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 = LHSInput: 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 Qvoid 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 codeint main(){ int X = 3; values_of_Q(X); return 0;} |
Java
// Java program for the above approachimport java.util.*; class GFG{ // Function to find all possible values of Qstatic 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 codepublic 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 = 3values_of_Q(X) # This code is contributed by divyeshrabadiya07 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find all possible // values of Qstatic 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 codepublic 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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