Check Neon Number
Check Neon Number in C++ (4 Programs)
C++ Check Neon Number Program
This program helps you to learn the fundamental structure and syntax of C++ programming.
#include <iostream>
using namespace std;
bool isNeon(int num) {
int square = num * num;
int sum = 0;
while (square > 0) {
sum += square % 10;
square /= 10;
}
return sum == num;
}
int main() {
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;
}Enter a number: 9 9 is a Neon number
Understanding Check Neon Number
This program teaches you how to check if a number is a Neon number in C++. A Neon number is a special number where the sum of the digits of its square equals the original number itself. This is an interesting number theory problem that helps students understand digit manipulation, squaring numbers, and accumulation patterns.
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1. What This Program Does
The program checks whether a given number is a Neon number. For example:
Example:
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2. Header File Used
This header provides:
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#include <iostream>3. Understanding Neon Numbers
Definition
:
A Neon number is a number where the sum of digits of its square equals the number itself.
Mathematical Expression
:
Examples
:
Known Neon Numbers
:
Only 0, 1, and 9 are Neon numbers in base-10 (decimal system).
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4. Function: isNeon()
bool isNeon(int num) {
int square = num * num;
int sum = 0;
while (square > 0) {
sum += square % 10;
square /= 10;
}
}
This function:
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return sum == num;5. Step-by-Step Algorithm
Let's trace through the algorithm for num = 9:
Step 1: Calculate Square
Step 2: Extract and Sum Digits
Iteration 1:
Iteration 2:
Step 3: Compare
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6. Understanding Digit Extraction
Modulo Operator (%)
:
Integer Division (/)
:
Accumulation
:
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7. Main Function
int num;
cin >> num;
if (isNeon(num)) {
cout << num << " is a Neon number" << endl;
} else {
cout << num << " is not a Neon number" << endl;
}
return 0;
}
int main() {Process Flow
:
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8. 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;
sum += c - '0';
}
return sum == num;Method 3: Using Recursion
int sumDigits(int n) {
if (n == 0) return 0;
}
bool isNeon(int num) {
return sumDigits(num * num) == num;
}
return (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];
}
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return sum == num;9. 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
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cout << num << " is a Neon number" << endl;10. Why Only 0, 1, and 9?
Mathematical Proof
:
Analysis
:
Conclusion
:
Only single-digit numbers can be Neon numbers, and only 0, 1, and 9 satisfy the condition.
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11. Common Use Cases
Number Theory
:
Algorithm Practice
:
Interview Questions
:
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12. Important Considerations
Edge Cases
:
Large Numbers
:
Efficiency
:
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13. return 0;
This ends the program successfully.
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Summary
This program is fundamental for beginners learning digit manipulation, understanding special number properties, and preparing for more complex number theory problems in C++ programs.
Let us now understand every line and the components of the above program.
Note: To write and run C++ programs, you need to set up the local environment on your computer. Refer to the complete article Setting up C++ Development Environment. If you do not want to set up the local environment on your computer, you can also use online IDE to write and run your C++ programs.
Practical Learning Notes for Check Neon Number
This C++ program is part of the "Advanced Number Programs" topic and is designed to help you build real problem-solving confidence, not just memorize syntax. Start by understanding the goal of the program in plain language, then trace the logic line by line with a custom input of your own. Once you can predict the output before running the code, your understanding becomes much stronger.
A reliable practice pattern is to run the original version first, then modify only one condition or variable at a time. Observe how that single change affects control flow and output. This deliberate style helps you understand loops, conditions, and data movement much faster than copying full solutions repeatedly.
For interview preparation, explain this solution in three layers: the high-level approach, the step-by-step execution, and the time-space tradeoff. If you can teach these three layers clearly, you are ready to solve close variations of this problem under time pressure.