# Sum of all N-digit palindromic numbers which doesn’t contains 0 and are divisible by 9

Given a number N, the task is to find the sum of all N-digit palindromic numbers which are divisible by 9 and the number doesn’t contain 0 in it.

Note: The sum can be very large, take modulo with 109+7.

Example:

Input: N = 2
Output: 99
Explanation:
There is only one 2-digit palindromic number divisible by 9 is 99.

Input: N = 3
Output: 4995

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

Naive Approach: The idea is to iterate over all the N-digit numbers and check whether the numbers are palindromic and divisible by 9 and doesn’t contain zero in it. If yes then the summation of all the numbers given the required result.

Efficient Approach: Below are the some observation for this problem statement:

1. If N is odd, then the number can be of the form “ab..x..ba” and if N is even then the number can be “ab..xx..ba”, where ‘a’, ‘b’ and ‘x’ can be replaced by digit from 1 to 9.
2. If number “ab..x..ba” divisible by 9, (2*(a+b) + x) must be divisible by 9.
3. For every possible pair of (a, b) there always exist one such ‘x’ so that the number is divisible by 9.
4. Then the count of N-digit palindromic numbers which are divisible by 9 can be calculated from below formula:
```Since we have 9 digits to fill at the position a and b.
Therefore, total possible combinations will be:
if N is Odd then, count = pow(9, N/2)
if N is Even then, count = pow(9, N/2 - 1)
as N/2 digits are repeated at the end to get the
number palindromic.
```

Proof of Uniqueness of ‘x’:

• Minimum value of a and b is 1, therefore minimum sum = 2.
• Maximum value of a and b is 9, therefore minimum sum = 18.
• As the numbers are palindromic, then the sum is given by:
```sum = 2*(a+b) + x;
```
• 2*(a+b) will be of the form 2, 4, 6, 8, …, 36. To make sum divisible by 9 the value of x can be:
```sum    one of the possible value of x    sum + x
2                   7                      9
4                   5                      9
6                   3                      9
8                   1                      9
.                   .                      .
.                   .                      .
.                   .                      .
36                  9                      45
```

Below are the steps:

