We are given a number N. The task is to find the count of numbers which have N digits and odd number of zeroes.

**Note**: The number can have preceding 0’s.

**Examples**:

Input : N = 2 Output : Count = 18 Input : N = 3 Output : Count = 244

Suppose a number with N digits which contains only single zero. So the digits in the number as zero can be filled in only 1 way and the rest of each of the positions can be filled in 9 different ways with numbers from 1 to 9. So count of all such numbers with N digits and only 1 zero = ** ^{N}C_{1}*(9^{N-1})**.

Similarily, count of all such numbers with N digits and 3 zeroes = ** ^{N}C_{3}*(9^{N-3})**.

and so on.

So, count of all such numbers with N digits and odd number of zeroes will be,

+^{N}C_{1}*(9^{N-1})+^{N}C_{3}*(9^{N-3})+…….+^{N}C_{5}*(9^{N-5})^{N}C_{K}*(9^{N-K})Where, K is an odd number less than N.

The above equation can be written as,

(9^{N})((^{N}C_{1}* (1/9)) + (^{N}C_{3}* (1/9^3)) + (^{N}C_{5}* (1/9^5)) +…

The above equation can be represented as subtraction of two series, (9^{N})*{(1+x)^{N}-(1-x)^{N}}/2, where **x = 1/9**

Which is equal to,

(10^{N}- 8^{N})/2

Below is the implementation of the above approach:

## C++

`// C++ program to count numbers with N digits ` `// which consists of odd number of 0's ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to count Numbers with N digits ` `// which consists of odd number of 0's ` `int` `countNumbers(` `int` `N) ` `{ ` ` ` `return` `(` `pow` `(10, N) - ` `pow` `(8, N)) / 2; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 5; ` ` ` ` ` `cout << countNumbers(n) << endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to count numbers ` `// with N digits which consists ` `// of odd number of 0's ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to count Numbers ` `// with N digits which consists ` `// of odd number of 0's ` `static` `int` `countNumbers(` `int` `N) ` `{ ` ` ` `return` `(` `int` `)(Math.pow(` `10` `, N) - ` ` ` `Math.pow(` `8` `, N)) / ` `2` `; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` `int` `n = ` `5` `; ` ` ` `System.out.println(countNumbers(n)); ` `} ` `} ` ` ` `// This code is contributed by Shashank ` |

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## Python 3

`# Python 3 program to count numbers ` `# with N digits which consists of ` `# odd number of 0's ` ` ` `# Function to count Numbers with ` `# N digits which consists of odd ` `# number of 0's ` `def` `countNumbers( N): ` ` ` ` ` `return` `(` `pow` `(` `10` `, N) ` `-` `pow` `(` `8` `, N)) ` `/` `/` `2` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `n ` `=` `5` ` ` ` ` `print` `(countNumbers(n)) ` ` ` `# This code is contributed ` `# by ChitraNayal ` |

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## C#

`// C# program to count numbers ` `// with N digits which consists ` `// of odd number of 0's ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to count Numbers ` `// with N digits which consists ` `// of odd number of 0's ` `static` `int` `countNumbers(` `int` `N) ` `{ ` ` ` `return` `(` `int` `)(Math.Pow(10, N) - ` ` ` `Math.Pow(8, N)) / 2; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main () ` `{ ` ` ` `int` `n = 5; ` ` ` `Console.WriteLine(countNumbers(n)); ` `} ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai(Abby_akku) ` |

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## PHP

`<?php ` `// PHP program to count numbers ` `// with N digits which consists ` `// of odd number of 0's ` ` ` `// Function to count Numbers ` `// with N digits which consists ` `// of odd number of 0's ` `function` `countNumbers(` `$N` `) ` `{ ` ` ` `return` `(pow(10, ` `$N` `) - ` ` ` `pow(8, ` `$N` `)) / 2; ` `} ` ` ` `// Driver code ` `$n` `= 5; ` ` ` `echo` `countNumbers(` `$n` `) ; ` ` ` `// This code is contributed ` `// by Shivi_Aggarwal ` `?> ` |

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**Output:**

33616

**Note**: Answer can be very large, so for N greater than 9, use modular exponentiation.

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