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1. Check Strong Number2. Calculate Power of a Number3. Calculate Compound Interest4. Reverse a Number5. Convert Binary to Decimal6. Convert Decimal to Binary7. Convert Octal to Decimal8. Convert Decimal to Octal9. Convert Binary to Octal10. Convert Octal to Binary11. Check Neon Number12. Check Narcissistic Number13. Convert Char to Int

Check Neon Number in C++

Check Neon Number in C++ (4 Programs).

IntermediateAdvanced Number ProgramsExample 11 of 13
check-neon-number.cpp
1#include <iostream>
2using namespace std;
3

4bool isNeon(int num) {
5    int square = num * num;
6    int sum = 0;
7    
8    while (square > 0) {
9        sum += square % 10;
10        square /= 10;
11    }
12    
13    return sum == num;
14}
15

16int main() {
17    int num;
18    cout << "Enter a number: ";
19    cin >> num;
20    
21    if (isNeon(num)) {
22        cout << num << " is a Neon number" << endl;
23    } else {
24        cout << num << " is not a Neon number" << endl;
25    }
26    
27    return 0;
28}

Output

Enter a number: 9
9 is a Neon number

What's going on

The program checks whether a given number is a Neon number. For example:

Input: 9
Calculation: 9² = 81, sum of digits of 81 = 8 + 1 = 9
Since 9 equals 9, it is a Neon number.

A Neon number must satisfy: sum of digits of (number²) = number

Example:

9 is a Neon number: 9² = 81, 8 + 1 = 9 ✓
1 is a Neon number: 1² = 1, 1 = 1 ✓
5 is NOT a Neon number: 5² = 25, 2 + 5 = 7 ≠ 5 ✗

Header File Used

This header provides:

cout for displaying output
cin for taking input from the user
1#include <iostream>

Understanding Neon Numbers

Definition

:

A Neon number is a number where the sum of digits of its square equals the number itself.

Mathematical Expression

:

If n is a Neon number, then: sum of digits of (n²) = n

Examples

:

0: 0² = 0, sum of digits = 0, 0 = 0 ✓ (Neon)
1: 1² = 1, sum of digits = 1, 1 = 1 ✓ (Neon)
9: 9² = 81, sum of digits = 8 + 1 = 9, 9 = 9 ✓ (Neon)
5: 5² = 25, sum of digits = 2 + 5 = 7, 7 ≠ 5 ✗ (Not Neon)

Known Neon Numbers

:

Only 0, 1, and 9 are Neon numbers in base-10 (decimal system).

Function: isNeon()

bool isNeon(int num) {

int square = num * num;

int sum = 0;

while (square > 0) {

sum += square % 10;

square /= 10;

}

}

This function:

Takes a number as input.
Calculates its square.
Sums the digits of the square.
Returns true if the sum equals the original number.
1return sum == num;

Step-by-Step Algorithm

Let's trace through the algorithm for num = 9:

Step 1: Calculate Square

square = 9 × 9 = 81

Step 2: Extract and Sum Digits

Initialize: sum = 0, square = 81

Iteration 1:

remainder = 81 % 10 = 1 (extract rightmost digit)
sum = 0 + 1 = 1
square = 81 / 10 = 8 (remove rightmost digit)

Iteration 2:

remainder = 8 % 10 = 8
sum = 1 + 8 = 9
square = 8 / 10 = 0 (stop condition)

Step 3: Compare

sum = 9, num = 9
9 == 9 → return true

Result: 9 is a Neon number.

Understanding Digit Extraction

Modulo Operator (%)

:

square % 10 extracts the rightmost digit.
Example: 81 % 10 = 1, 8 % 10 = 8

Integer Division (/)

:

square / 10 removes the rightmost digit.
Example: 81 / 10 = 8, 8 / 10 = 0

Accumulation

:

sum += square % 10 adds each digit to the sum.
Continues until all digits are processed.

Main Function

int num;

cout << "Enter a number: ";

cin >> num;

if (isNeon(num)) {

cout << num << " is a Neon number" << endl;

} else {

cout << num << " is not a Neon number" << endl;

}

return 0;

}

1int main() {

Process Flow

:

1.User enters a number (e.g., 9).
2.Call isNeon(9) to check.
3.Display result based on return value.

Other Methods (Mentioned but not shown in code)

Method 2: Using String Conversion

int square = num * num;

string squareStr = to_string(square);

int sum = 0;

for (char c : squareStr) {

sum += c - '0';

}

Converts square to string.
Iterates through each character.
Converts each character to digit and sums.
1return sum == num;

Method 3: Using Recursion

int sumDigits(int n) {

if (n == 0) return 0;

}

bool isNeon(int num) {

return sumDigits(num * num) == num;

}

Recursive approach for summing digits.
Base case: when n becomes 0.
Recursive case: extract digit and recurse.
1return (n % 10) + sumDigits(n / 10);

Method 4: Using Array

int square = num * num;

int digits[10], count = 0;

while (square > 0) {

digits[count++] = square % 10;

square /= 10;

}

int sum = 0;

for (int i = 0; i < count; i++) {

sum += digits[i];

}

Stores digits in an array.
Sums array elements.
1return sum == num;

Displaying the Result

The program prints:

Output (for input 9):

9 is a Neon number

Or if not a Neon number:

5 is not a Neon number

1cout << num << " is a Neon number" << endl;

Why Only 0, 1, and 9?

Mathematical Proof

:

For a number n to be Neon: sum of digits of (n²) = n

Analysis

:

For n ≥ 10, n² has at least 3 digits.
Maximum sum of digits for a k-digit number is 9k.
For n = 10, n² = 100, max sum = 9, but 9 < 10.
This pattern continues for larger numbers.

Conclusion

:

Only single-digit numbers can be Neon numbers, and only 0, 1, and 9 satisfy the condition.

Common Use Cases

Number Theory

:

Studying special number properties.
Mathematical puzzles and problems.
Educational programming exercises.

Algorithm Practice

:

Digit manipulation techniques.
Understanding modulo and division operations.
Building accumulation patterns.

Interview Questions

:

Common coding interview problem.
Tests understanding of digit operations.
Foundation for more complex number problems.

Important Considerations

Edge Cases

:

0 is a Neon number (0² = 0, sum = 0).
1 is a Neon number (1² = 1, sum = 1).
Negative numbers: squares are positive, but original is negative, so never Neon.

Large Numbers

:

For very large numbers, square may cause integer overflow.
Consider using long long for larger numbers.
Or use string-based methods for very large numbers.

Efficiency

:

Time complexity: O(log(n²)) = O(log n) - logarithmic.
Space complexity: O(1) - constant space.

return 0;

This ends the program successfully.

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