Legendre’s formula (Given p and n, find the largest x such that p^x divides n!)

Given an integer n and a prime number p, find the largest x such that p^x (p raised to power x) divides n! (factorial) 
Examples:

Input:  n = 7, p = 3
Output: x = 2
32 divides 7! and 2 is the largest such power of 3.

Input:  n = 10, p = 3
Output: x = 4
34 divides 10! and 4 is the largest such power of 3.

n! is multiplication of {1, 2, 3, 4, …n}.
How many numbers in {1, 2, 3, 4, ….. n} are divisible by p?
Every p’th number is divisible by p in {1, 2, 3, 4, ….. n}. Therefore in n!, there are ⌊n/p⌋ numbers divisible by p. So we know that the value of x (largest power of p that divides n!) is at-least ⌊n/p⌋. 
Can x be larger than ⌊n/p⌋ ?
Yes, there may be numbers which are divisible by p², p³, … 
How many numbers in {1, 2, 3, 4, ….. n} are divisible by p², p³, …?
There are ⌊n/(p²)⌋ numbers divisible by p² (Every p²‘th number would be divisible). Similarly, there are ⌊n/(p³)⌋ numbers divisible by p³ and so on.
What is the largest possible value of x?
So the largest possible power is ⌊n/p⌋ + ⌊n/(p²)⌋ + ⌊n/(p³)⌋ + …… 
Note that we add only ⌊n/(p²)⌋ only once (not twice) as one p is already considered by expression ⌊n/p⌋. Similarly, we consider ⌊n/(p³)⌋ (not thrice). 
Below is implementation of above idea. 

C++

// C++ program to find largest x such that p*x divides n!
#include <iostream>
using namespace std;

// Returns largest power of p that divides n!
int largestPower(int n, int p)
{
    // Initialize result
    int x = 0;

    // Calculate x = n/p + n/(p^2) + n/(p^3) + ....
    while (n)
    {
        n /= p;
        x += n;
    }
    return x;
}

// Driver code
int main()
{
    int n = 10, p = 3;
    cout << "The largest power of "<< p <<
            " that divides " << n << "! is "<<
            largestPower(n, p) << endl;
    return 0;
}

// This code is contributed by shubhamsingh10

C

// C program to find largest x such that p*x divides n!
#include <stdio.h>

// Returns largest power of p that divides n!
int largestPower(int n, int p)
{
    // Initialize result
    int x = 0;

    // Calculate x = n/p + n/(p^2) + n/(p^3) + ....
    while (n)
    {
        n /= p;
        x += n;
    }
    return x;
}

// Driver program
int main()
{
    int n = 10, p = 3;
    printf("The largest power of %d that divides %d! is %d\n",
           p, n, largestPower(n, p));
    return 0;
}

Java

// Java program to find largest x such that p*x divides n!
import java.io.*;

class GFG
{
    // Function that returns largest power of p
    // that divides n!
    static int Largestpower(int n, int p)
    {
        // Initialize result
        int ans = 0;

        // Calculate x = n/p + n/(p^2) + n/(p^3) + ....
        while (n > 0)
        {
            n /= p;
            ans += n;
        }
        return ans;
    }

    // Driver program
    public static void main (String[] args)
    {
        int n = 10;
        int p = 3;
        System.out.println(" The largest power of " + p + " that divides "
                + n + "! is " + Largestpower(n, p));


    }
}

Python3

# Python3 program to find largest
# x such that p*x divides n!

# Returns largest power of p that divides n!
def largestPower(n, p):

    # Initialize result
    x = 0

    # Calculate x = n/p + n/(p^2) + n/(p^3) + ....
    while n:
        n /= p
        x += n
    return x

# Driver program
n = 10; p = 3
print ("The largest power of %d that divides %d! is %d\n"%
                                (p, n, largestPower(n, p)))

# This code is contributed by Shreyanshi Arun.

C#

// C# program to find largest x
// such that p * x divides n!
using System;

public class GFG
{

    // Function that returns largest
    // power of p that divides n!
    static int Largestpower(int n, int p)
    {
        // Initialize result
        int ans = 0;

        // Calculate x = n / p + n / (p ^ 2) +
        // n / (p ^ 3) + ....
        while (n > 0)
        {
            n /= p;
            ans += n;
        }
        return ans;
    }

    // Driver Code
    public static void Main ()
    {
        int n = 10;
        int p = 3;
        Console.Write(" The largest power of " + p + " that divides "
                + n + "! is " + Largestpower(n, p));


    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to find largest
// x such that p*x divides n!

// Returns largest power
// of p that divides n!
function largestPower($n, $p)
{
    // Initialize result
    $x = 0;

    // Calculate x = n/p +
    // n/(p^2) + n/(p^3) + ....
    while ($n)
    {
        $n = (int)$n / $p;
        $x += $n;
    }
    return floor($x);
}

// Driver Code
$n = 10;
$p = 3;
echo "The largest power of ", $p ;
echo " that divides ",$n , "! is ";
echo largestPower($n, $p);

// This code is contributed by ajit
?>

Javascript

<script>
// Javascript program to find largest
// x such that p*x divides n!

// Returns largest power
// of p that divides n!
function largestPower(n, p)
{
    // Initialize result
    let x = 0;

    // Calculate x = n/p +
    // n/(p^2) + n/(p^3) + ....
    while (n)
    {
        n = parseInt(n / p);
        x += n;
    }
    return Math.floor(x);
}

// Driver Code
let n = 10;
let p = 3;
document.write("The largest power of " + p);
document.write(" that divides " + n + "! is ");
document.write(largestPower(n, p));

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

Output:

The largest power of 3 that divides 10! is 4

Time complexity: O(log_pn)

Auxiliary Space: O(1)
What to do if p is not prime?
We can find all prime factors of p and compute result for every prime factor. Refer Largest power of k in n! (factorial) where k may not be prime for details.
Source:
http://e-maxx.ru/algo/factorial_divisors
This article is contributed by Ankur. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.