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:

On solving above equations simultaneously, we get:

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++

// 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;
}

Java

// 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

Python3

# Python3 program for the above approach

# Function to find two numbers with
# difference and division both as N
def findAandB(N):

    # Condition if the answer
    # is not possible
    if (N == 1):
        print("No")
        return

    # Calculate a and b
    a = N * N / (N - 1)
    b = a / N

    # Print the values
    print("a = ", a)
    print("b = ", b)

# Driver Code

# Given N
N = 6

# Function call
findAandB(N)

# This code is contributed by sanjoy_62

C#

// 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

Javascript

<script>
// Javascript program for the above approach


    // Function to find two numbers with
    // difference and division both as N
    function findAandB( N) {

        // Condition if the answer
        // is not possible
        if (N == 1) {
            document.write("No");
            return;
        }

        // Calculate a and b
        let a = N * N / (N - 1);
        let b = a / N;

        // Print the values
        document.write("a = " + a + "<br/>");
        document.write("b = " + b + "<br/>");
    }

    // Driver Code


        // Given N
        let N = 6;

        // Function call
        findAandB(N);



// This code contributed by aashish1995
</script>

Output:

a = 7.2
b = 1.2

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