Java Program to Check If a Number is Neon Number or Not
A neon number is a number where the sum of digits of the square of the number is equal to the number. The task is to check and print neon numbers in a range.
Illustration:
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 :
- First, find the square of the given number.
- Find the sum of the digit of the square by using a loop.
- The condition checksum is equal to the given number
- Return true
- Else return false.
Pseudo code : Square =n*n; while(square>0) { int r=square%10; sum+=r; square=square/10; }
Example:
Java
// Java Program to Check If a Number is Neon number or not // Importing java input/output library import java.io.*; class GFG { // Method to check whether number is neon or not // Boolean type public static boolean checkNeon( int n) { // squarring the number to be checked int square = n * n; // Initializing current sum to 0 int sum = 0 ; // If product is positive while (square > 0 ) { // Step 1: Find remainder int r = square % 10 ; // Add remainder to the current sum sum += r; // Drop last digit of the product // and store the number square = square / 10 ; } // Condition check // Sum of digits of number obtained is // equal to original number if (sum == n) // number is neon return true ; else // numbe is not neon return false ; } // Main driver method public static void main(String[] args) { // Custom input int n = 9 ; // Calling above functon to check custom number or // if user entered number via Scanner class if (checkNeon(n)) // Print number considered is neon System.out.println( "Given number " + n + " is Neon number" ); else // Print number considered is not neon System.out.println( "Given number " + n + " is not a Neon number" ); } } |
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Output
Given number 9 is Neon number
Time Complexit: O(l) where l is the number of the digit in the square of the given number
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