# Count of all N digit numbers such that num + Rev(num) = 10^N – 1

Given an integer N, the task is to find the count of all N digit numbers such that num + Rev(num) = 10N – 1

Examples:

Input: N = 2
Output: 9
All possible numbers are
18 + 81 = 99
27 + 72 = 99
36 + 45 = 99
45 + 54 = 99
54 + 45 = 99
63 + 54 = 99
72 + 27 = 99
81 + 18 = 99
90 + 09 = 99

Input: N = 4
Output: 90

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

Approach There are 2 cases:
If n is odd then answer will be 0.

Let n = 3 then num = d1d2d3 and rev(num) = d3d2d1
num + rev(num) should be 999 since n = 3.
So the below equations must be satisfied
d1 + d3 = 9 … (1)
d2 + d2 = 9 … (2)
d3 + d1 = 9 … (3)
Considering equation 2:
d2 + d2 = 9
2 * d2 = 9
d2 = 4.5 which is not possible because the digit of a number should always be a whole number.
Therefore If n is odd then answer will be 0.

If n is even then answer will be 9 * 10(N / 2 – 1).

Let n = 4 then num = d1d2d3d4 and rev(num) = d4d3d2d1
So the below equations should be satisfied
d1 + d4 = 9 … (1)
d2 + d3 = 9 … (2)
d3 + d2 = 9 … (3)
d4 + d1 = 9 … (4)

Considering equation 1: d1 + d4 = 9. It can be true in 9 ways:
(1 + 8), (2 + 7), (3 + 6), (4 + 5), (5 + 4), (6 + 3), (7 + 2), (8 + 1) and (9 + 0)
Similarly other equations will also have 9 solutions + 1 more solution since the remaining digits are not the first and last digit of the number and we can take sum of the form (0 + 9).
And since half of the equations are same
Therefore, if n is even then answer will be 9 * 10(N / 2 – 1).

Below is the implementation of the above approach

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// count of such numbers ` `int` `countNumbers(``int` `n) ` `{ ` ` `  `    ``// If n is odd ` `    ``if` `(n % 2 == 1) ` `        ``return` `0; ` ` `  `    ``return` `(9 * ``pow``(10, n / 2 - 1)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 2; ` `    ``cout << countNumbers(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function to return the ` `    ``// count of such numbers ` `    ``static` `int` `countNumbers(``int` `n) ` `    ``{ ` ` `  `        ``// If n is odd ` `        ``if` `(n % ``2` `== ``1``) ` `            ``return` `0``; ` ` `  `        ``return` `(``9` `* (``int``)Math.pow(``10``, n / ``2` `- ``1``)); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``2``; ` `        ``System.out.print(countNumbers(n)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the  ` `# count of such numbers  ` `def` `countNumbers(n):  ` ` `  `    ``# If n is odd  ` `    ``if` `n ``%` `2` `=``=` `1``: ` `        ``return` `0` ` `  `    ``return` `(``9` `*` `pow``(``10``, n ``/``/` `2` `-` `1``))  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``n ``=` `2` `    ``print``(countNumbers(n))  ` ` `  `# This code is contributed  ` `# by Rituraj Jain `

## C#

 `// C# implementation of the approach ` `using` `System; ` `class` `GFG { ` ` `  `    ``// Function to return the ` `    ``// count of such numbers ` `    ``static` `int` `countNumbers(``int` `n) ` `    ``{ ` ` `  `        ``// If n is odd ` `        ``if` `(n % 2 == 1) ` `            ``return` `0; ` ` `  `        ``return` `(9 * (``int``)Math.Pow(10, n / 2 - 1)); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 2; ` `        ``Console.WriteLine(countNumbers(n)); ` `    ``} ` `} `

## PHP

 `

Output:

```9
```

Time Complexity: O(1)

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : rituraj_jain, Code_Mech