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Possible values of Q such that, for any value of R, their product is equal to X times their sum
  • Last Updated : 19 Mar, 2021
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Given an integer X, the task is to find the possible values of Q for the pair (Q, R) such that their product is equal to X times their sum, where Q ≤ R and X < 107. Print the total count of such values of Q along with values.

Examples:

Input: X = 3
Output: 
2
4, 6
Explanation: 
On taking Q = 4 and R = 12, 
LHS = 12 x 4 = 48 
RHS = 3(12 + 4) = 3 x 16 = 48 = LHS
Similarly, the equation also holds for value Q = 6 and R = 6. 
LHS = 6 x 6 = 36 
RHS = 3(6 + 6) = 3 x 12 = 36 = LHS

Input:  X = 16
Output: 
5
17, 18, 20, 24, 32 
Explanation: 
If Q = 17 and R = 272, 
LHS = 17 x 272 = 4624 
RHS = 16(17 + 272) = 16(289) = 4624 = LHS. 
Similarly, there exists a value R for all other values of Q given in the output. 

Approach: The idea is to understand the question to form an equation, that is (Q x R) = X(Q + R).



  • The idea is to iterate from 1 to X and check for every if ((( X + i ) * X) % i ) == 0.
  • Initialize a resultant vector, and iterate for all the values of X from 1.
  • Check if the above condition holds true. If it does then push the value X+i in the vector.
  • Let’s break the equation in order to understand it more clearly,

 
 

The given expression is (Q x R) = X(Q + R) 
On simplifying this we get,  

 

=>  QR – QX = RX
or,  QR – RX = QX
or,  R = QX / (Q – X)

 

  • Hence, observe that (X+i) is the possible value of Q and (X+i)*X is the possible value of R.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find all possible values of Q
void values_of_Q(int X)
{
    // Vector initialization
    // to store all numbers
    // satisfying the given condition
    vector<int> val_Q;
 
    // Iterate for all the values of X
    for (int i = 1; i <= X; i++) {
 
        // Check if condition satisfied
        // then push the number
        if ((((X + i) * X)) % i == 0) {
 
            // Possible value of Q
            val_Q.push_back(X + i);
        }
    }
 
    cout << val_Q.size() << endl;
 
    // Print all the numbers
    for (int i = 0; i < val_Q.size(); i++) {
        cout << val_Q[i] << " ";
    }
}
 
// Driver code
int main()
{
    int X = 3;
 
    values_of_Q(X);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG{
   
// Function to find all possible values of Q
static void values_of_Q(int X)
{
    // Vector initialization
    // to store all numbers
    // satisfying the given condition
    ArrayList<Integer> val_Q = new ArrayList<Integer>();
  
    // Iterate for all the values of X
    for (int i = 1; i <= X; i++)
    {
  
        // Check if condition satisfied
        // then push the number
        if ((((X + i) * X)) % i == 0)
        {
  
            // Possible value of Q
            val_Q.add(X + i);
        }
    }
  
    System.out.println(val_Q.size());
  
    // Print all the numbers
    for (int i = 0; i < val_Q.size(); i++)
    {
        System.out.print(val_Q.get(i)+" ");
    }
}
  
// Driver code
public static void main(String[] args)
{
    int X = 3;
  
    values_of_Q(X);
}
}
 
// This code is contributed by Ritik Bansal

Python3




# Python3 program for the above approach
 
# Function to find all possible values of Q
def values_of_Q(X):
 
    # Vector initialization
    # to store all numbers
    # satisfying the given condition
    val_Q = []
 
    # Iterate for all the values of X
    for i in range(1, X + 1):
 
        # Check if condition satisfied
        # then push the number
        if ((((X + i) * X)) % i == 0):
 
            # Possible value of Q
            val_Q.append(X + i)
 
    print(len(val_Q))
 
    # Print all the numbers
    for i in range(len(val_Q)):
        print(val_Q[i], end = " ")
 
# Driver Code       
X = 3
 
values_of_Q(X)
 
# This code is contributed by divyeshrabadiya07

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find all possible
// values of Q
static void values_of_Q(int X)
{
     
    // List initialization
    // to store all numbers
    // satisfying the given condition
    List<int> val_Q = new List<int>();
 
    // Iterate for all the values of X
    for(int i = 1; i <= X; i++)
    {
         
        // Check if condition satisfied
        // then push the number
        if ((((X + i) * X)) % i == 0)
        {
             
            // Possible value of Q
            val_Q.Add(X + i);
        }
    }
     
    Console.WriteLine(val_Q.Count);
 
    // Print all the numbers
    for(int i = 0; i < val_Q.Count; i++)
    {
        Console.Write(val_Q[i] + " ");
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int X = 3;
 
    values_of_Q(X);
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
    // Javascript program for the above approach
     
    // Function to find all possible values of Q
    function values_of_Q(X)
    {
     
        // Vector initialization
        // to store all numbers
        // satisfying the given condition
        let val_Q = [];
 
        // Iterate for all the values of X
        for (let i = 1; i <= X; i++)
        {
 
            // Check if condition satisfied
            // then push the number
            if ((((X + i) * X)) % i == 0)
            {
 
                // Possible value of Q
                val_Q.push(X + i);
            }
        }
 
        document.write(val_Q.length + "</br>");
 
        // Print all the numbers
        for (let i = 0; i < val_Q.length; i++) {
            document.write(val_Q[i] + " ");
        }
    }
     
    let X = 3;
    values_of_Q(X);
     
    // This code is contributed by divyesh072019.
</script>
Output: 
2
4 6

 

Time Complexity: O(N)

Auxiliary Space: O(X)

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