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Count of N digit palindromic numbers divisible by 9
  • Difficulty Level : Medium
  • Last Updated : 08 Jun, 2020

Given an integer N, the task is to count the number of N digit palindromic numbers containing digits from 1 to 9 and divisible by 9.

Examples:

Input: N = 1
Output: 1
Explanation:
Only 9 is 1 digit number which is palindrome and divisible by 9.

Input: N = 3
Output: 9
Explanation:
Three digit numbers those are palindrome and divisible by 9 are –
{171, 252, 333, 414, 585, 666, 747, 828, 999}

Approach: The key observation in the problem is if the number is divisible by 9 then sum of digits of the number is also divisible by 9. Therefore, the problem can be segreated on the basis of its parity.

  • If N is odd: We can put any number from 1 to 9 in position 1 to (N-1)/2, Similarly, the other digits are chosen in reverse order to form palindromic number and the middle element is chosen on to form the sum of digits divisible by 9. Therefore, there are 9 choices for each position of (N-1)/2 digits of the number due to which the count of such number will be:
    Count of N-digit Palindromic numbers =
                     9(N-1)/2
    
  • If N is even: We can put any number from 1 to 9 at the position from 1 to (N-2)/2, Similarly, the other digits are chosen in reverse order to form palindromic number and the middle element is chosen on to form the sum of digits divisible by 9. Therefore, there are 9 choices for each position of (N-2)/2 digits of the number due to which the count of such number will be:
    Count of N-digit Palindromic numbers =
                     9(N-2)/2
    

C++

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// C++ implementation to count the
// number of N digit palindromic
// numbers divisible by 9
  
#include <bits/stdc++.h>
  
using namespace std;
  
// Function to find the count of 
// N digits palindromic numbers 
// which are divisible by 9
int countPalindromic(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;
}
  
// Driver Code
int main()
{
    int n = 3;
    cout << countPalindromic(n);
    return 0;
}

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Java

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// Java implementation to count the
// number of N digit palindromic
// numbers divisible by 9
import java.util.*;
  
class GFG{
      
// Function to find the count of 
// N digits palindromic numbers 
// which are divisible by 9
static int countPalindromic(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;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 3;
    System.out.println(countPalindromic(n));
}
}
  
// This code is contributed by ANKITKUMAR34

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Python3

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# Python3 implementation to count the
# number of N digit palindromic
# numbers divisible by 9
  
# Function to find the count of 
# N digits palindromic numbers 
# which are divisible by 9
def countPalindromic(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
  
# Driver Code
n = 3
print(countPalindromic(n))
  
# This code is contributed by ANKITKUMAR34

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C#

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// C# implementation to count the
// number of N digit palindromic
// numbers divisible by 9
using System;
class GFG{
      
// Function to find the count of 
// N digits palindromic numbers 
// which are divisible by 9
static int countPalindromic(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;
}
  
// Driver Code
public static void Main()
{
    int n = 3;
    Console.Write(countPalindromic(n));
}
}
  
// This code is contributed by Nidhi_biet

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Output:

9

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