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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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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
?>

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Output:

2

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