# Number of divisors of a given number N which are divisible by K

Given a number and a number . The task is to find the number of divisors of N which are divisible by K. Here K is a number always less than or equal to Examples:

Input: N = 12, K = 3
Output: 3

Input: N = 8, K = 2
Output: 3


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

Simple Approach: A simple approach is to check all the numbers from 1 to N and check whether any number is a divisor of N and is divisible by K. Count such numbers less than N which satisfies both the conditions.

Below is the implementation of the above approach:

## C++

 // C++ program to count number of divisors  // of N which are divisible by K     #include  using namespace std;     // Function to count number of divisors  // of N which are divisible by K  int countDivisors(int n, int k)  {      // Variable to store      // count of divisors      int count = 0, i;         // Traverse from 1 to n      for (i = 1; i <= n; i++) {             // increase the count if both          // the conditions are satisfied          if (n % i == 0 && i % k == 0) {                 count++;          }      }         return count;  }     // Driver code  int main()  {      int n = 12, k = 3;         cout << countDivisors(n, k);         return 0;  }

## Java

 // Java program to count number of divisors  // of N which are divisible by K     import java.io.*;     class GFG {        // Function to count number of divisors  // of N which are divisible by K   static int countDivisors(int n, int k)  {      // Variable to store      // count of divisors      int count = 0, i;         // Traverse from 1 to n      for (i = 1; i <= n; i++) {             // increase the count if both          // the conditions are satisfied          if (n % i == 0 && i % k == 0) {                 count++;          }      }         return count;  }     // Driver code      public static void main (String[] args) {        int n = 12, k = 3;         System.out.println(countDivisors(n, k));      }  }  // This code is contributed by shashank..

## Python3

 # Python program to count number   # of divisors of N which are   # divisible by K      # Function to count number of divisors   # of N which are divisible by K   def countDivisors(n, k) :         # Variable to store       # count of divisors       count = 0        # Traverse from 1 to n       for i in range(1, n + 1) :             # increase the count if both           # the conditions are satisfied           if (n % i == 0 and i % k == 0) :                 count += 1                    return count     # Driver code       if __name__ == "__main__" :         n, k = 12, 3     print(countDivisors(n, k))     # This code is contributed by ANKITRAI1

## C#

 // C# program to count number   // of divisors of N which are  // divisible by K  using System;     class GFG  {         // Function to count number   // of divisors of N which  // are divisible by K  static int countDivisors(int n, int k)  {      // Variable to store      // count of divisors      int count = 0, i;         // Traverse from 1 to n      for (i = 1; i <= n; i++)       {             // increase the count if both          // the conditions are satisfied          if (n % i == 0 && i % k == 0)           {              count++;          }      }         return count;  }     // Driver code  public static void Main ()   {      int n = 12, k = 3;             Console.WriteLine(countDivisors(n, k));  }  }     // This code is contributed by Shashank

## PHP

 

Output:

3


Time Complexity : O(N)

Efficient Approach: The idea is to run a loop from 1 to < and check whether the number is a divisor of N and is divisible by K and we will also check whether ( N/i ) is divisible by K or not. As (N/i) will also be a factor of N if i is a factor of N.

Below is the implementation of the above approach:

## C++

 // C++ program to count number of divisors  // of N which are divisible by K  #include  using namespace std;     // Function to count number of divisors  // of N which are divisible by K  int countDivisors(int n, int k)  {      // integer to count the divisors      int count = 0, i;         // Traverse from 1 to sqrt(N)      for (i = 1; i < sqrt(n); i++) {             // Check if i is a factor          if (n % i == 0) {              // increase the count if i              // is divisible by k              if (i % k == 0) {                  count++;              }                 // (n/i) is also a factor              // check whether it is divisible by k              if ((n / i) % k == 0) {                  count++;              }          }      }         // If the number is a perfect square      // and it is divisible by k      if ((i * i == n) && (i % k == 0)) {          count--;      }         return count;  }     // Driver code  int main()  {      int n = 12, k = 3;         cout << countDivisors(n, k);         return 0;  } 

## Java

 // Java  program to count number of divisors   // of N which are divisible by K     import java.io.*;     class GFG {         // Function to count number of divisors   // of N which are divisible by K   static int countDivisors(int n, int k)   {       // integer to count the divisors       int count = 0, i;          // Traverse from 1 to sqrt(N)       for (i = 1; i < Math.sqrt(n); i++) {              // Check if i is a factor           if (n % i == 0) {               // increase the count if i               // is divisible by k               if (i % k == 0) {                   count++;               }                  // (n/i) is also a factor               // check whether it is divisible by k               if ((n / i) % k == 0) {                   count++;               }           }       }          // If the number is a perfect square       // and it is divisible by k       if ((i * i == n) && (i % k == 0)) {           count--;       }          return count;   }      // Driver code              public static void main (String[] args) {                 int n = 12, k = 3;           System.out.println( countDivisors(n, k));                  }  }  //This Code is Contributed by akt_mit 

## Python 3

 # Python 3 program to count number of   # divisors of N which are divisible by K  import math     # Function to count number of divisors  # of N which are divisible by K  def countDivisors(n, k):             # integer to count the divisors      count = 0        # Traverse from 1 to sqrt(N)      for i in range(1, int(math.sqrt(n)) + 1):             # Check if i is a factor          if (n % i == 0) :                             # increase the count if i              # is divisible by k              if (i % k == 0) :                  count += 1                # (n/i) is also a factor check              # whether it is divisible by k              if ((n // i) % k == 0) :                  count += 1        # If the number is a perfect square      # and it is divisible by k      # if i is sqrt reduce by 1      if ((i * i == n) and (i % k == 0)) :          count -= 1          return count     # Driver code  if __name__ == "__main__":      n = 12     k = 3        print(countDivisors(n, k))     # This code is contributed   # by ChitraNayal 

## C#

 // C# program to count number of divisors   // of N which are divisible by K  using System;     class GFG  {         // Function to count number of divisors   // of N which are divisible by K   static int countDivisors(int n, int k)   {       // integer to count the divisors       int count = 0, i;          // Traverse from 1 to sqrt(N)       for (i = 1; i < Math.Sqrt(n); i++)      {              // Check if i is a factor           if (n % i == 0)           {               // increase the count if i               // is divisible by k               if (i % k == 0)               {                   count++;               }                  // (n/i) is also a factor check              // whether it is divisible by k               if ((n / i) % k == 0)               {                   count++;               }           }       }          // If the number is a perfect square       // and it is divisible by k       if ((i * i == n) && (i % k == 0))      {           count--;       }          return count;   }      // Driver code   static public void Main ()  {      int n = 12, k = 3;       Console.WriteLine( countDivisors(n, k));   }  }     // This code is contributed by ajit 

## PHP

  

Output:

3
`

Time Complexity : My Personal Notes arrow_drop_up Second year Department of Information Technology Jadavpur University

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