# Find any pair with given GCD and LCM

Given gcd G and lcm L. The task is to print any pair which has gcd G and lcm L.

Examples:

```Input: G = 3, L = 12
Output: 3, 12

Input: G = 1, L = 10
Output: 1, 10
```

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

A normal solution will be to perform iteration over all the factor pairs of g*l and check if any pair has gcd g and lcm as l. If they have, then the pair will be the answer.

Below is the implementation of the above approach.

## C++

 `// C++ program to print any pair ` `// with a given gcd G and lcm L ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print the pairs ` `void` `printPair(``int` `g, ``int` `l) ` `{ ` `    ``int` `n = g * l; ` ` `  `    ``// iterate over all factor pairs ` `    ``for` `(``int` `i = 1; i * i <= n; i++) { ` ` `  `        ``// check if a factor ` `        ``if` `(n % i == 0) { ` `            ``int` `first = i; ` `            ``int` `second = n / i; ` ` `  `            ``// find gcd ` `            ``int` `gcd = __gcd(first, second); ` ` `  `            ``// check if gcd is same as given g ` `            ``// and lcm is same as lcm l ` `            ``if` `(gcd == g && l % first == 0 && l % second == 0) { ` `                ``cout << first << ``" "` `<< second; ` `                ``return``; ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `g = 3, l = 12; ` `    ``printPair(g, l); ` `    ``return` `0; ` `} `

## Java

 `// Java program to print any pair  ` `// with a given gcd G and lcm L  ` ` `  `import` `java.math.BigInteger; ` ` `  `class` `GFG { ` ` `  `// Function to print the pairs  ` `    ``static` `void` `printPair(``int` `g, ``int` `l) { ` `        ``int` `n = g * l; ` ` `  `        ``// iterate over all factor pairs  ` `        ``for` `(``int` `i = ``1``; i * i <= n; i++) { ` ` `  `            ``// check if a factor  ` `            ``if` `(n % i == ``0``) { ` `                ``int` `first = i; ` `                ``int` `second = n / i; ` ` `  `                ``// find gcd  ` `                ``int` `gcd = __gcd(first, second); ` ` `  `                ``// check if gcd is same as given g  ` `                ``// and lcm is same as lcm l  ` `                ``if` `(gcd == g && l % first == ``0` `&& l % second == ``0``) { ` `                    ``System.out.println(first + ``" "` `+ second); ` `                    ``return``; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `//Function return GCD of two give number ` ` `  `    ``private` `static` `int` `__gcd(``int` `a, ``int` `b) { ` `        ``// there's a better way to do this. I forget. ` `        ``BigInteger b1 = ``new` `BigInteger(``""` `+ a); ` `        ``BigInteger b2 = ``new` `BigInteger(``""` `+ b); ` `        ``BigInteger gcd = b1.gcd(b2); ` `        ``return` `gcd.intValue(); ` `    ``} ` `// Driver function  ` ` `  `    ``public` `static` `void` `main(String[] args) { ` `        ``int` `g = ``3``, l = ``12``; ` `        ``printPair(g, l); ` ` `  `    ``} ` `} ` `// This code is contributed by RAJPUT-JI `

## Python3

 `# Python program to print any pair  ` `# with a given gcd G and lcm L  ` `  `  `# Function to print the pairs  ` `def` `printPair(g, l):  ` `    ``n ``=` `g ``*` `l;  ` `  `  `    ``# iterate over all factor pairs  ` `    ``for` `i ``in` `range``(``1``,n``+``1``):  ` `  `  `        ``# check if a factor  ` `        ``if` `(n ``%` `i ``=``=` `0``): ` `            ``first ``=` `i;  ` `            ``second ``=` `n ``/``/` `i;  ` `  `  `            ``# find gcd  ` `            ``gcd ``=` `__gcd(first, second);  ` `  `  `            ``# check if gcd is same as given g  ` `            ``# and lcm is same as lcm l  ` `            ``if` `(gcd ``=``=` `g ``and` `l ``%` `first ``=``=` `0` `and` `                              ``l ``%` `second ``=``=` `0``): ` `                ``print``(first , ``" "` `, second);  ` `                ``return``;  ` ` `  `  `  `# Function return GCD of two give number  ` `def` `__gcd(a, b):  ` `    ``if``(b``=``=``0``): ` `        ``return` `a; ` `    ``else``: ` `        ``return` `__gcd(b, a ``%` `b);  ` `  `  `# Driver Code  ` `g ``=` `3``; ` `l ``=` `12``;  ` `printPair(g, l); ` ` `  `# This code is contributed by Princi Singh `

