# 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

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.