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

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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. 

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C#

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// 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.

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PHP

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<?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.
?>

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Output :

25

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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