# Pair of integers with difference K having an element as the K-th multiple of the other

Given an integer K, the task is to find a pair of numbers (A, B) such that A – B = K and A / B = K. If no such pair can be generated, print “No”.
Examples:

Input: K = 6
Output: 7.2 1.2
Explanation:
Since 7.2 – 1.2 = 6 and 7.2 / 1.2 = 6, The pair {7.2, 1.2} satisfy the necessary conditions.

Input: K = 1
Output: No

Approach:
The following observations are needed to be made to solve the problem:

• The given conditions can be written in the form of equations as:
1. Equation (1): A – B = K => A – B – K = 0
2. Equation (2): A / B = K => A – (K * B) = 0
• Solving both these equations, we obtain:

( K * Equation (1) ) – Equation(2) = 0
=> K*A – K*B – K2 – (A – K*B) = 0
=> K*A – K2 – A – K*B + K*B = 0
=> A*(K-1) – K2 = 0
=> A*(K-1) = K2
=> A = K2 / (K -1)

• Replacing the value of A in Equation (1), the value of B can be obtained to be K / (K – 1)
• It can be observed that if the value of K is 1, then no such pair can be found as the denominator of both A and B becomes 0.

Follow the steps below to solve the problem:

• If K is equal to 1, then print “NO”.
• Otherwise, if K is equal to any value other than 0, compute the values (K*K) / (K -1) and K / (K – 1) and print them.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement ` `// the above problem ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the ` `// required pair ` `void` `computePair(``double` `K) ` `{ ` `    ``// No pair possible ` `    ``if` `(K == 1) { ` `        ``cout << ``"No"``; ` `        ``return``; ` `    ``} ` `    ``else` `{ ` ` `  `        ``cout << K * K / (K - 1) << ``" "``; ` `        ``cout << K / (K - 1) << endl; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``double` `K = 6; ` `    ``computePair(K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to implement ` `// the above problem ` `class` `GFG{ ` ` `  `// Function to find the ` `// required pair ` `static` `void` `computePair(``double` `K) ` `{ ` `     `  `    ``// No pair possible ` `    ``if` `(K == ``1``)  ` `    ``{ ` `        ``System.out.print(``"No"``); ` `        ``return``; ` `    ``} ` `    ``else` `    ``{ ` `        ``System.out.print(K * K / (K - ``1``) + ``" "``); ` `        ``System.out.print(K / (K - ``1``) + ``"\n"``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``double` `K = ``6``; ` `     `  `    ``computePair(K); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## C#

 `// C# program to implement ` `// the above problem ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find the ` `// required pair ` `static` `void` `computePair(``double` `K) ` `{ ` `     `  `    ``// No pair possible ` `    ``if` `(K == 1)  ` `    ``{ ` `        ``Console.Write(``"No"``); ` `        ``return``; ` `    ``} ` `    ``else` `    ``{ ` `        ``Console.Write(K * K / (K - 1) + ``" "``); ` `        ``Console.Write(K / (K - 1) + ``"\n"``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``double` `K = 6; ` `     `  `    ``computePair(K); ` `} ` `} ` ` `  `// This code is contributed by gauravrajput1 `

Output:

```7.2 1.2
```

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

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Improved By : Rajput-Ji, GauravRajput1