Smallest number with at least n trailing zeroes in factorial

Given a number n. The task is to find the smallest number whose factorial contains at least n trailing zeroes.

Examples :

Input : n = 1
Output : 5 
1!, 2!, 3!, 4! does not contain trailing zero.
5! = 120, which contains one trailing zero.

Input : n = 6
Output : 25

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

In the article for Count trailing zeroes in factorial of a number, we have discussed number of zeroes is equal to number of 5’s in prime factors of x!. We have discussed below formula to count number of 5’s.

Trailing 0s in x! = Count of 5s in prime factors of x!
                  = floor(x/5) + floor(x/25) + floor(x/125) + ....

Let us take few examples to observe pattern

5!  has 1 trailing zeroes 
[All numbers from 6 to 9
 have 1 trailing zero]

10! has 2 trailing zeroes
[All numbers from 11 to 14
 have 2 trailing zeroes]

15! to 19! have 3 trailing zeroes

20! to 24! have 4 trailing zeroes

25! to 29! have 6 trailing zeroes

We can notice that, the minimum value whose factorial contain n trailing zeroes is 5*n.

So, to find minimum value whose factorial contains n trailing zeroes, use binary search on range from 0 to 5*n. And, find the smallest number whose factorial contains n trailing zeroes.

C++

// C++ program tofind smallest number whose 
// factorial contains at least n trailing 
// zeroes. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Return true if number's factorial contains 
// at least n trailing zero else false. 
bool check(int p, int n) 
{ 
    int temp = p, count = 0, f = 5; 
    while (f <= temp) 
    { 
        count += temp/f; 
        f = f*5; 
    } 
    return (count >= n); 
} 
  
// Return smallest number whose factorial 
// contains at least n trailing zeroes 
int findNum(int n) 
{ 
    // If n equal to 1, return 5. 
    // since 5! = 120. 
    if (n==1) 
        return 5; 
  
    // Initalising low and high for binary 
    // search. 
    int low = 0; 
    int high = 5*n; 
  
    // Binary Search. 
    while (low <high) 
    { 
        int mid = (low + high) >> 1; 
  
        // Checking if mid's factorial contains 
        // n trailing zeroes. 
        if (check(mid, n)) 
            high = mid; 
        else
            low = mid+1; 
    } 
  
    return low; 
} 
  
// driver code 
int main() 
{ 
    int n = 6; 
    cout << findNum(n) << endl; 
    return 0; 
} 

Java

// Java program tofind smallest number whose 
// factorial contains at least n trailing 
// zeroes. 
  
class GFG 
{ 
    // Return true if number's factorial contains 
    // at least n trailing zero else false. 
    static boolean check(int p, int n) 
    { 
        int temp = p, count = 0, f = 5; 
        while (f <= temp) 
        { 
            count += temp / f; 
            f = f * 5; 
        } 
        return (count >= n); 
    }  
      
    // Return smallest number whose factorial 
    // contains at least n trailing zeroes 
    static int findNum(int n) 
    { 
        // If n equal to 1, return 5. 
        // since 5! = 120. 
        if (n==1) 
            return 5; 
      
        // Initalising low and high for binary 
        // search. 
        int low = 0; 
        int high = 5 * n; 
      
        // Binary Search. 
        while (low < high) 
        { 
            int mid = (low + high) >> 1; 
      
            // Checking if mid's factorial  
            // contains n trailing zeroes. 
            if (check(mid, n)) 
                high = mid; 
            else
                low = mid + 1; 
        } 
      
        return low; 
    } 
      
    // Driver code  
    public static void main (String[] args) 
    { 
        int n = 6; 
        System.out.println(findNum(n)); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python3 program tofind smallest 
# number whose 
# factorial contains at least 
# n trailing zeroes 
  
# Return true if number's factorial contains 
# at least n trailing zero else false. 
def check(p,n): 
  
    temp = p 
    count = 0
    f = 5
    while (f <= temp): 
        count += temp/f 
        f = f*5
  
    return (count >= n) 
  
# Return smallest number whose factorial 
# contains at least n trailing zeroes 
def findNum(n): 
  
    # If n equal to 1, return 5. 
    # since 5! = 120. 
    if (n==1): 
        return 5
   
    # Initalizing low and high for binary 
    # search. 
    low = 0
    high = 5*n 
   
    # Binary Search. 
    while (low <high): 
  
        mid = (low + high) >> 1
   
        # Checking if mid's factorial contains 
        # n trailing zeroes. 
        if (check(mid, n)): 
            high = mid 
        else: 
            low = mid+1
      
   
    return low 
  
  
# driver code 
n = 6
print(findNum(n)) 
  
# This code is contributed 
# by Anant Agarwal.  

C#

// C# program tofind smallest number whose 
// factorial contains at least n trailing 
// zeroes. 
using System; 
  
class GFG 
{ 
    // Return true if number's factorial contains 
    // at least n trailing zero else false. 
    static bool check(int p, int n) 
    { 
        int temp = p, count = 0, f = 5; 
        while (f <= temp) 
        { 
            count += temp / f; 
            f = f * 5; 
        } 
        return (count >= n); 
    }  
      
    // Return smallest number whose factorial 
    // contains at least n trailing zeroes 
    static int findNum(int n) 
    { 
        // If n equal to 1, return 5. 
        // since 5! = 120. 
        if (n == 1) 
            return 5; 
      
        // Initalising low and high for binary 
        // search. 
        int low = 0; 
        int high = 5 * n; 
      
        // Binary Search. 
        while (low < high) 
        { 
            int mid = (low + high) >> 1; 
      
            // Checking if mid's factorial  
            // contains n trailing zeroes. 
            if (check(mid, n)) 
                high = mid; 
            else
                low = mid + 1; 
        } 
      
        return low; 
    } 
      
    // Driver code  
    public static void Main () 
    { 
        int n = 6; 
          
        Console.WriteLine(findNum(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find smallest 
// number whose factorial contains 
// at least n trailing zeroes. 
  
// Return true if number's  
// factorial contains at  
// least n trailing zero  
// else false. 
function check($p, $n) 
{ 
    $temp = $p; $count = 0; $f = 5; 
    while ($f <= $temp) 
    { 
        $count += $temp / $f; 
        $f = $f * 5; 
    } 
    return ($count >= $n); 
} 
  
// Return smallest number  
// whose factorial contains  
// at least n trailing zeroes 
function findNum($n) 
{ 
    // If n equal to 1, return 5. 
    // since 5! = 120. 
    if ($n == 1) 
        return 5; 
  
    // Initalising low and high 
    // for binary search. 
    $low = 0; 
    $high = 5 * $n; 
  
    // Binary Search. 
    while ($low < $high) 
    { 
        $mid = ($low + $high) >> 1; 
  
        // Checking if mid's factorial  
        // contains n trailing zeroes. 
        if (check($mid, $n)) 
            $high = $mid; 
        else
            $low = $mid + 1; 
    } 
  
    return $low; 
} 
  
// Driver Code 
$n = 6; 
echo(findNum($n)); 
  
// This code is contributed by Ajit. 
?> 

Output :

This article is contributed by Anuj Chauhan. 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.

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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