# Represent (2 / N) as the sum of three distinct positive integers of the form (1 / m)

Given a positive integer N, the task is to represent the fraction 2 / N as the sum of three distinct positive integers of the form 1 / m i.e. (2 / N) = (1 / x) + (1 / y) + (1 / z) and print x, y and z.

Examples:

Input: N = 3
Output: 3 4 12
(1 / 3) + (1 / 4) + (1 / 12) = ((4 + 3 + 1) / 12)
= (8 / 12) = (2 / 3) i.e. 2 / N

Input: N = 28
Output: 28 29 812

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: It can be easily inferred that for N = 1, there will be no solution. For N > 1, (2 / N) can be represented as (1 / N) + (1 / N) and the problem gets reduced to representing it as a sum of two fractions. Now, find the difference between (1 / N) and 1 / (N + 1) and get the fraction 1 / (N * (N + 1)). Therefore, the solution is (2 / N) = (1 / N) + (1 / (N + 1)) + (1 / (N * (N + 1))) where x = N, y = N + 1 and z = N * (N + 1).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the required fractions ` `void` `find_numbers(``int` `N) ` `{ ` `    ``// Base condition ` `    ``if` `(N == 1) { ` `        ``cout << -1; ` `    ``} ` ` `  `    ``// For N > 1 ` `    ``else` `{ ` `        ``cout << N << ``" "` `<< N + 1 << ``" "` `             ``<< N * (N + 1); ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 5; ` ` `  `    ``find_numbers(N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Function to find the required fractions ` `static` `void` `find_numbers(``int` `N) ` `{ ` `    ``// Base condition ` `    ``if` `(N == ``1``)  ` `    ``{ ` `        ``System.out.print(-``1``); ` `    ``} ` ` `  `    ``// For N > 1 ` `    ``else`  `    ``{ ` `        ``System.out.print(N + ``" "` `+ (N + ``1``) +  ` `                             ``" "` `+ (N * (N + ``1``))); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args)  ` `{ ` `    ``int` `N = ``5``; ` ` `  `    ``find_numbers(N); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to find the required fractions  ` `def` `find_numbers(N) : ` ` `  `    ``# Base condition  ` `    ``if` `(N ``=``=` `1``) : ` `        ``print``(``-``1``, end ``=` `"");  ` ` `  `    ``# For N > 1  ` `    ``else` `: ` `        ``print``(N, N ``+` `1` `, N ``*` `(N ``+` `1``));  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``N ``=` `5``;  ` ` `  `    ``find_numbers(N); ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System;  ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to find the required fractions ` `static` `void` `find_numbers(``int` `N) ` `{ ` `    ``// Base condition ` `    ``if` `(N == 1)  ` `    ``{ ` `        ``Console.Write(-1); ` `    ``} ` ` `  `    ``// For N > 1 ` `    ``else` `    ``{ ` `        ``Console.Write(N + ``" "` `+ (N + 1) +  ` `                          ``" "` `+ (N * (N + 1))); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args)  ` `{ ` `    ``int` `N = 5; ` ` `  `    ``find_numbers(N); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```5 6 30
```

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