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Find the count of natural Hexadecimal numbers of size N

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  • Last Updated : 19 Mar, 2022

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 
 

 

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 <bits/stdc++.h>
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


Javascript




<script>
 
// JavaScript implementation of the above approach
 
// Function to return the count of n-digit
// natural hexadecimal numbers
function count(n)
{
    return 15 * Math.pow(16, n - 1);
}
 
// Driver code
var n = 2;
document.write(count(n));
 
</script>


Output: 

240

 

Time Complexity: O(log n)
 


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