Sum of all N digit palindromic numbers divisible by 9 formed using digits 1 to 9

Given a number N, the task is to find the sum of all N digits palindromic numbers (formed by digits from 1 to 9) that are divisible by 9. Examples:

Input: N = 1 Output: 9 Explanation: Only 9 is a palindromic number of 1 digit divisible by 9 Input: N = 3 Output: 4995 Explanation: Three-digit Palindromic Numbers divisible by 9 are – 171, 252, 333, 414, 585, 666, 747, 828, 999

Approach: The key observation in the problem is that if a number is divisible by 9, then the sum of digits of that number is also divisible by 9. Another observation is if we count the number of N-digit palindromic numbers using the digits from 1 to 9, then it can be observed that

Occurrence of each digit = (count of N-digit numbers / 9)

Therefore,

1. First find the count of N-digit Palindromic numbers divisible by 9, as: 
2. Then if N is 1 or 2, the sum will be simply 9 and 99 respectively, as they are the only palindromic numbers of 1 and 2 digits.
3. If N > 2, then the sum for Nth digit palindromic numbers divisible by 9 is \text{Sum of Nth digit palindromic numbers divisible by 9 }= (\text{sum of }(N-1)^{th}\text{ digit } * 10) + (5*\text{ count of N digit palindromic numbers divisible by 9}) 

Below is the implementation of the above approach:

C++

// C++ implementation to find the sum
// of all the N digit palindromic
// numbers divisible by 9

#include <bits/stdc++.h>

using namespace std;

// Function for finding count of
// N digits palindrome which
// are divisible by 9
int countPalindrome(int n)
{
    int count;

    // if N is odd
    if (n % 2 == 1) {
        count = pow(9, (n - 1) / 2);
    }
    // if N is even
    else {
        count = pow(9, (n - 2) / 2);
    }
    return count;
}

// Function for finding sum of N
// digits palindrome which are
// divisible by 9
int sumPalindrome(int n)
{
    // count the possible
    // number of palindrome
    int count = countPalindrome(n);

    int res = 0;

    if (n == 1)
        return 9;
    if (n == 2)
        return 99;

    for (int i = 0; i < n; i++) {
        res = res * 10 + count * 5;
    }

    return res;
}

// Driver Code
int main()
{
    int n = 3;
    cout << sumPalindrome(n);
    return 0;
}

Java

// Java implementation to find the sum
// of all the N digit palindromic
// numbers divisible by 9
import java.util.*;

class GFG{

// Function for finding count of
// N digits palindrome which
// are divisible by 9
static int countPalindrome(int n)
{
    int count;

    // If N is odd
    if (n % 2 == 1)
    {
        count = (int)Math.pow(9, (n - 1) / 2);
    }

    // If N is even
    else
    {
        count = (int)Math.pow(9, (n - 2) / 2);
    }
    return count;
}

// Function for finding sum of N
// digits palindrome which are
// divisible by 9
static int sumPalindrome(int n)
{

    // Count the possible
    // number of palindrome
    int count = countPalindrome(n);

    int res = 0;

    if (n == 1)
        return 9;
    if (n == 2)
        return 99;

    for(int i = 0; i < n; i++)
    {
    res = res * 10 + count * 5;
    }

    return res;
}

// Driver Code
public static void main(String[] args)
{
    int n = 3;

    System.out.println(sumPalindrome(n));
}
}

// This code is contributed by ANKITKUMAR34

Python3

# Python3 implementation to find the
# sum of all the N digit palindromic
# numbers divisible by 9

# Function for finding count of
# N digits palindrome which
# are divisible by 9
def countPalindrome(n):

    count = 0

    # If N is odd
    if (n % 2 == 1):
        count = pow(9, (n - 1) // 2)

    # If N is even
    else:
        count = pow(9, (n - 2) // 2)

    return count

# Function for finding sum of N
# digits palindrome which are
# divisible by 9
def sumPalindrome(n):

    # Count the possible
    # number of palindrome
    count = countPalindrome(n)

    res = 0

    if (n == 1):
        return 9
    if (n == 2):
        return 99

    for i in range(n):
        res = res * 10 + count * 5

    return res

# Driver Code
n = 3

print(sumPalindrome(n))

# This code is contributed by ANKITKUMAR34

C#

// C# implementation to find the sum
// of all the N digit palindromic
// numbers divisible by 9
using System;

class GFG{

// Function for finding count of
// N digits palindrome which
// are divisible by 9
static int countPalindrome(int n)
{
    int count;

    // If N is odd
    if (n % 2 == 1)
    {
        count = (int)Math.Pow(9, (n - 1) / 2);
    }

    // If N is even
    else
    {
        count = (int)Math.Pow(9, (n - 2) / 2);
    }
    return count;
}

// Function for finding sum of N
// digits palindrome which are
// divisible by 9
static int sumPalindrome(int n)
{

    // Count the possible
    // number of palindrome
    int count = countPalindrome(n);

    int res = 0;

    if (n == 1)
        return 9;
    if (n == 2)
        return 99;

    for(int i = 0; i < n; i++)
    {
    res = res * 10 + count * 5;
    }

    return res;
}

// Driver Code
public static void Main(String[] args)
{
    int n = 3;

    Console.WriteLine(sumPalindrome(n));
}
}

// This code is contributed by Amit Katiyar

Javascript

// JavaScript implementation to find the sum
// of all the N digit palindromic
// numbers divisible by 9

// Function for finding count of
// N digits palindrome which
// are divisible by 9
function countPalindrome(n)
{
    let count;

    // if N is odd
    if (n % 2 == 1) {
        count = Math.pow(9, (n - 1) / 2);
    }
    // if N is even
    else {
        count = Math.pow(9, (n - 2) / 2);
    }
    return count;
}

// Function for finding sum of N
// digits palindrome which are
// divisible by 9
function sumPalindrome(n)
{
    // count the possible
    // number of palindrome
    let count = countPalindrome(n);

    let res = 0;

    if (n == 1)
        return 9;
    if (n == 2)
        return 99;

    for (var i = 0; i < n; i++) {
        res = res * 10 + count * 5;
    }

    return res;
}

// Driver Code
let n = 3;
console.log(sumPalindrome(n));

// This code is contributed by phasing17

Output:

4995

Time Complexity: O(log₉n)

Auxiliary Space: O(1)