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 the implementation of the 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.

Approach 2: Using recursion
Another way to implement Legendre’s formula is to use recursion. Here is the code implementation:

•     Start the program by including the required header files and the standard namespace.
•    Define a function named largestPower that takes two integers n and p as input and returns an integer.
•    Check if n is equal to 0. If it is, return 0 as the largest power of p that divides n! is 0.
•    Calculate the largest power of p that divides n! recursively by dividing n by p and adding the quotient to the result of largestPower(n/p, p).
•    Return the result of the calculation in step 4.
•    Define the main function, which sets n equal to 10 and p equal to 3, calls the largestPower function with these values, and outputs the result.
•    End the program by returning 0 from the main function.

C++

#include<bits/stdc++.h>
using namespace std;

int largestPower(int n, int p) {
    if (n == 0) {
        return 0;
    }
    return n/p + largestPower(n/p, p);
}

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

Java

public class Main {
    public static int largestPower(int n, int p) {
        if (n == 0) {
            return 0;
        }
        return n / p + largestPower(n / p, p);
    }

    public static void main(String[] args) {
        int n = 10, p = 3;
        System.out.printf("The largest power of %d that divides %d! is %d", p, n, largestPower(n, p));
    }
}

Python3

def largestPower(n, p):
    if n == 0:
        return 0
    return n // p + largestPower(n // p, p)

n, p = 10, 3
print(f"The largest power of {p} that divides {n}! is {largestPower(n, p)}")

C#

using System;

public class MainClass {
    public static int largestPower(int n, int p) {
        if (n == 0) {
            return 0;
        }
        // recursively calculate the largest power of p that divides n!
        return n / p + largestPower(n / p, p);
    }

    public static void Main(string[] args) {
        int n = 10, p = 3;
        Console.WriteLine("The largest power of {0} that divides {1}! is {2}", p, n, largestPower(n, p));
    }
}

Javascript

function largestPower(n, p) {
    if (n == 0) {
        return 0;
    }
    let x=Math.floor(n/p);
    return x + largestPower(x, p);
}

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

Output

The largest power of 3 that divides 10! is 4

Time complexity: O(logpn)

The time complexity of the largestPower function is O(log_p(n)), where n is the input integer and p is the divisor. This is because the function divides n by p in each recursive call, which reduces the size of n by a factor of p. The number of times the function can be called before n becomes less than p is proportional to the logarithm base p of n. Therefore, the time complexity is O(log_p(n)).

Auxiliary Space: O(logpn)

The space complexity of the function is O(log_p(n)), because each recursive call to the function adds a new stack frame to the call stack, and the depth of the call stack is proportional to the number of times the function is called, which is proportional to the logarithm base p of n. Therefore, the space complexity is O(log_p(n)).
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.