# 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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