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.

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 <bits/stdc++.h> ` `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; ` `} ` |

## 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

**Output:**

YES

**Time Complexity: O(r – l + 1)**

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

`// Program to count the numbers divisible ` `// by k in a given range ` `#include <bits/stdc++.h> ` `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; ` `} ` |

**Output:**

YES

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