# Check if there is any pair in a given range with GCD is divisible by k

Given a range, we need to check if is their any pairs in the segment whose GCD is divisible by k.

Examples:

```Input : l=4, r=6, k=2
Output : YES
There are two numbers 4 and 6 whose GCD is 2 which is divisible by 2.

Input : l=3 r=5 k=4
Output : NO
Their is no such pair whose gcd is divisible by 5.
```

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

Basically we need to count such numbers in range l to r such that they are divisible by k. Because if we choose any such two numbers then their gcd is also a multiple of k. Now if count of such numbers is greater than one then we can form pair, otherwise it’s impossible to form a pair (x, y) such that gcd(x, y) is divisible by k.

Below is the implementation of above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// function to count such possible numbers ` `bool` `Check_is_possible(``int` `l, ``int` `r, ``int` `k) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = l; i <= r; i++) { ` ` `  `        ``// if i is divisible by k ` `        ``if` `(i % k == 0) ` `            ``count++; ` `    ``} ` ` `  `    ``// if count of such numbers ` `    ``// is greater than one ` `    ``return` `(count > 1); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `l = 4, r = 12; ` `    ``int` `k = 5; ` ` `  `    ``if` `(Check_is_possible(l, r, k)) ` `        ``cout << ``"YES\n"``; ` `    ``else` `        ``cout << ``"NO\n"``; ` `    ``return` `0; ` `} `

## Java

 `// function to count such  ` `// possible numbers  ` ` `  `class` `GFG { ` ` `  `    ``public` `boolean` `Check_is_possible(``int` `l, ``int` `r, ` `            ``int` `k) { ` `        ``int` `count = ``0``; ` ` `  `        ``for` `(``int` `i = l; i <= r; i++) { ` ` `  `            ``// if i is divisible by k  ` `            ``if` `(i % k == ``0``) { ` `                ``count++; ` `            ``} ` `        ``} ` ` `  `        ``// if count of such numbers  ` `        ``// is greater than one  ` `        ``return` `(count > ``1``); ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) { ` `        ``GFG g = ``new` `GFG(); ` `        ``int` `l = ``4``, r = ``12``; ` `        ``int` `k = ``5``; ` ` `  `        ``if` `(g.Check_is_possible(l, r, k)) { ` `            ``System.out.println(``"YES"``); ` `        ``} ``else` `{ ` `            ``System.out.println(``"NO"``); ` `        ``} ` ` `  `    ``} ` `} ` `// This code is contributed by RAJPUT-JI `

## Python3

 `# function to count such possible numbers ` `def` `Check_is_possible(l, r, k): ` ` `  `    ``count ``=` `0``; ` ` `  `    ``for` `i ``in` `range``(l, r ``+` `1``): ` ` `  `        ``# if i is divisible by k ` `        ``if` `(i ``%` `k ``=``=` `0``): ` `            ``count ``+``=` `1``; ` ` `  `    ``# if count of such numbers ` `    ``# is greater than one ` `    ``return` `(count > ``1``); ` ` `  `# Driver code ` `l ``=` `4``; ` `r ``=` `12``; ` `k ``=` `5``; ` `if` `(Check_is_possible(l, r, k)): ` `    ``print``(``"YES"``); ` `else``: ` `    ``print``(``"NO"``); ` ` `  `# This code is contributed by mits `

## C#

 `using` `System; ` ` `  `// function to count such  ` `// possible numbers  ` `class` `GFG ` `{ ` `public` `bool` `Check_is_possible(``int` `l, ``int` `r, ` `                              ``int` `k)  ` `{  ` `    ``int` `count = 0;  ` ` `  `    ``for` `(``int` `i = l; i <= r; i++) ` `    ``{  ` ` `  `        ``// if i is divisible by k  ` `        ``if` `(i % k == 0)  ` `            ``count++;  ` `    ``}  ` ` `  `    ``// if count of such numbers  ` `    ``// is greater than one  ` `    ``return` `(count > 1);  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `    ``GFG g = ``new` `GFG(); ` `    ``int` `l = 4, r = 12;  ` `    ``int` `k = 5;  ` ` `  `    ``if` `(g.Check_is_possible(l, r, k))  ` `        ``Console.WriteLine(``"YES\n"``);  ` `    ``else` `    ``Console.WriteLine(``"NO\n"``);  ` `}  ` `} ` ` `  `// This code is contributed ` `// by Soumik `

## PHP

 ` 1); ` `} ` ` `  `// Driver code ` `\$l` `= 4; ``\$r` `= 12; ` `\$k` `= 5; ` ` `  `if` `(Check_is_possible(``\$l``, ``\$r``, ``\$k``)) ` `    ``echo` `"YES\n"``; ` `else` `    ``echo` `"NO\n"``; ` ` `  `// This code is contributed ` `// by Akanksha Rai ` `?> `

Output:

```YES
```

Time Complexity: O(r – l + 1)

An efficient solution is based on the efficient approach discussed here.

