# 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) = 10 ^{N} – 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

**Approach** There are 2 cases:

If **n is odd** then answer will be 0.

Let

n = 3thennum = d1d2d3andrev(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 = 4thennum = d1d2d3d4andrev(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 be9 * 10.^{(N / 2 – 1)}

Below is the implementation of the above approach

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `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; ` `} ` |

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## 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)); ` ` ` `} ` `} ` |

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## 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 ` |

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## 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)); ` ` ` `} ` `} ` |

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## PHP

**Output:**

9

**Time Complexity:** O(1)

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