Neon Number in Java

Last Updated on July 11, 2023 by Mayank Dham

In the realm of mathematics, numbers often hold hidden patterns and fascinating properties. One such intriguing numerical concept is the Neon Number. Neon Numbers, also known as the "Pluperfect Digital Invariant," captivates mathematicians and programming enthusiasts alike, offering a glimpse into the mesmerizing world of numeric patterns.

In this article, we delve into the realm of Neon Numbers and their implementation in Java. We will explore what Neon Numbers are, how they are identified, and how to efficiently determine if a given number qualifies as a Neon Number using Java programming techniques. Let’s discuss, what is neon number in Java first.

What is Neon Number in Java?

A Neon Number, also known as a Pluperfect Digital Invariant, is a special type of number in mathematics. In the context of Java programming, a Neon Number is defined as a number where the sum of the digits of its square is equal to the original number itself.

To determine if a given number is a Neon Number, the following steps are typically followed in Java:

• Square the given number.
• Calculate the sum of the digits in the squared number.
• Compare the sum with the original number. If they are equal, the number is a Neon Number; otherwise, it is not.

For example, let’s consider the number 9. When squared, we get 81. The sum of the digits in 81 is 8 + 1 = 9, which matches the original number. Therefore, 9 is a Neon Number.

Neon Numbers possess unique properties and are an interesting topic for exploration in number theory and programming. Implementing a Neon Number checker in Java can help identify and study these special numbers, allowing for further investigation into their patterns and properties.

Example of Neon Number in Java

Case 1:

Input : 9
Output : Given number 9 is Neon number

Explanation : square of 9=9*9=81;
sum of digit of square : 8+1=9(which is equal to given number)

Case 2:

Input : 8
Output : Given number is not a Neon number

Explanation : square of 8=8*8=64
sum of digit of square : 6+4=10(which is not equal to given number)

Algorithm to Find Neon Number in Java

1. Read the input number num.
2. Square the value of num and store it in a variable square.
   • square = num * num
3. Initialize a variable sum as 0 to store the sum of the digits.
4. Iterate through each digit in the square number until it becomes 0.
   • Extract the last digit using the modulus operator % and add it to the sum.
   • Reduce the value of square by dividing it by 10 using the integer division operator //.
   • Repeat until the square becomes 0.
   • Check if the value of sum is equal to the original num.
5. If sum equals num, then num is a Neon Number.
   • Otherwise, num is not a Neon Number.
   • Output the result accordingly.

Dry Run To Find Neon Number Program in Java

Input: 9

• The user enters the number 9.
• The code squares the number: square = 9 * 9 = 81.
• The variable sum is initialized to 0.
• The code starts iterating through each digit in the squared number (81).
• In the first iteration, the last digit is 1. sum is incremented by 1: sum = 0 + 1 = 1.
• The value of the square becomes 8 after dividing it by 10 (81 // 10).
• In the second iteration, the last digit is 8. sum is incremented by 8: sum = 1 + 8 = 9.
• The value of the square becomes 0 after dividing it by 10 (8 // 10).
• The code compares the value of sum (9) with the original number (9).
• Since the sum is equal to the original number, it prints: "9 is a Neon Number."
• The program ends.

Output:

9 is a Neon Number.

In this example, the number 9 is a Neon Number because the sum of the digits in its square (81) is equal to the original number itself.

Pseudo code For Checking Neon Number in Java

Square =n*n;
              while(square>0)
              {
              int r=square%10;
              sum+=r;
              square=square/10;
              }

Code Implementation

import java.util.Scanner;

class NeonNumber {
    public static void main(String[] args) {
        
        int num = 9;

        // Step 1: Square the number
        int square = num * num;

        // Step 2: Calculate the sum of the digits
        int sum = 0;
        int temp = square;
        while (temp != 0) {
            int digit = temp % 10;
            sum += digit;
            temp /= 10;
        }

        // Step 3: Check if it is a Neon Number
        if (sum == num) {
            System.out.println(num + " is a Neon Number.");
        } else {
            System.out.println(num + " is not a Neon Number.");
        }
    }
}

Output

9 is a Neon Number.

Explanation
In the code, the number is squared, resulting in 81. Then, the sum of the digits in the squared number (81) is calculated, which is 9.

Since the sum of the digits (9) is equal to the original number (9), the code determines that the given number is a Neon Number.

Hence, the output states that "9 is a Neon Number."

Time Complexity: The time complexity to find neon number in Java is O(l), where l represents the number of digits in the square of the given number.

Recursive Approach For Finding Neon Number in Java

• In this approach, we use a recursive function isNeonNumber to check if the input number is a neon number.
• The function takes two arguments: the square of the input number and the input number itself.
• At each recursive call, we extract the last digit of the square number and subtract it from the input number.
• We then call the function recursively with the remaining digits of the square number and the updated input number.
• If the square number has no more digits left (i.e., square == 0), then we check if the input number is zero (i.e., number == 0).
• If the input number is zero, then the original input number is a neon number; otherwise, it is not.

Code Implementation

import java.util.Scanner;
 
class NeonNumber {
    public static void main(String[] args) {
        int number=9;
        int square = number * number;
        if (isNeonNumber(square, number)) {
            System.out.println(number + " is a neon number");
        } else {
            System.out.println(number + " is not a neon number");
        }
    }
 
    private static boolean isNeonNumber(int square, int number) {
        if (square == 0) {
            return number == 0;
        } else {
            int digit = square % 10;
            return isNeonNumber(square / 10, number - digit);
        }
    }
}

Output

9 is a neon number

Time Complexity: O(logn) is the time complexity of the recursive approach to find neon number in Java.

Auxiliary Space: O(logn) is the space complexity of the recursive approach to find the neon number in Java.

Conclusion:
In this article, we explored the fascinating concept of Neon Numbers in Java. We discussed how Neon Numbers are defined as numbers where the sum of the digits in their squares is equal to the original number itself. By leveraging Java programming techniques, we learned how to determine if a given number qualifies as a Neon Number.

Throughout our exploration, we examined both iterative and recursive approaches to identify Neon Numbers. We provided detailed explanations and code implementations to showcase how these methods can be utilized. Additionally, we highlighted the importance of understanding number theory concepts and their application in programming challenges.

Frequently Asked Questions (FAQ) Related to Neon Number in Java:

Here are some frequently asked questions on neon number in Java.

Q: Can a negative number be a Neon Number?
A: No, Neon Numbers are defined for positive integers only. Negative numbers and zero cannot be Neon Numbers.

Q: Are all perfect squares Neon Numbers?
A: No, not all perfect squares are Neon Numbers. To be a Neon Number, the sum of the digits in the square must be equal to the original number.

Q: What is the significance of Neon Numbers in real-world applications?
A: While Neon Numbers may not have direct real-world applications, they serve as intriguing mathematical concepts that expand our understanding of number theory and provide interesting programming challenges.

Q: Can we optimize the algorithm to find Neon Numbers?
A: The algorithm to determine Neon Numbers has a time complexity of O(log(n)), which is already quite efficient. Further optimization may not be necessary for most practical scenarios.

Q: Can we find all Neon Numbers within a given range?
A: Yes, it is possible to find all Neon Numbers within a given range by iterating through the numbers and applying the Neon Number check algorithm to each one. This approach can help identify and analyze patterns among Neon Numbers.