1. After the above observations, the sum of the digit of all N-digit palindromic number which is divisible by 9 at any kth position is given by:
```sum at kth position = 45*(number of combinations)/9
```
2. Iterate a loop over [1, N] and find the number of possible combination to placed a digit at current index.
3. Find the sum of all the digits at current digit using the above mentioned formula.
4. The summation of all the sum calculated at each iteration given the required result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `long` `long` `int` `MOD = 1000000007; ` ` `  `// Function to find a^b efficiently ` `long` `long` `int` `power(``long` `long` `int` `a, ` `                    ``long` `long` `int` `b) ` `{ ` ` `  `    ``// Base Case ` `    ``if` `(b == 0) ` `        ``return` `1; ` ` `  `    ``// To store the value of a^b ` `    ``long` `long` `int` `temp = power(a, b / 2); ` ` `  `    ``temp = (temp * temp) % MOD; ` ` `  `    ``// Multiply temp by a until b is not 0 ` `    ``if` `(b % 2 != 0) { ` `        ``temp = (temp * a) % MOD; ` `    ``} ` ` `  `    ``// Return the final ans a^b ` `    ``return` `temp; ` `} ` ` `  `// Function to find sum of all N-digit ` `// palindromic number divisible by 9 ` `void` `palindromicSum(``int` `N) ` `{ ` ` `  `    ``long` `long` `int` `sum = 0, res, ways; ` ` `  `    ``// Base Case ` `    ``if` `(N == 1) { ` `        ``cout << ``"9"` `<< endl; ` `        ``return``; ` `    ``} ` ` `  `    ``if` `(N == 2) { ` `        ``cout << ``"99"` `<< endl; ` `        ``return``; ` `    ``} ` ` `  `    ``ways = N / 2; ` ` `  `    ``// If N is even, decrease ways by 1 ` `    ``if` `(N % 2 == 0) ` `        ``ways--; ` ` `  `    ``// Find the total number of ways ` ` `  `    ``res = power(9, ways - 1); ` ` `  `    ``// Iterate over [1, N] and find the ` `    ``// sum at each index ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``sum = sum * 10 + 45 * res; ` `        ``sum %= MOD; ` `    ``} ` ` `  `    ``// Print the final Sum ` `    ``cout << sum << endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `N = 3; ` `    ``palindromicSum(N); ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `class` `GFG{ ` ` `  `static` `int` `MOD = ``1000000007``; ` ` `  `// Function to find a^b efficiently ` `static` `int` `power(``int` `a, ``int` `b) ` `{ ` ` `  `    ``// Base Case ` `    ``if` `(b == ``0``) ` `        ``return` `1``; ` ` `  `    ``// To store the value of a^b ` `    ``int` `temp = power(a, b / ``2``); ` `    ``temp = (temp * temp) % MOD; ` ` `  `    ``// Multiply temp by a until b is not 0 ` `    ``if` `(b % ``2` `!= ``0``)  ` `    ``{ ` `        ``temp = (temp * a) % MOD; ` `    ``} ` ` `  `    ``// Return the final ans a^b ` `    ``return` `temp; ` `} ` ` `  `// Function to find sum of all N-digit ` `// palindromic number divisible by 9 ` `static` `void` `palindromicSum(``int` `N) ` `{ ` `    ``int` `sum = ``0``, res, ways; ` ` `  `    ``// Base Case ` `    ``if` `(N == ``1``) ` `    ``{ ` `        ``System.out.print(``"9"` `+ ``"\n"``); ` `        ``return``; ` `    ``} ` ` `  `    ``if` `(N == ``2``) ` `    ``{ ` `        ``System.out.print(``"99"` `+ ``"\n"``); ` `        ``return``; ` `    ``} ` `    ``ways = N / ``2``; ` ` `  `    ``// If N is even, decrease ways by 1 ` `    ``if` `(N % ``2` `== ``0``) ` `        ``ways--; ` ` `  `    ``// Find the total number of ways ` `    ``res = power(``9``, ways - ``1``); ` ` `  `    ``// Iterate over [1, N] and find the ` `    ``// sum at each index ` `    ``for` `(``int` `i = ``0``; i < N; i++)  ` `    ``{ ` `        ``sum = sum * ``10` `+ ``45` `* res; ` `        ``sum %= MOD; ` `    ``} ` ` `  `    ``// Print the final Sum ` `    ``System.out.print(sum + ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``3``; ` `    ``palindromicSum(N); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `static` `int` `MOD = 1000000007; ` ` `  `// Function to find a^b efficiently ` `static` `int` `power(``int` `a, ``int` `b) ` `{ ` ` `  `    ``// Base Case ` `    ``if` `(b == 0) ` `        ``return` `1; ` ` `  `    ``// To store the value of a^b ` `    ``int` `temp = power(a, b / 2); ` `    ``temp = (temp * temp) % MOD; ` ` `  `    ``// Multiply temp by a until b is not 0 ` `    ``if` `(b % 2 != 0)  ` `    ``{ ` `        ``temp = (temp * a) % MOD; ` `    ``} ` ` `  `    ``// Return the readonly ans a^b ` `    ``return` `temp; ` `} ` ` `  `// Function to find sum of all N-digit ` `// palindromic number divisible by 9 ` `static` `void` `palindromicSum(``int` `N) ` `{ ` `    ``int` `sum = 0, res, ways; ` ` `  `    ``// Base Case ` `    ``if` `(N == 1) ` `    ``{ ` `        ``Console.Write(``"9"` `+ ``"\n"``); ` `        ``return``; ` `    ``} ` ` `  `    ``if` `(N == 2) ` `    ``{ ` `        ``Console.Write(``"99"` `+ ``"\n"``); ` `        ``return``; ` `    ``} ` `    ``ways = N / 2; ` ` `  `    ``// If N is even, decrease ways by 1 ` `    ``if` `(N % 2 == 0) ` `        ``ways--; ` ` `  `    ``// Find the total number of ways ` `    ``res = power(9, ways - 1); ` ` `  `    ``// Iterate over [1, N] and find the ` `    ``// sum at each index ` `    ``for``(``int` `i = 0; i < N; i++)  ` `    ``{ ` `       ``sum = sum * 10 + 45 * res; ` `       ``sum %= MOD; ` `    ``} ` ` `  `    ``// Print the readonly sum ` `    ``Console.Write(sum + ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `N = 3; ` `     `  `    ``palindromicSum(N); ` `} ` `} ` ` `  `// This code is contributed by AbhiThakur `

Output:

```4995
```

Time Complexity: O(N), where N is the given number. My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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