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# 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 ``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

 ``

## Javascript

 `  ```
Output:
`8`

Time Complexity: O(1).

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