# Probability that a N digit number is palindrome

Given an integer N, the task is to find the probability that a number with a number of digits as N is a palindrome.
The number may have leading zeros.

Examples:

Input: N = 5
Output: 1 / 100

Input: N = 6
Output: 1 / 1000

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

Solution:

• As leading zeroes are allowed total number of N digit number is 10N.
• A number is a palindrome when first N/2 digits match with last N/2 digits in reverse order.
• For even number of digits, we can pick first N/2 digits and then duplicate them to form the rest of N/2 digits so we can choose (N)/2 digits.
• For an odd number of digits we can pick first (N-1)/2 digits and then duplicate them to form the rest of (N-1)/2 digits so we can choose (N+1)/2 digits.
• So the probability that an N digit number is palindrome is 10ceil( N / 2 ) / 10N or 1 / 10floor( N / 2 )

Below is the implementation of the approach:

## C++

 `// C++ code of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Find the probability that a ` `// n digit number is palindrome ` `void` `solve(``int` `n) ` `{ ` `    ``int` `n_2 = n / 2; ` ` `  `    ``// Denominator ` `    ``string den; ` `    ``den = ``"1"``; ` ` `  `    ``// Assign 10^(floor(n/2)) to ` `    ``// denominator ` `    ``while` `(n_2--) ` `        ``den += ``'0'``; ` ` `  `    ``// Display the answer ` `    ``cout << 1 << ``"/"` `<< den << ``"\n"``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `N = 5; ` ` `  `    ``solve(N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java code of above approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Find the probability that a ` `// n digit number is palindrome ` `static` `void` `solve(``int` `n) ` `{ ` `    ``int` `n_2 = n / ``2``; ` ` `  `    ``// Denominator ` `    ``String den; ` `    ``den = ``"1"``; ` ` `  `    ``// Assign 10^(floor(n/2)) to ` `    ``// denominator ` `    ``while` `(n_2-- > ``0``) ` `        ``den += ``'0'``; ` ` `  `    ``// Display the answer ` `    ``System.out.println(``1` `+ ``"/"` `+ den); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `N = ``5``; ` ` `  `    ``solve(N); ` `} ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 code of above approach  ` ` `  `# Find the probability that a  ` `# n digit number is palindrome  ` `def` `solve(n) :  ` ` `  `    ``n_2 ``=` `n ``/``/` `2``;  ` ` `  `    ``# Denominator  ` `    ``den ``=` `"1"``;  ` ` `  `    ``# Assign 10^(floor(n/2)) to  ` `    ``# denominator  ` `    ``while` `(n_2) :  ` `        ``den ``+``=` `'0'``;  ` `         `  `        ``n_2 ``-``=` `1` `         `  `    ``# Display the answer ` `    ``print``(``str``(``1``) ``+` `"/"` `+` `str``(den)) ` `     `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``N ``=` `5``;  ` ` `  `    ``solve(N);  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Find the probability that a ` `// n digit number is palindrome ` `static` `void` `solve(``int` `n) ` `{ ` `    ``int` `n_2 = n / 2; ` ` `  `    ``// Denominator ` `    ``String den; ` `    ``den = ``"1"``; ` ` `  `    ``// Assign 10^(floor(n/2)) to ` `    ``// denominator ` `    ``while` `(n_2-- > 0) ` `        ``den += ``'0'``; ` ` `  `    ``// Display the answer ` `    ``Console.WriteLine(1 + ``"/"` `+ den); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `N = 5; ` ` `  `    ``solve(N); ` `} ` `}  ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```1/100
```

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