Related Articles

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

• Difficulty Level : Expert
• Last Updated : 09 Jul, 2020

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,

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   using namespace std;  // Function for finding count of// N digits palindrome which// are divisible by 9int 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 9int 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 Codeint 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 9import java.util.*;  class GFG{      // Function for finding count of// N digits palindrome which// are divisible by 9static 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 9static 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 Codepublic 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 9def 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 9def 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 Coden = 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 9using System;  class GFG{       // Function for finding count of// N digits palindrome which// are divisible by 9static 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 9static 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 Codepublic static void Main(String[] args){    int n = 3;           Console.WriteLine(sumPalindrome(n));}}  // This code is contributed by Amit Katiyar
Output:
4995


Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up