Skip to content
Related Articles

Related Articles

Improve Article
Last digit of Product of two Large or Small numbers (a * b)
  • Difficulty Level : Medium
  • Last Updated : 25 Mar, 2021

Given two large or small numbers, the task is to find the last digit of the product of these two numbers.
Examples: 
 

Input: a = 1234567891233789, b = 567891233156156
Output: 4

Input: a = 123, b = 456
Output: 8

 

Approach: In general, the last digit of multiplication of 2 numbers a and b is the last digit of the product of the LSB of these two numbers. 
For example: For a = 123 and b = 456, 
the last digit of a*b 
= Last digit of ((LSB of a)*(LSB of b)) 
= Last digit of ((3)*(6)) 
= Last digit of (18) 
= 8
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Fthe unction to print last digit of product a * b
void lastDigit(string a, string b)
{
    int lastDig = (a[a.length() - 1] - '0')
                  * (b[b.length() - 1] - '0');
    cout << lastDig % 10;
}
 
// Driver code
int main()
{
    string a = "1234567891233", b = "1234567891233156";
    lastDigit(a, b);
    return 0;
}

Java




// Java implementation of the above approach
 
class Solution {
 
    // Function to print last digit of product a * b
    public static void lastDigit(String a, String b)
    {
        int lastDig = (a.charAt(a.length() - 1) - '0')
                      * (b.charAt(b.length() - 1) - '0');
        System.out.println(lastDig % 10);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String a = "1234567891233", b = "1234567891233156";
        lastDigit(a, b);
    }
}

Python3




# Python3 implementation of the above approach
 
# Function to print the last digit
# of product a * b
def lastDigit(a, b):
    lastDig = ((int(a[len(a) - 1]) - int('0')) *
               (int(b[len(b) - 1]) - int('0')))
    print(lastDig % 10)
 
# Driver code
if __name__ == '__main__':
    a, b = "1234567891233", "1234567891233156"
    lastDigit(a, b)
 
# This code has been contributed
# by 29AjayKumar

C#




// CSharp implementation of the above approach
 
using System;
class Solution {
 
    // Function to print last digit of product a * b
    public static void lastDigit(String a, String b)
    {
        int lastDig = (a[a.Length - 1] - '0')
                      * (b[b.Length - 1] - '0');
 
        Console.Write(lastDig % 10);
    }
 
    // Driver code
    public static void Main()
    {
        String a = "1234567891233", b = "1234567891233156";
        lastDigit(a, b);
    }
}

PHP




<?php
// PHP implementation of the above approach
 
// Function to print last digit of product a * b
function lastDigit($a, $b)
{
    $lastDig = (ord($a[strlen($a) - 1]) - 48) *
               (ord($b[strlen($b) - 1]) - 48);
    echo $lastDig % 10;
}
 
// Driver code
$a = "1234567891233";
$b = "1234567891233156";
lastDigit($a, $b);
 
// This code is contributed by ihritik
?>

Javascript




  <script>
    // Javascript implementation of the above approach
 
    // Fthe unction to print last digit of product a * b
    function lastDigit(a, b) {
      var lastDig = (a[a.length - 1] - '0')
        * (b[b.length - 1] - '0');
      document.write(lastDig % 10);
    }
 
    // Driver code
    var a = "1234567891233", b = "1234567891233156";
    lastDigit(a, b);
 
// This code is contributed by rrrtnx.
  </script>
Output: 
8

 

Time Complexity: O(1).
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes




My Personal Notes arrow_drop_up
Recommended Articles
Page :