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 = 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. |
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; } |
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