Power of a prime number ‘r’ in n!

Difficulty Level : Medium
Last Updated : 10 May, 2018

Given an integer n, find the power of a given prime number(r) in n!

Examples :

Input  : n = 6  r = 3
         Factorial of 6 is 720 = 2^4 * 3^2 *5 (prime factor 2,3,5)
         and power of 3 is 2 
Output : 2

Input  : n = 2000 r = 7
Output : 330

A simple method is to first calculate factorial of n, then factorize it to find the power of a prime number.
The above method can cause overflow for a slightly bigger numbers as factorial of a number is a big number. The idea is to consider prime factors of a factorial n.

Legendre Factorization of n!
For any prime number p and any integer n, let Vp(n) be the exponent of the largest power of p that divides n (that is, the p-adic valuation of n). Then
Vp(n) = summation of floor(n / p^i) and i goes from 1 to infinity

While the formula on the right side is an infinite sum, for any particular values of n and p it has only finitely many nonzero terms: for every i large enough that p^i > n, one has floor(n/p^i) = 0 . since, the sum is convergent.

Power of ‘r’ in n! = floor(n/r) + floor(n/r^2) + floor(n/r^3) + ....

Program to count power of a no. r in n! based on above formula.

C++

// C++ program to find power of  
// a prime number ‘r’ in n! 
#include <bits/stdc++.h> 
using  namespace std; 
  
// Function to return power of a  
// no. 'r' in factorial of n 
int power(int n, int r) 
{            
    // Keep dividing n by powers of  
    // 'r' and update count 
    int count = 0; 
    for (int i = r; (n / i) >= 1; i = i * r)     
        count += n / i; 
    return count; 
} 
  
// Driver program to 
// test above function 
int main()  
{ 
    int n = 6,  r = 3;    
    printf(" %d ", power(n, r));     
    return 0; 
} 

Java

// Java program to find power of 
// a prime number 'r' in n! 
  
class GFG { 
      
// Function to return power of a 
// no. 'r' in factorial of n 
static int power(int n, int r) { 
      
    // Keep dividing n by powers of 
    // 'r' and update count 
    int count = 0; 
    for (int i = r; (n / i) >= 1; i = i * r) 
    count += n / i; 
    return count; 
} 
  
// Driver code 
public static void main(String[] args)  
{ 
    int n = 6, r = 3; 
    System.out.print(power(n, r)); 
} 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

   
# Python3 program to find power  
# of a prime number ‘r’ in n! 
  
# Function to return power of a  
# no. 'r' in factorial of n 
def power(n, r): 
          
    # Keep dividing n by powers of  
    # 'r' and update count 
    count = 0; i = r 
      
    while((n / i) >= 1): 
        count += n / i 
        i = i * r 
          
    return int(count) 
  
# Driver Code 
n = 6; r = 3
print(power(n, r))  
  
# This code is contributed by Smitha Dinesh Semwal. 

C#

// C# program to find power of 
// a prime number 'r' in n! 
using System; 
  
class GFG { 
      
// Function to return power of a 
// no. 'r' in factorial of n 
static int power(int n, int r) { 
      
    // Keep dividing n by powers of 
    // 'r' and update count 
    int count = 0; 
    for (int i = r; (n / i) >= 1; i = i * r) 
    count += n / i; 
    return count; 
} 
  
// Driver code 
public static void Main()  
{ 
    int n = 6, r = 3; 
    Console.WriteLine(power(n, r)); 
} 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
//PHP program to find power of  
// a prime number ‘r’ in n! 
  
// Function to return power of a  
// no. 'r' in factorial of n 
function power($n, $r) 
{          
      
    // Keep dividing n by powers   
    // of 'r' and update count 
    $count = 0; 
    for ($i = $r; ($n / $i) >= 1; 
                   $i = $i * $r)  
        $count += $n / $i; 
    return $count; 
} 
  
    // Driver Code 
    $n = 6; $r = 3;  
    echo power($n, $r);  
  
// This code is contributed by ajit 
?> 

Output:

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