# Neon Number in C++

C++Server Side ProgrammingProgramming

#### C in Depth: The Complete C Programming Guide for Beginners

45 Lectures 4.5 hours

#### Practical C++: Learn C++ Basics Step by Step

Most Popular

50 Lectures 4.5 hours

#### Master C and Embedded C Programming- Learn as you go

66 Lectures 5.5 hours

A neon number is a number where the sum of digits of the square of the number is equal to the number. Let's take an example.

n = 9

square = 81

sum of digits of square = 8 + 1 = 9

So, the number 9 is a neon number.

We need to check whether the given number is a neon number or not. If the given number is a neon number, then print Yes else print No.

## Algorithm

• Initialise the number n.
• Find the square of the number.
• Find the sum of the digits of the square
• If the sum of digits of the square is equal to the given number then result true else false.

## Implementation

Following is the implementation of the above algorithm in C++

#include <bits/stdc++.h>
using namespace std;
int isNeonNumber(int x) {
int square = x * x;
int digitsSum = 0;
while (square != 0) {
digitsSum += (square % 10);
square = square / 10;
}
return digitsSum == x;
}
int main(void) {
string result;
result = isNeonNumber(1) ? "Yes" : "No";
cout << 1 << "->" << result << endl;
result = isNeonNumber(3) ? "Yes" : "No";
cout << 3 << "->" << result << endl;
result = isNeonNumber(9) ? "Yes" : "No";
cout << 9 << "->" << result << endl;
}

## Output

If you run the above code, then you will get the following result.

1->Yes
3->No
9->Yes