## C++

 `// C++ program to count the numbers divisible ` `// by k in a given range ` `#include ` `using` `namespace` `std; ` ` `  `// Returns count of numbers in [l r] that ` `// are divisible by k. ` `int` `Check_is_possible(``int` `l, ``int` `r, ``int` `k) ` `{ ` `    ``int` `div_count = (r / k) - (l / k); ` ` `  `    ``// Add 1 explicitly as l is divisible by k ` `    ``if` `(l % k == 0) ` `        ``div_count++; ` ` `  `    ``// l is not divisible by k ` `    ``return` `(div_count > 1); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `l = 30, r = 70, k = 10; ` ` `  `    ``if` `(Check_is_possible(l, r, k)) ` `        ``cout << ``"YES\n"``; ` `    ``else` `        ``cout << ``"NO\n"``; ` `    ``return` `0; ` `} `

## Java

 `// Java program to count the numbers divisible ` `// by k in a given range ` ` `  `class` `GFG { ` ` `  `// Returns count of numbers in [l r] that ` `// are divisible by k. ` `    ``static` `boolean` `Check_is_possible(``int` `l, ``int` `r, ``int` `k) { ` `        ``int` `div_count = (r / k) - (l / k); ` ` `  `        ``// Add 1 explicitly as l is divisible by k ` `        ``if` `(l % k == ``0``) { ` `            ``div_count++; ` `        ``} ` ` `  `        ``// l is not divisible by k ` `        ``return` `(div_count > ``1``); ` `    ``} ` ` `  `// Driver Code ` `    ``public` `static` `void` `main(String[] args) { ` `        ``int` `l = ``30``, r = ``70``, k = ``10``; ` `        ``if` `(Check_is_possible(l, r, k)) { ` `            ``System.out.println(``"YES"``); ` `        ``} ``else` `{ ` `            ``System.out.println(``"NO"``); ` `        ``} ` ` `  `    ``} ` `} ` `// This code is contributed by RAJPUT-JI `

## Python3

 `# Python3 program to count the numbers  ` `# divisible by k in a given range  ` ` `  `# Returns count of numbers in [l r]  ` `# that are divisible by k.  ` `def` `Check_is_possible(l, r, k):  ` ` `  `    ``div_count ``=` `(r ``/``/` `k) ``-` `(l ``/``/` `k)  ` ` `  `    ``# Add 1 explicitly as l is ` `    ``# divisible by k  ` `    ``if` `l ``%` `k ``=``=` `0``: ` `        ``div_count ``+``=` `1` ` `  `    ``# l is not divisible by k  ` `    ``return` `div_count > ``1` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``l, r, k ``=` `30``, ``70``, ``10` ` `  `    ``if` `Check_is_possible(l, r, k) ``=``=` `True``:  ` `        ``print``(``"YES"``)  ` `    ``else``: ` `        ``print``(``"NO"``)  ` `     `  `# This code is contributed  ` `# by Rituraj Jain `

## C#

 `// C# program to count the numbers divisible ` `// by k in a given range  ` `using` `System; ` `  `  ` `  `public` `class` `GFG { ` ` `  `// Returns count of numbers in [l r] that ` `// are divisible by k. ` `    ``static` `bool` `Check_is_possible(``int` `l, ``int` `r, ``int` `k) { ` `        ``int` `div_count = (r / k) - (l / k); ` ` `  `        ``// Add 1 explicitly as l is divisible by k ` `        ``if` `(l % k == 0) { ` `            ``div_count++; ` `        ``} ` ` `  `        ``// l is not divisible by k ` `        ``return` `(div_count > 1); ` `    ``} ` ` `  `// Driver Code ` `    ``public` `static` `void` `Main() { ` `        ``int` `l = 30, r = 70, k = 10; ` `        ``if` `(Check_is_possible(l, r, k)) { ` `            ``Console.WriteLine(``"YES"``); ` `        ``} ``else` `{ ` `            ``Console.WriteLine(``"NO"``); ` `        ``} ` ` `  `    ``} ` `} ` `// This code is contributed by RAJPUT-JI `

## PHP

 ` 1); ` `} ` ` `  `// Driver Code ` `\$l` `= 30; ` `\$r` `= 70; ` `\$k` `= 10; ` ` `  `if` `(Check_is_possible(``\$l``, ``\$r``, ``\$k``)) ` `    ``echo` `"YES\n"``; ` `else` `    ``echo` `"NO\n"``; ` ` `  `// This Code is contributed by mits ` `?> `

Output:

```YES
```

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