Find the last digit when factorial of A divides factorial of B

We are given two numbers A and B such that B >= A. We need to compute the last digit of this resulting F such that F = B!/A! where 1 = A, B <= 10^18 (A and B are very large).

Examples:

Input : A = 2, B = 4
Output : 2
Explanation : A! = 2 and B! = 24. 
F = 24/2 = 12 --> last digit = 2

Input : 107 109
Output : 2

As we know, factorial function grows on an exponential rate. Even the largest data type
cannot hold factorial of numbers like 100. To compute factorial of moderately large numbers, refer this.
Here the given constraints are very large. Thus, calculating the two factorials and later
dividing them and computing the last digit is practically an impossible task.



Thus we have to find an alternate approach to break down our problem. It is known that the last digit of factorial always belongs to the set {0, 1, 2, 4, 6}
The approach is as follows: –
1) We evaluate the difference between B and A
2) If the (B – A) >= 5, then the answer is always 0
3) If the difference (B – A) < 5, then we iterate from (A+1) to B, multiply and store them. multiplication_answer % 10 shall be our answer.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to find last digit of a number 
// obtained by dividing factorial of a number
// with factorial of another number.
#include <iostream>
using namespace std;
  
// Function which computes the last digit
// of resultant of B!/A!
int computeLastDigit(long long int A, long long int B)
{
  
    int variable = 1;
    if (A == B) // If A = B, B! = A! and B!/A! = 1
        return 1;
  
    // If difference (B - A) >= 5, answer = 0
    else if ((B - A) >= 5) 
        return 0;
  
    else {
  
        // If non of the conditions are true, we
        // iterate from  A+1 to B and multiply them. 
        // We are only concerned for the last digit,
        // thus we take modulus of 10
        for (long long int i = A + 1; i <= B; i++)
            variable = (variable * (i % 10)) % 10;
  
        return variable % 10;
    }
}
  
// driver function
int main()
{
    cout << computeLastDigit(2632, 2634);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find last digit of a number 
// obtained by dividing factorial of a number
// with factorial of another number.
import java.io.*;
  
class GFG {
  
// Function which computes the last digit
// of resultant of B!/A!
static int computeLastDigit(long A, long B)
{
  
    int variable = 1;
    if (A == B) // If A = B, B! = A! and B!/A! = 1
        return 1;
  
    // If difference (B - A) >= 5, answer = 0
    else if ((B - A) >= 5
        return 0;
  
    else {
  
        // If non of the conditions are true, we
        // iterate from A+1 to B and multiply them. 
        // We are only concerned for the last digit,
        // thus we take modulus of 10
        for (long i = A + 1; i <= B; i++)
            variable = (int)(variable * (i % 10)) % 10;
  
        return variable % 10;
    }
}
  
// driver function
public static void main(String[] args)
{
    System.out.println(computeLastDigit(2632, 2634));
}
}
  
// This article is contributed by Prerna Saini

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python program to find
# last digit of a number 
# obtained by dividing
# factorial of a number
# with factorial of another number.
  
# Function which computes
# the last digit
# of resultant of B!/A!
def computeLastDigit(A,B):
  
    variable = 1
    if (A == B): # If A = B, B! = A! and B!/A! = 1
        return 1
   
    # If difference (B - A) >= 5, answer = 0
    elif ((B - A) >= 5): 
        return 0
   
    else
   
        # If non of the conditions
        # are true, we
        # iterate from  A+1 to B
        # and multiply them. 
        # We are only concerned
        # for the last digit,
        # thus we take modulus of 10
        for i in range(A + 1, B + 1):
            variable = (variable * (i % 10)) % 10
   
        return variable % 10
      
# driver function
  
print(computeLastDigit(2632, 2634))
  
# This code is contributed
# by Anant Agarwal.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find last digit of
// a number obtained by dividing 
// factorial of a number with
// factorial of another number.
using System;
  
class GFG {
  
// Function which computes the last 
// digit of resultant of B!/A!
static int computeLastDigit(long A, long B)
{
  
    int variable = 1;
     // If A = B, B! = A! 
     // and B!/A! = 1
     if (A == B)
        return 1;
  
    // If difference (B - A) >= 5,
    // answer = 0
    else if ((B - A) >= 5) 
        return 0;
  
    else {
  
        // If non of the conditions are true, we
        // iterate from A+1 to B and multiply them. 
        // We are only concerned for the last digit,
        // thus we take modulus of 10
        for (long i = A + 1; i <= B; i++)
            variable = (int)(variable * 
                       (i % 10)) % 10;
  
        return variable % 10;
    }
}
  
// Driver Code
public static void Main()
{
    Console.WriteLine(computeLastDigit(2632, 2634));
}
}
  
// This code is contributed by vt_m.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find last digit of a number 
// obtained by dividing factorial of a number
// with factorial of another number.
  
// Function which computes the last 
// digit of resultant of B!/A!
function computeLastDigit($A, $B)
{
  
    $variable = 1;
      
    // If A = B, B! = A! 
    // and B!/A! = 1
    if ($A == $B
        return 1;
  
    // If difference (B - A) >= 5,
    // answer = 0
    else if (($B - $A) >= 5) 
        return 0;
  
    else
    {
  
        // If non of the conditions 
        // are true, we iterate from
        // A+1 to B and multiply them. 
        // We are only concerned for
        // the last digit, thus we 
        // take modulus of 10
        for ($i = $A + 1; $i <= $B; $i++)
            $variable = ($variable * ($i % 10)) % 10;
  
        return $variable % 10;
    }
}
  
    // Driver Code
    echo computeLastDigit(2632, 2634);
  
// This code is contributed by ajit
?>

chevron_right



Output:

2

This article is contributed by Shivani Mittal. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : vt_m, jit_t



Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.