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 squareInput: 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
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
61
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