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 3^{2}divides 7! and 2 is the largest such power of 3. Input: n = 10, p = 3 Output: x = 4 3^{4}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 p^{2}, p^{3}, …

**How many numbers in {1, 2, 3, 4, ….. n} are divisible by p ^{2}, p^{3}, …?**

There are ⌊n/(p

^{2})⌋ numbers divisible by p

^{2}(Every p

^{2}‘th number would be divisible). Similarly, there are ⌊n/(p

^{3})⌋ numbers divisible by p

^{3}and so on.

**What is the largest possible value of x?**

So the largest possible power is ⌊n/p⌋ + ⌊n/(p^{2})⌋ + ⌊n/(p^{3})⌋ + ……

Note that we add only ⌊n/(p^{2})⌋ only once (not twice) as one p is already considered by expression ⌊n/p⌋. Similarly, we consider ⌊n/(p^{3})⌋ (not thrice).

Following is implementation based on above idea.

## C/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.

Output:

The largest power of 3 that divides 10! is 4

Time complexity of the above solution is Log_{p}n.

**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.