Find maximum value of x such that n! % (k^x) = 0

Given two integers and . The task is to find the maximum value of x, such that, n! % (k^x) = 0.

Examples: 

Input : n = 5, k = 2
Output : 3
Explanation : Given n = 5 and k = 2. So, n! = 120. 
Now for different values of x:
n! % 2^0 = 0,
n! % 2^1 = 0,
n! % 2^2 = 0, 
n! % 2^3 = 0, 
n! % 2^4 = 8, 
n! % 2^5 = 24, 
n! % 2^6 = 56, 
n! % 2^7 = 120. 
So, the answer should be 3.

Input : n = 1000, x = 2
Output : 994

Approach: 

1. First take the squareroot of and store it in a variable say, .
2. Run the loop from i=2 to m.
3. If i = m then copy k to i.
4. If k is divisible by i then divide k by i.
5. Run a loop to n and add the quotient to a variable say, .
6. Store the minimum value of r after every loop.

Below is the implementation of the above approach: 

C++

// C++ program to maximize the value
// of x such that n! % (k^x) = 0
#include<iostream>
#include<math.h>
using namespace std;

class GfG
{

    // Function to maximize the value
    // of x such that n! % (k^x) = 0
    public:
    int findX(int n, int k)
    {
        int r = n, v, u;

        // Find square root of k and add 1 to it
        int m = sqrt(k) + 1;

        // Run the loop from 2 to m and k
        // should be greater than 1
        for (int i = 2; i <= m && k > 1; i++) {
            if (i == m) {
                i = k;
            }

            // optimize the value of k
            for (u = v = 0; k % i == 0; v++) {
                k /= i;
            }

            if (v > 0) {
                int t = n;
                while (t > 0) {
                    t /= i;
                    u += t;
                }

                // Minimum store
                r = min(r, u / v);
            }
        }

        return r;
    }
};

    // Driver Code
    int main()
    {
        GfG g;
        int n = 5;
        int k = 2;
        cout<<g.findX(n, k);
    }

//This code is contributed by Soumik

Java

// Java program to maximize the value
// of x such that n! % (k^x) = 0

import java.util.*;

public class GfG {

    // Function to maximize the value
    // of x such that n! % (k^x) = 0
    private static int findX(int n, int k)
    {
        int r = n, v, u;

        // Find square root of k and add 1 to it
        int m = (int)Math.sqrt(k) + 1;

        // Run the loop from 2 to m and k
        // should be greater than 1
        for (int i = 2; i <= m && k > 1; i++) {
            if (i == m) {
                i = k;
            }

            // optimize the value of k
            for (u = v = 0; k % i == 0; v++) {
                k /= i;
            }

            if (v > 0) {
                int t = n;
                while (t > 0) {
                    t /= i;
                    u += t;
                }

                // Minimum store
                r = Math.min(r, u / v);
            }
        }

        return r;
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 5;
        int k = 2;

        System.out.println(findX(n, k));
    }
}

Python 3

# Python 3 program to maximize the value
# of x such that n! % (k^x) = 0
import math

# Function to maximize the value
# of x such that n! % (k^x) = 0
def findX(n, k):
    r = n

    # Find square root of k
    # and add 1 to it
    m = int(math.sqrt(k)) + 1

    # Run the loop from 2 to m
    # and k should be greater than 1
    i = 2
    while i <= m and k > 1 :
        if (i == m) :

            i = k

        # optimize the value of k
        u = 0
        v = 0
        while k % i == 0 :
            k //= i
            v += 1

        if (v > 0) :
            t = n
            while (t > 0) :
                t //= i
                u += t

            # Minimum store
            r = min(r, u // v)

        i += 1

    return r

# Driver Code
if __name__ == "__main__":

    n = 5
    k = 2

    print(findX(n, k))

# This code is contributed
# by ChitraNayal

C#

// C# program to maximize the value
// of x such that n! % (k^x) = 0

using System;

class GfG
{

    // Function to maximize the value
    // of x such that n! % (k^x) = 0
    public int findX(int n, int k)
    {
        int r = n, v, u;

        // Find square root of k and add 1 to it
        int m = (int)Math.Sqrt(k) + 1;

        // Run the loop from 2 to m and k
        // should be greater than 1
        for (int i = 2; i <= m && k > 1; i++) {
            if (i == m) {
                i = k;
            }

            // optimize the value of k
            for (u = v = 0; k % i == 0; v++) {
                k /= i;
            }

            if (v > 0) {
                int t = n;
                while (t > 0) {
                    t /= i;
                    u += t;
                }

                // Minimum store
                r = Math.Min(r, u / v);
            }
        }

        return r;
    }
}

    // Driver Code
class geek
{
    public static void Main()
    {
        GfG g = new GfG();
        int n = 5;
        int k = 2;

        Console.WriteLine(g.findX(n, k));
    }
}

PHP

<?php
// PHP program to maximize the value
// of x such that n! % (k^x) = 0

// Function to maximize the value
// of x such that n! % (k^x) = 0
function findX($n, $k)
{
    $r = $n;

    // Find square root of k and add 1 to it
    $m = (int)sqrt($k) + 1;

    // Run the loop from 2 to m and k
    // should be greater than 1
    for ($i = 2; $i <= $m && $k > 1; $i++)
    {
        if ($i == $m)
        {
            $i = $k;
        }

        // optimize the value of k
        for ($u = $v = 0; $k % $i == 0; $v++)
        {
            $k = (int)($k / $i);
        }

        if ($v > 0)
        {
            $t = $n;
            while ($t > 0)
            {
                $t = (int)($t / $i);
                $u = $u + $t;
            }

            // Minimum store
            $r = min($r, (int)($u / $v));
        }
    }
    return $r;
}


// Driver Code
$n = 5;
$k = 2;
echo findX($n, $k);

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

Javascript

<script>
    // Javascript program to maximize the value
// of x such that n! % (k^x) = 0



    // Function to maximize the value
    // of x such that n! % (k^x) = 0
    function findX(n, k)
    {
        let r = n;
        let v = 0;
        let u = 0;

        // Find square root of k and add 1 to it
        let m = Math.floor(Math.sqrt(k) + 1);


        // Run the loop from 2 to m and k
        // should be greater than 1
        for (let i = 2; i <= m && k > 1; i++) {
            if (i == m) {
                i = k;
            }

            // optimize the value of k
            for (let u = v = 0; k % i == 0; v++) {
                k = Math.floor(k / i);
            }

            if (v > 0) {
                let t = n;
                while (t > 0) {
                    t = Math.floor(t / i);
                    u += t;
                }

                // Minimum store
                r = Math.min(r, Math.floor(u / v));
            }
        }

        return r;
    }

    // Driver Code


        let n = 5;
        let k = 2;
        document.write(findX(n, k));

//This code is contributed by gfgking
</script>

Output:

3

Time complexity: O(sqrt(k)log(n)), where n is the input parameter n and k is the input parameter k.
Auxiliary space: O(1)