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

Given a number N and a number K. 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 $\sqrt(N)$

Examples:

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

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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count number of divisors
// of N which are divisible by K
  
#include <iostream>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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..

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP 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
function countDivisors($n, $k)
{
    // Variable to store
    // count of divisors
    $count = 0;
  
    // 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
$n = 12; $k = 3;
  
echo countDivisors($n, $k);
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

chevron_right


Output:

3

Time Complexity : O(N)

Efficient Approach: The idea is to run a loop from 1 to <$\sqrt(N)$ 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count number of divisors
// of N which are divisible by K
#include <bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP 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
function countDivisors($n, $k)
{
    // integer to count the divisors
    $count = 0;
  
    // 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
$n = 12;
$k = 3;
  
echo (countDivisors($n, $k));
  
// This code is contributed
// by Shivi_Aggarwal
?>

chevron_right


Output:

3

Time Complexity :$O(\sqrt(n))



My Personal Notes arrow_drop_up

Second year Department of Information Technology Jadavpur University

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.