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

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

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,

First find the count of N-digit Palindromic numbers divisible by 9, as:
$\text{Count of N-digit Palindromic numbers divisible by 9} = \begin{cases} 9^{\frac{N-1}{2}} & \text{ if N is odd} \\ 9^{\frac{N-2}{2}} & \text{ if N is even} \end{cases}$
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.
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 

Output:

My Personal Notes

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

C++

Java

Python3

C#

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