**What is a Neon Number?**

If you are given a number and the summation of the square of the individual digits of the number equal to the number itself, then the number is called a neon number.

Example: For the number 9: 9*9 = 81 and 8 + 1 = 9

**Illustration:**

Case 1:

Input: 9

Output: Given number is a Neon Number

Explanation : square of 9 = 9 * 9 = 81;

sum of digits of square : 8 + 1 = 9 (same as the original number )

Case 2:

Input: 8

Output: Given number is not a Neon Number

Explanation: Square of 8 = 8 * 8 = 64

sum of digits of square : 6 + 4 = 10 (same as the original number)

**Algorithm**

- Start
- The first step is to find the square of the number.
- The next step is to implement a loop that will give the sum of the digits of the square of the number.
- Implement a condition such that when the condition checksum is true, True will be returned, which means that a number is a neon number. It will return false if the condition is false, which means that the number is not a neon number.

**Pseudo Code**

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

**Sample Code**

import java.io.*; class Demo { //the function below will check if the inputted number is a neon number public static boolean checkNeon (int n) { //in this step, we are finding the square of the original number int square = n * n; //using a flag variable called sum int sum = 0; //If the square of the number is > 0 while (square > 0) { //Finding the remainder int r = square % 10; //Summing the current value of the sum variable to the remainder sum + = r; //Extracting the last digit square = square / 10; } //If the number obtained after the calculation is equal to the original number, if (sum == n) //returns true for a neon number return true; else //returns false for a neon number return false; } public static void main (String args []) { int n = 9; //Invoking the above function created that will check for a neon number if (checkNeon (n)) //if a neon number System.out.println (“Given number ” + n + “is Neon Number”); else //if not, a neon number System.out.println (“The given number” + n + “is not a neon number”); } }

**OUTPUT**

The given number is a Neon Number

**Time Complexity**

O (l) where the value of l depends on the number of digits that is contained in the square of the original number.

## 0 Comments