# Find two distinct numbers such that their LCM lies in given range

Given two numbers L and R, the task is to find two distinct minimum positive integers X and Y such that whose LCM lies in the range [L, R]. If there doesn’t exist any value of X and Y then print “-1”.

Examples:

Input: L = 3, R = 8
Output: x = 3, y=6
Explanation:
LCM of 3 and 6 is 6 which is in range 3, 8

Input: L = 88, R = 90
Output: -1
Explanation:
Minimum possible x and y are 88 and 176 respectively, but 176 is greater than 90.

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

Approach: The idea is to choose the value of X and Y in such a way that their LCM lies in the given range [L, R]. Below are the steps:

1. For the minimum value of X choose L as the minimum value as this is the minimum value in the given range.
2. Now for the value of Y choose 2*L as this is the minimum value of Y whose LCM is L.
3. Now if the above two values of X and Y lie in the range [L, R], then this is required pair of integers with minimum possible values of X and Y.
4. Otherwise, print “-1” as there doesn’t exist any other pair.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find two distinct numbers ` `// X and Y s.t. their LCM lies between ` `// L and R  and X, Y are minimum possible ` `void` `answer(``int` `L, ``int` `R) ` `{ ` ` `  `    ``// Check if 2*L lies in range L, R ` `    ``if` `(2 * L <= R) ` ` `  `        ``// Print the answer ` `        ``cout << L << ``", "` `             ``<< 2 * L << ``"\n"``; ` `    ``else` `        ``cout << -1; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given value of ranges ` `    ``int` `L = 3, R = 8; ` ` `  `    ``// Function call ` `    ``answer(L, R); ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `import` `java.io.*; ` ` `  `class` `GFG{ ` ` `  `// Function to find two distinct numbers ` `// X and Y s.t. their LCM lies between ` `// L and R and X, Y are minimum possible ` `static` `void` `answer(``int` `L, ``int` `R) ` `{ ` ` `  `    ``// Check if 2*L lies in range L, R ` `    ``if` `(``2` `* L <= R) ` ` `  `        ``// Print the answer ` `        ``System.out.println(L + ``", "` `+ (``2` `* L)); ` `     `  `    ``else` `        ``System.out.println(``"-1"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `     `  `    ``// Given value of ranges ` `    ``int` `L = ``3``, R = ``8``; ` ` `  `    ``// Function call ` `    ``answer(L, R); ` `} ` `}  ` ` `  `// This code is contributed by sanjoy_62 `

## Python3

 `# Python3 program for the above approach ` ` `  `# Function to find two distinct numbers ` `# X and Y s.t. their LCM lies between ` `# L and R and X, Y are minimum possible ` `def` `answer(L, R): ` ` `  `    ``# Check if 2*L lies in range L, R ` `    ``if` `(``2` `*` `L <``=` `R): ` ` `  `        ``# Print the answer ` `        ``print``(L, ``","``, ``2` `*` `L) ` ` `  `    ``else``: ` `        ``print``(``-``1``) ` ` `  `# Driver Code ` ` `  `# Given value of ranges ` `L ``=` `3` `R ``=` `8` ` `  `# Function call ` `answer(L, R) ` ` `  `# This code is contributed by sanjoy_62 `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find two distinct numbers ` `// X and Y s.t. their LCM lies between ` `// L and R and X, Y are minimum possible ` `static` `void` `answer(``int` `L, ``int` `R) ` `{ ` ` `  `    ``// Check if 2*L lies in range L, R ` `    ``if` `(2 * L <= R) ` `     `  `        ``// Print the answer ` `        ``Console.WriteLine(L + ``", "` `+ (2 * L)); ` `     `  `    ``else` `        ``Console.WriteLine(``"-1"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `     `  `    ``// Given value of ranges ` `    ``int` `L = 3, R = 8; ` ` `  `    ``// Function call ` `    ``answer(L, R); ` `} ` `} ` ` `  `// This code is contributed by sanjoy_62 `

Output:

```3, 6
```

Time Complexity: O(1)
Auxiliary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : sanjoy_62