Number of trailing zeroes in base 16 representation of N!

Given an integer N, the task is to find the number of trailing zeroes in the base 16 representation of the factorial of N.

Examples:

Input: N = 6
Output: 1
6! = 720 (base 10) = 2D0 (base 16)

Input: N = 100
Output: 24

Approach:

  • Number of trailing zeroes would be the highest power of 16 in the factorial of N in base 10.
  • We know that 16 = 24. So, the highest power of 16 is equal to the highest power 2 in the factorial of N divided by 4.
  • To calculate the highest power of 2 in N!, we can use Legendre’s Formula.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
  
// Function to return the count of trailing zeroes
ll getTrailingZeroes(ll n)
{
    ll count = 0;
    ll val, powerTwo = 2;
  
    // Implementation of the Legendre's formula
    do {
        val = n / powerTwo;
        count += val;
        powerTwo *= 2;
    } while (val != 0);
  
    // Count has the highest power of 2
    // that divides n! in base 10
    return (count / 4);
}
  
// Driver code
int main()
{
    int n = 6;
    cout << getTrailingZeroes(n);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach 
class GfG 
{
  
// Function to return the count of trailing zeroes 
static long getTrailingZeroes(long n) 
    long count = 0
    long val, powerTwo = 2
  
    // Implementation of the Legendre's formula 
    do 
    
        val = n / powerTwo; 
        count += val; 
        powerTwo *= 2
    } while (val != 0); 
  
    // Count has the highest power of 2 
    // that divides n! in base 10 
    return (count / 4); 
  
// Driver code 
public static void main(String[] args) 
    int n = 6
    System.out.println(getTrailingZeroes(n)); 
}
  
// This code is contributed by 
// Prerna Saini.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
  
# Function to return the count of
# trailing zeroes
def getTrailingZeroes(n):
  
    count = 0
    val, powerTwo = 1, 2
  
    # Implementation of the Legendre's 
    # formula
    while (val != 0):
        val = n //powerTwo
        count += val
        powerTwo *= 2
  
    # Count has the highest power of 2
    # that divides n! in base 10
    return (count // 4)
  
# Driver code
n = 6
print(getTrailingZeroes(n))
  
# This code is contributed 
# by Mohit Kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach 
using System;
class GFG 
{
  
// Function to return the count of
// trailing zeroes 
static long getTrailingZeroes(long n) 
    long count = 0; 
    long val, powerTwo = 2; 
  
    // Implementation of the 
    // Legendre's formula 
    do
    
        val = n / powerTwo; 
        count += val; 
        powerTwo *= 2; 
    } while (val != 0); 
  
    // Count has the highest power of 2 
    // that divides n! in base 10 
    return (count / 4); 
  
// Driver code 
public static void Main() 
    int n = 6; 
    Console.Write(getTrailingZeroes(n)); 
}
  
// This code is contributed by 
// Akanksha Rai

chevron_right


PHP

Output:

1


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.