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Last digit of Product of two Large or Small numbers (a * b)

Last Updated : 22 Jun, 2022
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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;
 
// Function 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
 
    // Function 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).

Auxiliary Space: O(1)
 



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