## C#

 `// C# program to print any pair  ` `// with a given gcd G and lcm L  ` `using` `System; ` `public` `class` `GFG { ` ` `  `// Function to print the pairs  ` `    ``static` `void` `printPair(``int` `g, ``int` `l) { ` `        ``int` `n = g * l; ` ` `  `        ``// iterate over all factor pairs  ` `        ``for` `(``int` `i = 1; i * i <= n; i++) { ` ` `  `            ``// check if a factor  ` `            ``if` `(n % i == 0) { ` `                ``int` `first = i; ` `                ``int` `second = n / i; ` ` `  `                ``// find gcd  ` `                ``int` `gcd = __gcd(first, second); ` ` `  `                ``// check if gcd is same as given g  ` `                ``// and lcm is same as lcm l  ` `                ``if` `(gcd == g && l % first == 0 && l % second == 0) { ` `                    ``Console.WriteLine(first + ``" "` `+ second); ` `                    ``return``; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `//Function return GCD of two give number ` ` `  `    ``private` `static` `int` `__gcd(``int` `a, ``int` `b) { ` `        ``return` `b == 0 ? a : __gcd(b, a % b); ` `    ``} ` `// Driver function  ` ` `  `    ``public` `static` `void` `Main() { ` `        ``int` `g = 3, l = 12; ` `        ``printPair(g, l); ` ` `  `    ``} ` `} ` ` `  `// This code is contributed by RAJPUT-JI `

## PHP

 `

Output:

`3 12`

Time Complexity: O(sqrt(g*l))

An efficient solution will be to observe that the lcm is always divisible by gcd, hence the answer can be obtained in O(1). One of the numbers will be the gcd G itself and the other will be the lcm L.

Below is the implementation of the above approach.

## C++

 `// C++ program to print any pair ` `// with a given gcd G and lcm L ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print the pairs ` `void` `printPair(``int` `g, ``int` `l) ` `{ ` `    ``cout << g << ``" "` `<< l; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `g = 3, l = 12; ` `    ``printPair(g, l); ` `    ``return` `0; ` `} `

## Java

 `// Java program to print any pair ` `// with a given gcd G and lcm L ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  ` `  ` `  `// Function to print the pairs ` ` ``static` `void` `printPair(``int` `g, ``int` `l) ` `{ ` `    ``System.out.print( g + ``" "` `+ l); ` `} ` ` `  `// Driver Code ` `    ``public` `static` `void` `main (String[] args) { ` `    ``int` `g = ``3``, l = ``12``; ` `    ``printPair(g, l); ` `    ``} ` `} ` `// This code is contributed by inder_verma. `

## Python 3

 `# Python 3 program to print any pair ` `# with a given gcd G and lcm L ` ` `  `# Function to print the pairs ` `def` `printPair(g, l): ` `    ``print``(g, l) ` ` `  `# Driver Code ` `g ``=` `3``; l ``=` `12``; ` `printPair(g, l); ` ` `  `# This code is contributed  ` `# by Akanksha Rai `

## C#

 `// C# program to print any pair  ` `// with a given gcd G and lcm L  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `// Function to print the pairs  ` `static` `void` `printPair(``int` `g, ``int` `l)  ` `{  ` `    ``Console.Write( g + ``" "` `+ l);  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main ()  ` `{  ` `    ``int` `g = 3, l = 12;  ` `    ``printPair(g, l);  ` `}  ` `}  ` ` `  `// This code is contributed  ` `// by Subhadeep `

## PHP

 ` `

Output:

`3 12`

Time Complexity: O(1)

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