
02, May 20
Find smallest number n such that n XOR n+1 equals to given k.
11, Sep 17
Modify given string such that odd and even indices is lexicographically largest and smallest
10, Sep 21
Count possible values of K such that X - 1 and Y - 1 modulo K is same
08, Oct 20
Maximize sum of profits for N items such that profit for ith item is product of its weight and count of distinct chosen values
20, Sep 21
Split the number into N parts such that difference between the smallest and the largest part is minimum
02, Nov 18
Smallest element with K set bits such that sum of Bitwise AND of each array element with K is maximum
12, Apr 21
Partition N into M parts such that difference between Max and Min part is smallest
28, Nov 18
Number of values of b such that a = b + (a^b)
16, Oct 18
Count of all possible values of X such that A % X = B
21, May 19
Count of all values of N in [L, R] such that count of primes upto N is also prime
03, Jul 20
Possible values of Q such that, for any value of R, their product is equal to X times their sum
09, Aug 20
Count all possible values of K less than Y such that GCD(X, Y) = GCD(X+K, Y)
13, Jan 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
GeeksforGeeks
Vote for difficulty
Current difficulty :
Easy





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!