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
   

   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

   Output:

   4995
   

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