Possible values of Q such that, for any value of R, their product is equal to X times their sum

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 

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Output: 

2
4 6

Time Complexity: O(N)

Auxiliary Space: O(X)

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