Program to find the last digit of X in base Y

Given a positive integer X and Y, the task is to find the last digit of X in the given base Y.

**Examples:**

Input:X = 10, Y = 7Output:3 10 is 13 in base 9 with last digit 3Input:X = 55, Y = 3Output:1 55 is 3 in base 601 with last digit 1

**Approach:**

- When we try to convert X into the base Y
- We repeatedly divide X by base Y and store the remainder.
- So final result comprises of the remainders in the order of division steps.
- Lets say the remainder of division step 1 is p, step 2 is q, step 3 is r
- Then the resultant number in base Y will be rqp
- And the last digit will be p
- Therefore, we just need to find the first remainder of X when divided by Y to get the last digit in X in base Y.

last digit = X % Y

Below is the implementation of the above approach:

## C++

`// C++ Program to find` `// the last digit of X in base Y` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the last` `// digit of X in base Y` `void` `last_digit(` `int` `X, ` `int` `Y)` `{` ` ` `cout << X % Y;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `X = 55, Y = 3;` ` ` `last_digit(X, Y);` ` ` `return` `0;` `}` |

## Java

`// Java Program to find` `// the last digit of X in base Y` `class` `GFG` `{` `// Function to find the last` `// digit of X in base Y` `static` `void` `last_digit(` `int` `X, ` `int` `Y)` `{` ` ` `System.out.print(X % Y);` `}` `// Driver code` `public` `static` `void` `main(String []args)` `{` ` ` `int` `X = ` `55` `, Y = ` `3` `;` ` ` `last_digit(X, Y);` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 Program to find` `# the last digit of X in base Y` `# Function to find the last` `# digit of X in base Y` `def` `last_digit(X, Y) :` ` ` `print` `(X ` `%` `Y);` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `X ` `=` `55` `; Y ` `=` `3` `;` ` ` `last_digit(X, Y);` `# This code is contributed` `# by AnkitRai01` |

## C#

`// C# Program to find the last digit` `// of X in base Y` `using` `System;` `class` `GFG` `{` `// Function to find the last` `// digit of X in base Y` `static` `void` `last_digit(` `int` `X, ` `int` `Y)` `{` ` ` `Console.Write(X % Y);` `}` `// Driver code` `public` `static` `void` `Main(String []args)` `{` ` ` `int` `X = 55, Y = 3;` ` ` `last_digit(X, Y);` `}` `}` `// This code is contributed by Rajput-Ji` |

## PHP

`<?php` `// PHP Program to find the last digit` `// of X in base Y` ` ` ` ` `// Function to find the last` `// digit of X in base Y` `function` `last_digit(` `$X` `,` `$Y` `){` ` ` ` ` `echo` `( ` `$X` `% ` `$Y` `);` ` ` `}` ` ` `// Driver code` `$X` `= 55;` `$Y` `= 3;` `last_digit(` `$X` `,` `$Y` `);` ` ` `?>` |

## Javascript

`<script>` `// Javascript Program to find` `// the last digit of X in base Y` `// Function to find the last` `// digit of X in base Y` `function` `last_digit(X, Y)` `{` ` ` `document.write(X % Y);` `}` `// Driver code` `var` `X = 55, Y = 3;` `last_digit(X, Y);` `</script>` |

**Output:**

1

**Time Complexity:** O(1)

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