# Find the count of natural Hexadecimal numbers of size N

Given an integer N, the task is to find the count of natural Hexadecimal numbers with N digits.

Examples:

Input: N = 1
Output: 15

Input: N = 2
Output: 240

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

Approach: It can be observed that for the values of N = 1, 2, 3, …, a series will be formed as 15, 240, 3840, 61440, 983040, 15728640, … which is a GP series whose common ratio is 16 and a = 15.

Hence the nth term will be 15 * pow(16, n – 1).

So, the count of n-digit natural hexadecimal numbers will be 15 * pow(16, n – 1).

Below is the implementation of the above approach:

## C++

 // C++ implementation of the above approach #include using namespace std;    // Function to return the count of n-digit  // natural hexadecimal numbers int count(int n) {     return 15 * pow(16, n - 1); }    // Driver code int main() {     int n = 2;     cout << count(n);     return 0; }

## Java

 // Java implementation of the approach class GFG  {    // Function to return the count of n-digit  // natural hexadecimal numbers static int count(int n) {     return (int) (15 * Math.pow(16, n - 1)); }    // Driver code public static void main(String args[])  {     int n = 2;     System.out.println(count(n)); } }    // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation of the above approach     # Function to return the count of n-digit  # natural hexadecimal numbers  def count(n) :         return 15 * pow(16, n - 1);     # Driver code  if __name__ == "__main__" :         n = 2;      print(count(n));        # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approach using System;        class GFG  {        // Function to return the count of n-digit      // natural hexadecimal numbers     static int count(int n)     {         return (int) (15 * Math.Pow(16, n - 1));     }            // Driver code     public static void Main(String []args)      {         int n = 2;         Console.WriteLine(count(n));     } }    // This code is contributed by 29AjayKumar

Output:

240

Time Complexity: O(1)

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Improved By : AnkitRai01, 29AjayKumar

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