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Smallest number with at least n trailing zeroes in factorial

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  • Difficulty Level : Medium
  • Last Updated : 18 Jun, 2022

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 maximum 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 to find 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;
 
    // Initialising 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 to find 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;
     
        // Initialising 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 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.
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
  
    # Initializing 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 to find 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;
     
        // Initialising 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;
 
    // Initialising 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.
?>
Javascript



<script>
// JavaScript 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)
{
    let temp = p, count = 0, f = 5;
    while (f <= temp)
    {
        count += Math.floor(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;
 
    // Initialising low and high for binary
    // search.
    let low = 0;
    let high = 5*n;
 
    // Binary Search.
    while (low <high)
    {
        let 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
    let n = 6;
    document.write(findNum(n) + "<br>");
 
 
// This code is contributed by Surbhi Tyagi
 
</script>
Output : 
 
25
Time Complexity: O(log2N)
We take log2N in binary search and our check() function takes log5N time so the overall time complexity beacomes log2N * log5N which in a more general sense can be written as (logN)2 which can also be written as log2N.
Auxiliary Space: O(1)
As constant extra space is used.
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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