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 √(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++

// 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;
}

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

Javascript

<script>
// Javascript implementation of above approach

// Function to count number of divisors
// of N which are divisible by K
function countDivisors(n, k)
{
    // Variable to store
    // count of divisors
    var 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;
}

var n = 12, k = 3;
document.write(countDivisors(n, k));

// This code is contributed by SoumikMondal.
</script>

PHP

<?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)

Output

3

Time Complexity : O(N)

Auxiliary Space: O(1)
Efficient Approach: The idea is to run a loop from 1 to < √(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++

// 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++;
            }
        }
    }

    i--;

    // 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 = 16, k = 4;

    // Function Call
    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++;
            }
        }
    }

    i--;

    // 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 = 16, k = 4;
        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 = 16
    k = 4

    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++;
            }
        }
    }

    i--;

    // 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 = 16, k = 4;
    Console.WriteLine( countDivisors(n, k));
}
}

// This code is contributed by ajit

Javascript

<script>

    // Javascript 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
        let 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++;
                }
            }
        }

        i--;

        // If the number is a perfect square
        // and it is divisible by k
        if ((i * i == n) && (i % k == 0))
        {
            count--;
        }

        return count;
    }

    let n = 16, k = 4;
    document.write( countDivisors(n, k));

</script>

PHP

<?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++;
            }
        }
    }

    i--;

    // 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 = 16;
$k = 4;

echo (countDivisors($n, $k));

// This code is contributed
// by Shivi_Aggarwal
?>

Output

3

Time Complexity: O(√(n))
Auxiliary Space: O(1) as using constant space