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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 ` `

Output:

```8
```

Time Complexity: O(1).

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Improved By : 29AjayKumar, ihritik