# 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.

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

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++

 `// C++ program to find the count of ` `// natural octal numbers upto N digits ` ` `  `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

Output:

```4095
```

Time Complexity: O(N)

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