Largest N digit Octal number which is a Perfect square

Given a natural number N, the task is to find the largest N digit Octal number which is a perfect square.

Examples:

Input: N = 1
Output: 4
Explanation:
4 is the largest 1 digit Octal number which is also perfect square

Input: N = 2
Output: 61
Explanation:
49 is the largest number which is a 2-Digit Octal Number and also a perfect square.
Therefore 49 in Octal = 61

Approach:
It can be observed that the series of largest numbers which is also a perfect square in Octal is:



4, 61, 744, 7601, 77771, 776001 …..

As we know the digits in the octal system increases when a number Greater than 8k where k denotes the number of digits in the number. So for any N digit number in the octal number system must be less than the value of 8N+1. So, the general term that can be derived using this observation is –

N-Digit Octal Number = octal(pow(ceil(sqrt(pow(8, N))) -1, 2))

Below is the implementation of the above approach:

CPP

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// C++ implementation to find the maximum
// N-digit octal number which is perfect square
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to convert decimal number
// to a octal number
void decToOctal(int n)
{
  
    // Array to store octal number
    int octalNum[100];
  
    // Counter for octal number array
    int i = 0;
    while (n != 0) {
  
        // Store remainder in 
        // octal array
        octalNum[i] = n % 8;
        n = n / 8;
        i++;
    }
  
    // Print octal number array
    // in reverse order
    for (int j = i - 1; j >= 0; j--)
        cout << octalNum[j];
    cout << "\n";
}
  
void nDigitPerfectSquares(int n)
{
    // Largest n-digit perfect square
    int decimal = pow(
        ceil(sqrt(pow(8, n))) - 1, 2
        );
    decToOctal(decimal);
}
  
// Driver Code
int main()
{
    int n = 2;
    nDigitPerfectSquares(n);
  
    return 0;
}

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Java














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// Java implementation to find the maximum 


// N-digit octal number which is perfect square 


import java.util.*; 


import java.lang.*; 


import java.io.*; 


  


class GFG 


{ 


      


    // Function to convert decimal number  


    // to a octal number  


    static void decToOctal(int n)  


    { 


          


        // Array to store octal number  


        int octalNum[] = new int[100]; 


          


        // Counter for octal number array  


        int i = 0;  


        while (n != 0)  


        {  


      


            // Store remainder in  


            // octal array  


            octalNum[i] = n % 8;  


            n = n / 8;  


            i++;  


        }  


      


        // Print octal number array  


        // in reverse order  


        for (int j = i - 1; j >= 0; j--)  


            System.out.print(octalNum[j]);  


        System.out.println("\n"); 


    }  


      


    static void nDigitPerfectSquares(int n)  


    {  


        // Largest n-digit perfect square  


        int decimal = (int) Math.pow(Math.ceil(Math.sqrt(Math.pow(8, n))) - 1, 2);  


        decToOctal(decimal);  


    }  


  


    // Driver code 


    public static void main (String[] args)  


    { 


        int n = 2;  


        nDigitPerfectSquares(n);  


    } 


} 


  


// This code is contributed by nidhiva 



















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Python3














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# Python3 implementation to find the maximum  


# N-digit octal number which is perfect square  


from math import sqrt,ceil 


  


# Function to convert decimal number  


# to a octal number  


def decToOctal(n) :  


  


    # Array to store octal number  


    octalNum = [0]*100;  


  


    # Counter for octal number array  


    i = 0;  


    while (n != 0) : 


  


        # Store remainder in  


        # octal array  


        octalNum[i] = n % 8;  


        n = n // 8;  


        i += 1;  


  


    # Print octal number array  


    # in reverse order  


    for j in range(i - 1, -1, -1) : 


        print(octalNum[j], end= "");  


    print(); 


  


def nDigitPerfectSquares(n) : 


  


    # Largest n-digit perfect square  


    decimal = pow(ceil(sqrt(pow(8, n))) - 1, 2);  


    decToOctal(decimal);  


  


# Driver Code  


if __name__ == "__main__" :  


  


    n = 2;  


    nDigitPerfectSquares(n);  


  


# This code is contributed by AnkitRai01 



















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C#














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// C# implementation to find the maximum 


// N-digit octal number which is perfect square 


using System; 


  


class GFG 


{ 


      


    // Function to convert decimal number  


    // to a octal number  


    static void decToOctal(int n)  


    { 


          


        // Array to store octal number  


        int []octalNum = new int[100]; 


          


        // Counter for octal number array  


        int i = 0;  


        while (n != 0)  


        {  


      


            // Store remainder in  


            // octal array  


            octalNum[i] = n % 8;  


            n = n / 8;  


            i++;  


        }  


      


        // Print octal number array  


        // in reverse order  


        for (int j = i - 1; j >= 0; j--)  


            Console.Write(octalNum[j]);  


    Console.WriteLine(); 


    }  


      


    static void nDigitPerfectSquares(int n)  


    {  


        // Largest n-digit perfect square  


        int _decimal = (int) Math.Pow(Math.Ceiling(Math.Sqrt(Math.Pow(8, n))) - 1, 2);  


        decToOctal(_decimal);  


    }  


  


    // Driver code 


    public static void Main()  


    { 


        int n = 2;  


        nDigitPerfectSquares(n);  


    } 


} 


  


// This code is contributed by AnkitRai01 



















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Output:


61







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