Find an integer such that if it is multiplied by any of the given integers they form G.P.

Given three integers A, B, and C, the task is to find out a positive integer K (if exists) such that if k is multiplied by any of the given integers A, B, or C they form G.P.(Geometric Progression) in the given order. If there exists no K print -1.

Examples:

Input: A = 1, B = 9, C = 27
Output: 3
Explanation: If A is multiplied by 3 then the integers 3, 9, and 27 forms a G.P. with a common ratio of 3.

Input: A = 3, B = 2, C = 12
Output:  3
Explanation: If B is multiplied by 3 then the integers 3, 6, and 12 form a G.P. with a common ratio of 2.

Input: A = 2, B = 6, C = 8
Output: -1
Explanation: There exists no positive integer K

Approach: Implement the idea below to solve the problem:

Three numbers A, B, and C form a G.P. if and only if C / B = B / A = r. ( where r is the common ratio ). For three integers we have 3 cases: 

• Case 1:  A is multiplied by K, Now the integers become A*K, B, and C.  Using the G.P. property we have C / B = B / A*K   =>  K = ( B*B ) / ( A*C )
• Case 2: B is multiplied by K, Now the integers become A, B*K, and C. Using the G.P. property we have C / B*k = B*K / A  => K = sqrt(A*C)/ B 
• Case 3: C is multiplied by K, Now the integers become A, B, and C*K.  Using the G.P. property we have C / B = B / A*K   =>  K = ( B*B ) / ( A*C )

If any of the above 3 three cases yield a positive integer K then the answer(s) is/are found else return -1.

Follow these steps to solve the above problem:

• Check the cases mentioned above.
• If any of the cases return true, Print respectively K value for that case.
• Otherwise, return -1.

Below is the implementation of the above approach:

C++

// C++ code for the given approach
#include <cmath>
#include <iostream>
using namespace std;

// Utility function to check if given a number is a
// positive integer
bool ifposInt(double num)
{
  int a = (int)num;
  if (num - a > 0) {
    return false;
  }
  else {
    return true;
  }
}

// Function to find any integer M if exists
void findk(int A, int B, int C)
{
  double k1 = (double)((B * B) / (A * C));
  double k2 = (double)(sqrt(C * A) / B);
  double k3 = (double)((B * B) / (A * C));
  // Checks if possible values k are positive
  // integers or not
  if (k1 > 1 && ifposInt(k1)) {
    cout << (int)k1 << endl;
  }
  else if (k2 > 1 && ifposInt(k2)) {
    cout << (int)k2 << endl;
  }
  else if (k3 > 1 && ifposInt(k3)) {
    cout << (int)k3 << endl;
  }
  else {
    cout << "-1" << endl;
  }
}

int main()
{

  int A = 3;
  int B = 2;
  int C = 12;

  // Function call
  findk(A, B, C);
  return 0;
}

// This code is contributed by lokeshmvs21.

Java

// Java code for the given approach
import java.io.*;

class GFG {

  // Utility function to check if given a number is a
  // positive integer
  static boolean ifposInt(double num)
  {
    int a = (int)num;
    if (num - a > 0) {
      return false;
    }
    else {
      return true;
    }
  }

  // Function to find any integer M if exists
  static void findk(int A, int B, int C)
  {
    double k1 = (double)((B * B) / (A * C));
    double k2 = (double)(Math.sqrt(C * A) / B);
    double k3 = (double)((B * B) / (A * C));

    // Checks if possible values k are positive
    // integers or not
    if (k1 > 1 && ifposInt(k1)) {
      System.out.println((int)k1);
    }
    else if (k2 > 1 && ifposInt(k2)) {
      System.out.println((int)k2);
    }
    else if (k3 > 1 && ifposInt(k3)) {
      System.out.println((int)k3);
    }
    else {
      System.out.println("-1");
    }
  }

  public static void main(String[] args)
  {
    int A = 3;
    int B = 2;
    int C = 12;

    // Function call
    findk(A, B, C);
  }
}

// This code is contributed by lokesh.

Python3

# Python3 code for the given approach
import math

# Utility function to check if given a
# number is a positive integer
def ifposInt(num):
    a = int(num)
    if num - a > 0:
        return False
    else:
        return True

# Function to find any integer
# M if exists
def findk(A, B, C):
    k1 = (B * B) / (A * C)
    k2 = math.sqrt(C * A) / B
    k3 = (B * B) / (A * C)

    # Checks if possible values k are
    # positive integers or not
    if k1 > 1 and ifposInt(k1):
        k1 = int(k1)
        print(k1)
    elif k2 > 1 and ifposInt(k2):
        k2 = int(k2)
        print(k2)
    elif k3 > 1 and ifposInt(k3):
        k3 = int(k3)
        print(k3)
    else:
        print(-1)

# Driver code
A = 3
B = 2
C = 12

#Function call
findk(A, B, C)

#This code is contributed by nikhilsainiofficial546

C#

// C# code for the given approach
using System;

class MainClass {

    // Utility function to check if given a
    // number is a positive integer
    static bool ifposInt(double num)
    {
        int a = (int)num;

        if (num - a > 0)
            return false;
        else
            return true;
    }

    // Function to find any integer
    // M if exists
    static void findk(int A, int B, int C)
    {
        double k1 = (double)((B * B) / (A * C));
        double k2 = (double)(Math.Sqrt(C * A) / B);
        double k3 = (double)((B * B) / (A * C));

        // Checks if possible values k are
        // positive integers or not
        if (k1 > 1 && ifposInt(k1))
            Console.WriteLine(k1);

        else if (k2 > 1 && ifposInt(k2))
            Console.WriteLine(k2);

        else if (k3 > 1 && ifposInt(k3))
            Console.WriteLine(k3);

        else
            Console.WriteLine("-1");
    }

    //Drive Code
    public static void Main(string[] args)
    {
        int A = 3;
        int B = 2;
        int C = 12;

        // Function call
        findk(A, B, C);
    }
}

// This code is contributed by nikhilsainiofficial546

Javascript

// Javascript code for the given approach

// Utility function to check if given a number is a
// positive integer
function ifposInt(num)
{
  let a = parseInt(num);
  if (num - a > 0) {
    return false;
  }
  else {
    return true;
  }
}

// Function to find any integer M if exists
function findk( A,  B,  C)
{
  let k1 = ((B * B) / (A * C));
  let k2 = (Math.sqrt(C * A) / B);
  let k3 = ((B * B) / (A * C));

  // Checks if possible values k are positive
  // integers or not
  if (k1 > 1 && ifposInt(k1)) {
    console.log(parseInt(k1));
  }
  else if (k2 > 1 && ifposInt(k2)) {
    console.log(parseInt(k2));
  }
  else if (k3 > 1 && ifposInt(k3)) {
    console.log(parseInt(k3));
  }
  else {
    console.log("-1");
  }
}

  let A = 3;
  let B = 2;
  let C = 12;

  // Function call
  findk(A, B, C);

 // This code is contributed by poojaagarwal2.

Output

3

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