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 < 10 ^{7}*. Print the total count of such values of

**Q**along with values.

**Examples:**

Input:X = 3Output:

2

4, 6Explanation:

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 = 16Output:

5

17, 18, 20, 24, 32Explanation:

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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