Find two numbers with difference and division both same as N

Given an integer N, the task is to find two numbers a and b such that a / b = N and a – b = N. Print “No” if no such numbers are possible.
Examples: 
 

Input: N = 6 
Output: 
a = 7.2
b = 1.2
Explanation:
For the given two numbers a and b, a/b = 6 = N and a-b = 6 = N

Input: N = 1  
Output: No
Explanation:
There are no values of a and b that satisfy the condition.

Approach: To solve the problem observe the equations derived below:
 

\begin{cases} a - b &= N \\ a - Nb& = 0 \end{cases}



On solving above equations simultaneously, we get:
 

A=\dfrac{N^2}{N-1}
B=\dfrac{N}{N-1}

Since the denominator is N – 1, so the answer will not be possible when N = 1. For all other cases, the answer is possible. Hence find the values of a and b respectively.

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 two numbers with
// difference and division both as N
void findAandB(double N)
{
    // Condition if the answer
    // is not possible
  
    if (N == 1) {
        cout << "No";
        return;
    }
  
    // Calculate a and b
    double a = N * N / (N - 1);
    double b = a / N;
  
    // Print the values
    cout << "a = " << a << endl;
    cout << "b = " << b << endl;
}
  
// Driver Code
int main()
{
    // Given N
    double N = 6;
  
    // Function Call
    findAandB(N);
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
  
// Function to find two numbers with
// difference and division both as N
static void findAandB(double N)
{
      
    // Condition if the answer
    // is not possible
    if (N == 1
    {
        System.out.print("No");
        return;
    }
  
    // Calculate a and b
    double a = N * N / (N - 1);
    double b = a / N;
  
    // Print the values
    System.out.print("a = " + a + "\n");
    System.out.print("b = " + b + "\n");
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given N
    double N = 6;
  
    // Function call
    findAandB(N);
}
}
  
// This code is contributed by Rajput-Ji

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C#

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// C# program for the above approach
using System;
  
class GFG{
  
// Function to find two numbers with
// difference and division both as N
static void findAandB(double N)
{
  
    // Condition if the answer
    // is not possible
    if (N == 1) 
    {
        Console.Write("No");
        return;
    }
  
    // Calculate a and b
    double a = N * N / (N - 1);
    double b = a / N;
  
    // Print the values
    Console.Write("a = " + a + "\n");
    Console.Write("b = " + b + "\n");
}
  
// Driver Code
public static void Main(String[] args)
{
  
    // Given N
    double N = 6;
  
    // Function call
    findAandB(N);
}
}
  
// This code is contributed by amal kumar choubey

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Output: 

a = 7.2
b = 1.2

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

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