Find a triplet (X, Y, Z) such that all are divisible by A, exactly one is divisible by both A and B, and X + Y = Z

Given two integers A and B, the task is to find a triplet (X, Y, Z) such that all of them are divisible by A, exactly one of them is divisible by both A and B, and X + Y = Z.

Example:

Input: A = 5, B = 3
Output: 10 50 60
Explanation: For the triplet (10, 50, 60), all of them are divisible by 5, 60 is divisible by both 5 and 3, and 10 + 50 = 60. Therefore, (10, 50, 60) is valid triplet. Other possible triplets are (5, 25, 30), (5, 15, 20)

Input: A = 7, B = 11
Output: 28 154 182

Approach:  The given problem is an observation-based problem that can be solved using basic mathematics. It can be observed that the triplet (A, A * B, A * (B + 1)), satisfies all of the given conditions except when the value of B is 1. In that case, it can be seen that the condition that exactly one of them should be divisible by both A and B will always be violated. So, if B = 1, no valid triplet exists, otherwise print (A, A * B, A * (B + 1)).

Below is the implementation of the above approach:

// C++ program of the above approach
#include <iostream>
using namespace std;

// Function to find a triplet (X, Y, Z)
// such that all of them are divisible
// by A, exactly one of them is divisible
// by both A and B, and X + Y = Z
void findTriplet(int A, int B)
{
    // If the value of B is 1
    if (B == 1) {
        cout << -1;
        return;
    }

    // Print Answer
    cout << A << " " << A * B
         << " " << A * (B + 1);
}

// Driver Code
int main()
{
    int A = 5;
    int B = 3;
    findTriplet(A, B);

    return 0;
}

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to find a triplet (X, Y, Z)
  // such that all of them are divisible
  // by A, exactly one of them is divisible
  // by both A and B, and X + Y = Z
  static void findTriplet(int A, int B)
  {

    // If the value of B is 1
    if (B == 1) {
      System.out.println(-1);
      return;
    }

    // Print Answer
    System.out.println(A + " " + A * B + " " +  A * (B + 1));
  }

  // Driver Code
  public static void main (String[] args) {
    int A = 5;
    int B = 3;
    findTriplet(A, B);
  }
}

// This code is contributed by hrithikgarg03188.

# Python code for the above approach

# Function to find a triplet (X, Y, Z)
# such that all of them are divisible
# by A, exactly one of them is divisible
# by both A and B, and X + Y = Z
def findTriplet(A, B):

    # If the value of B is 1
    if (B == 1):
        print(-1)
        return

    # Print Answer
    print(f"{A} {A * B} {A * (B + 1)}")

# Driver Code
A = 5
B = 3
findTriplet(A, B)

# This code is contributed by Saurabh Jaiswal

// C# program of the above approach
using System;
class GFG
{

  // Function to find a triplet (X, Y, Z)
  // such that all of them are divisible
  // by A, exactly one of them is divisible
  // by both A and B, and X + Y = Z
  static void findTriplet(int A, int B)
  {

    // If the value of B is 1
    if (B == 1) {
      Console.Write(-1);
      return;
    }

    // Print Answer
    Console.Write(A + " " + A * B + " " + A * (B + 1));
  }

  // Driver Code
  public static int Main()
  {
    int A = 5;
    int B = 3;
    findTriplet(A, B);
    return 0;
  }
}

// This code is contributed by Taranpreet

<script>
        // JavaScript code for the above approach 

        // Function to find a triplet (X, Y, Z)
        // such that all of them are divisible
        // by A, exactly one of them is divisible
        // by both A and B, and X + Y = Z
        function findTriplet(A, B)
        {
        
            // If the value of B is 1
            if (B == 1) {
                document.write(-1);
                return;
            }

            // Print Answer
            document.write(A + " " + A * B
                + " " + A * (B + 1));
        }

        // Driver Code
        let A = 5;
        let B = 3;
        findTriplet(A, B);

       // This code is contributed by Potta Lokesh
    </script>

5 15 20

Time Complexity: O(1)
Auxiliary Space: O(1)