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

Last Updated : 22 Jun, 2022

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 ` `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

 ``

## Javascript

 `  ```

Output:

`8`

Time Complexity: O(1).

Auxiliary Space: O(1)