Count of Octal numbers upto N digits

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

Examples:
Input: N = 1
Output: 7
Explanation:
1, 2, 3, 4, 5, 6, 7 are 1 digit Natural Octal numbers.

Input: N = 2
Output: 63
Explanation:
There are a total of 56 two digit octal numbers and 7 one digit octal numbers. Therefore, 56 + 7 = 63.

Approach: On observing carefully, a geometric progression series is formed [ 7 56 448 3584 28672 229376… ] with the first term being 7 and a common ratio of 8.

Therefore,



Nth term = Number of Octal numbers of N digits = 7 * 8N - 1

Finally, count of all octal numbers up to N digits can be found out by iterating a loop from 1 to N and calculating the sum of ith term using the above formula.

Below is the implementation of the above approach:

C++

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// C++ program to find the count of
// natural octal numbers upto N digits
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of
// natural octal numbers upto N digits
int count(int N)
{
    int sum = 0;
  
    // Loop to iterate from 1 to N
    // and calculating number of
    // octal numbers for every 'i'th digit.
    for (int i = 1; i <= N; i++) {
        sum += 7 * pow(8, i - 1);
    }
    return sum;
}
  
// Driver code
int main()
{
    int N = 4;
    cout << count(N);
  
    return 0;
}

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Java

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// Java program to find the count of 
// natural octal numbers upto N digits 
  
public class GFG {
      
    // Function to return the count of 
    // natural octal numbers upto N digits 
    static int count(int N) 
    
        int sum = 0
      
        // Loop to iterate from 1 to N 
        // and calculating number of 
        // octal numbers for every 'i'th digit. 
        for (int i = 1; i <= N; i++) { 
            sum += 7 * Math.pow(8, i - 1); 
        
        return sum; 
    
      
    // Driver code 
    public static void main (String[] args) 
    
        int N = 4
        System.out.println(count(N)); 
      
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 program to find the count of 
# natural octal numbers upto N digits 
  
# Function to return the count of 
# natural octal numbers upto N digits 
def count(N) : 
  
    sum = 0
  
    # Loop to iterate from 1 to N 
    # and calculating number of 
    # octal numbers for every 'i'th digit. 
    for i in range(N + 1) :
        sum += 7 * (8 **(i - 1)); 
      
    return int(sum);
  
# Driver code 
if __name__ == "__main__"
  
    N = 4
    print(count(N)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# program to find the count of 
// natural octal numbers upto N digits 
using System;
  
class GFG
{
      
    // Function to return the count of 
    // natural octal numbers upto N digits 
    static int count(int N) 
    
        int sum = 0; 
      
        // Loop to iterate from 1 to N 
        // and calculating number of 
        // octal numbers for every 'i'th digit. 
        for (int i = 1; i <= N; i++)
        
            sum += (int)(7 * Math.Pow(8, i - 1)); 
        
        return sum; 
    
      
    // Driver code 
    public static void Main ()
    
        int N = 4; 
        Console.WriteLine(count(N)); 
    
}
  
// This code is contributed by AnkitRai01

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Output:

4095

Time Complexity: O(N)

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