Given a number **A** in decimal base, the task is to find the **N ^{th}** digit from last in base

**B**

**Examples:**

Input:A = 100, N = 3, B = 4Output:2Explanation:

(100)_{4}= 1210

3^{rd}digit from last is 2Input:A = 50, N = 3, B = 5Output:2Explanation:

(50)_{5}= 200

3^{rd}digit from last is 2

**Approach:** The idea is to skip (N-1) digits of the given number in base B by dividing the number with B, (N – 1) times and then return the modulo of the current number by the B to get the N^{th} digit from the right.

Below is the implementation of the above approach:

## C++

`// C++ Implementation to find Nth digit` `// from right in base B` `#include <iostream>` `using` `namespace` `std;` `// Function to compute Nth digit` `// from right in base B` `int` `nthDigit(` `int` `a, ` `int` `n, ` `int` `b)` `{` ` ` `// Skip N-1 Digits in Base B` ` ` `for` `(` `int` `i = 1; i < n; i++)` ` ` `a = a / b;` ` ` `// Nth Digit from right in Base B` ` ` `return` `a % b;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `a = 100;` ` ` `int` `n = 3;` ` ` `int` `b = 4;` ` ` `cout << nthDigit(a, n, b);` ` ` `return` `0;` `}` |

## Java

`// Java Implementation to find Nth digit` `// from right in base B` `import` `java.util.*;` `class` `GFG` `{` `// Function to compute Nth digit` `// from right in base B` `static` `int` `nthDigit(` `int` `a, ` `int` `n, ` `int` `b)` `{` ` ` `// Skip N-1 Digits in Base B` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++)` ` ` `a = a / b;` ` ` `// Nth Digit from right in Base B` ` ` `return` `a % b;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `a = ` `100` `;` ` ` `int` `n = ` `3` `;` ` ` `int` `b = ` `4` `;` ` ` `System.out.print(nthDigit(a, n, b));` `}` `}` `// This code is contributed by PrinciRaj1992` |

## Python3

`# Ptyhon3 Implementation to find Nth digit` `# from right in base B` `# Function to compute Nth digit` `# from right in base B` `def` `nthDigit(a, n, b):` ` ` `# Skip N-1 Digits in Base B` ` ` `for` `i ` `in` `range` `(` `1` `, n):` ` ` `a ` `=` `a ` `/` `/` `b` ` ` `# Nth Digit from right in Base B` ` ` `return` `a ` `%` `b` `# Driver Code` `a ` `=` `100` `n ` `=` `3` `b ` `=` `4` `print` `(nthDigit(a, n, b))` `# This code is contributed by ApurvaRaj` |

## C#

`// C# Implementation to find Nth digit` `// from right in base B` `using` `System;` `class` `GFG` `{` ` ` `// Function to compute Nth digit` ` ` `// from right in base B` ` ` `static` `int` `nthDigit(` `int` `a, ` `int` `n, ` `int` `b)` ` ` `{` ` ` ` ` `// Skip N-1 Digits in Base B` ` ` `for` `(` `int` `i = 1; i < n; i++)` ` ` `a = a / b;` ` ` ` ` `// Nth Digit from right in Base B` ` ` `return` `a % b;` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `a = 100;` ` ` `int` `n = 3;` ` ` `int` `b = 4;` ` ` `Console.Write(nthDigit(a, n, b));` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` `// Javascript Implementation to find Nth digit` `// from right in base B` `// Function to compute Nth digit` `// from right in base B` `function` `nthDigit(a, n, b)` `{` ` ` `// Skip N-1 Digits in Base B` ` ` `for` `(` `var` `i = 1; i < n; i++)` ` ` `a = parseInt(a / b);` ` ` `// Nth Digit from right in Base B` ` ` `return` `a % b;` `}` `// Driver Code` `var` `a = 100;` `var` `n = 3;` `var` `b = 4;` `document.write(nthDigit(a, n, b));` `// This code is contributed by rutvik_56.` `</script>` |

**Output:**

2

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