# 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 = 567891233156156Output:4Input:a = 123, b = 456Output: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. Participate in the **Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students**.