# Find the total count of numbers up to N digits in a given base B

• Last Updated : 04 Jun, 2022

Given two integers N and B, the task is to find the count of natural numbers of Base B up to N digits.

Examples:

Input: N = 2, B = 10
Output: 99
Explanation:
1, 2, 3, 4, 5, 6, 7, 8, 9 are 1 digit Natural numbers of Base 10.
10, 11, 12………99 are 2 digit Natural numbers of Base 10
So, total = 9 + 90 = 99

Input: N = 2, B = 16
Output: 255
Explanation:
There are a total of 240 two digit hexadecimal numbers and 15 one digit hexadecimal numbers.
Therefore, 240 + 15 = 255.

Approach: On observing carefully the count of numbers with N digits in base B is a geometric progression formed with the first term being (B – 1) and a common ratio of B.
Therefore,

Nth term = Number of natural numbers of N digits in Base B = (B – 1) * BN – 1

Finally, count of all natural numbers in Base B 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++ implementation to find the count``// of natural numbers upto N digits` `#include ` `using` `namespace` `std;` `// Function to return the count of``// natural numbers upto N digits``int` `count(``int` `N, ``int` `B)``{``    ``int` `sum = 0;` `    ``// Loop to iterate from 1 to N``    ``// and calculating number of``    ``// natural numbers for every 'i'th digit.``    ``for` `(``int` `i = 1; i <= N; i++) {``        ``sum += (B - 1) * ``pow``(B, i - 1);``    ``}``    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``int` `N = 2, B = 10;``    ``cout << count(N, B);` `    ``return` `0;``}`

## Java

 `// Java implementation to find the count``// of natural numbers upto N digits` `class` `GFG{` `// Function to return the count of``// natural numbers upto N digits``static` `int` `count(``int` `N, ``int` `B)``{``    ``int` `sum = ``0``;` `    ``// Loop to iterate from 1 to N``    ``// and calculating number of``    ``// natural numbers for every 'i'th digit.``    ``for` `(``int` `i = ``1``; i <= N; i++){``        ``sum += (B - ``1``) * Math.pow(B, i - ``1``);``    ``}``    ``return` `sum;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``2``, B = ``10``;``    ``System.out.print(count(N, B));``}``}` `// This code is contributed by gauravrajput1`

## Python3

 `# Python3 implementation to find the count``# of natural numbers up to N digits` `from` `math ``import` `pow` `# Function to return the count of``# natural numbers upto N digits``def` `count(N, B):``    ``sum` `=` `0` `    ``# Loop to iterate from 1 to N``    ``# and calculating number of``    ``# natural numbers for every 'i'th digit.``    ``for` `i ``in` `range``(``1``, N``+``1``):``        ``sum` `+``=` `(B ``-` `1``) ``*` `pow``(B, i ``-` `1``)``    ``return` `sum` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `2``    ``B ``=` `10``    ``print``(``int``(count(N, B)))` `# This code is contributed by Bhupendra_Singh`

## C#

 `// C# implementation to find the count``// of natural numbers upto N digits``using` `System;``using` `System.Collections.Generic;``class` `GFG{` `// Function to return the count of``// natural numbers upto N digits``static` `int` `count(``int` `N, ``int` `B)``{``    ``int` `sum = 0;` `    ``// Loop to iterate from 1 to N``    ``// and calculating number of``    ``// natural numbers for every``    ``// 'i'th digit.``    ``for``(``int` `i = 1; i <= N; i++)``    ``{``       ``sum += (``int``)((B - 1) * Math.Pow(B, i - 1));``    ``}``    ``return` `sum;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 2, B = 10;``    ` `    ``Console.Write(count(N, B));``}``}` `// This code is contributed by amal kumar choubey`

## Javascript

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Output:

`99`

Time Complexity: O(N)

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