Find two numbers whose sum and GCD are given Last Updated : 29 Jun, 2022 Summarize Comments Improve Suggest changes Share Like Article Like Report Given sum and gcd of two numbers a and b . The task is to find both the numbers a and b. If the numbers do not exist then print -1 .Examples: Input: sum = 6, gcd = 2 Output: a = 4, b = 2 4 + 2 = 6 and GCD(4, 2) = 2Input: sum = 7, gcd = 2 Output: -1 There are no such numbers whose sum is 7 and GCD is 2 Approach: As GCD is given then it is known that both the numbers will be multiples of it. Choose the first number as gcd then the other number will be sum - gcd.If the sum of both the numbers chosen in the previous step equals to sum then print both the numbers.Else the numbers do not exist and print -1 instead. Below is the implementation of the above approach: C++ // C++ program to find two numbers // whose sum and GCD is given #include <bits/stdc++.h> using namespace std; // Function to find two numbers // whose sum and gcd is given void findTwoNumbers(int sum, int gcd) { // sum != gcd checks that both the // numbers are positive or not if (__gcd(gcd, sum - gcd) == gcd && sum != gcd) cout << "a = " << min(gcd, sum - gcd) << ", b = " << sum - min(gcd, sum - gcd) << endl; else cout << -1 << endl; } // Driver code int main() { int sum = 8; int gcd = 2; findTwoNumbers(sum, gcd); return 0; } Java // Java program to find two numbers // whose sum and GCD is given import java.util.*; class Solution{ //function to find gcd of two numbers static int __gcd(int a,int b) { if (b==0) return a; return __gcd(b,a%b); } // Function to find two numbers // whose sum and gcd is given static void findTwoNumbers(int sum, int gcd) { // sum != gcd checks that both the // numbers are positive or not if (__gcd(gcd, sum - gcd) == gcd && sum != gcd) System.out.println( "a = " + Math.min(gcd, sum - gcd) + ", b = " + (int)(sum - Math.min(gcd, sum - gcd)) ); else System.out.println( -1 ); } // Driver code public static void main(String args[]) { int sum = 8; int gcd = 2; findTwoNumbers(sum, gcd); } } //contributed by Arnab Kundu Python3 # Python 3 program to find two numbers # whose sum and GCD is given from math import gcd as __gcd # Function to find two numbers # whose sum and gcd is given def findTwoNumbers(sum, gcd): # sum != gcd checks that both the # numbers are positive or not if (__gcd(gcd, sum - gcd) == gcd and sum != gcd): print("a =", min(gcd, sum - gcd), ", b =", sum - min(gcd, sum - gcd)) else: print(-1) # Driver code if __name__ == '__main__': sum = 8 gcd = 2 findTwoNumbers(sum, gcd) # This code is contributed by # Surendra_Gangwar C# // C# program to find two numbers // whose sum and GCD is given using System; class GFG { // function to find gcd of two numbers static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); } // Function to find two numbers // whose sum and gcd is given static void findTwoNumbers(int sum, int gcd) { // sum != gcd checks that both the // numbers are positive or not if (__gcd(gcd, sum - gcd) == gcd && sum != gcd) Console.WriteLine("a = " + Math.Min(gcd, sum - gcd) + ", b = " + (int)(sum - Math.Min(gcd, sum - gcd))); else Console.WriteLine( -1 ); } // Driver code public static void Main() { int sum = 8; int gcd = 2; findTwoNumbers(sum, gcd); } } // This code is contributed by anuj_67.. PHP <?php // PHP program to find two numbers // whose sum and GCD is given // Function to find gcd of two numbers function __gcd($a, $b) { if ($b == 0) return $a; return __gcd($b, $a % $b); } // Function to find two numbers // whose sum and gcd is given function findTwoNumbers($sum, $gcd) { // sum != gcd checks that both the // numbers are positive or not if (__gcd($gcd, $sum - $gcd) == $gcd && $sum != $gcd) echo "a = " , min($gcd, $sum - $gcd), " b = ", (int)($sum - min($gcd, $sum - $gcd)); else echo (-1); } // Driver code $sum = 8; $gcd = 2; findTwoNumbers($sum, $gcd); // This code is contributed by Sachin ?> JavaScript <script> // Javascript program to find two numbers // whose sum and GCD is given //function to find gcd of two numbers function __gcd(a, b) { if (b==0) return a; return __gcd(b,a%b); } // Function to find two numbers // whose sum and gcd is given function findTwoNumbers(sum, gcd) { // sum != gcd checks that both the // numbers are positive or not if (__gcd(gcd, sum - gcd) == gcd && sum != gcd) document.write( "a = " + Math.min(gcd, sum - gcd) + ", b = " + (sum - Math.min(gcd, sum - gcd)) ); else document.write( -1 ); } // Driver code let sum = 8; let gcd = 2; findTwoNumbers(sum, gcd); </script> Output: a = 2, b = 6 Time Complexity: O(log(min(a, b))), where a and b are two parameters of gcd. 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