In this post, we will learn** how to find neon numbers using the C** Programming language. But before that, let’s learn about neon numbers.

A number is said to be a **neon number** if the sum of the digits of the square of the number is equal to the same number. For example, **9** is a neon number because **9 * 9 = 81 **and the sum of **81** is **8 + 1 = 9** which is equal to the original number.

We will use following approaches to find neon number:

- Using For Loop
- Using While Loop

So, without further ado, let’s begin this tutorial.

**C Program to Find Neon Number Using For Loop**

// C Program to Find Neon Number Using For Loop #include <stdio.h> #include <math.h> int main(){ int num, sqr, rem, sum; // Asking for input printf("Enter a number: "); scanf("%d", &num); // Finding the square root sqr = pow(num, 2); // Finding the sum for (sum = 0; sqr > 0; sqr /= 10){ rem = sqr % 10; sum = sum + rem; } // Checking whether sum is equals to original number if (sum == num){ printf("%d is a neon number.", num); } else{ printf("%d is not a neon number.", num); } return 0; }

**Output **

```
Enter a number: 9
9 is a neon number.
```

**How Does This Program Work ?**

int num, sqr, rem, sum;

In this program, we have declared four integer data type variables named** num**, **sqr**, **rem** and **sum**.

// Asking for input printf("Enter a number: "); scanf("%d", &num);

Now, the user is asked to enter a number.

// Finding the square root sqr = pow(num, 2);

We find the square of the entered number using the** pow()** function.

// Finding the sum for (sum = 0; sqr > 0; sqr /= 10){ rem = sqr % 10; sum = sum + rem; }

Now, we calculate the sum of the digits of the square number using for loop. Suppose, the entered number is **9**, then it’s square will be** 81**. Therefore,

**1st For Loop Iteration: for (sum = 0; 81 > 0; 81 /= 10)**

rem = sqr % 10 = 81 % 10 = 1

sum = sum + rem = 0 + 1 = 1

**2nd For Loop Iteration: for (sum = 1; 8 > 0; 8 /= 10)**

rem = sqr % 10 = 8 % 10 = 8

sum = sum + rem = 1 + 8 = 9

We get **sum = 9**.

// Checking whether sum is equals to original number if (sum == num){ printf("%d is a neon number.", num); } else{ printf("%d is not a neon number.", num); }

After that, we check whether the sum is equal to the original number. If the condition is** True**, then the entered number is a neon number otherwise it’s not a neon number.

**C Program to Find Neon Number Using While Loop**

// C program to Find Neon Number #include <stdio.h> int main(){ int num, rem, sqr, sum = 0; // Asking for input printf("Enter a number: "); scanf("%d", &num); // Calculating square sqr = num * num; while (sqr != 0){ rem = sqr % 10; sum = sum + rem; sqr = sqr / 10; } // Checking for Neon Number if (sum == num){ printf("%d is a neon number.", num); } else{ printf("%d is not a neon number.", num); } return 0; }

**Output 1**

```
Enter a number: 9
9 is a neon number.
```

**Output 2**

```
Enter a number: 7
7 is not a neon number.
```

**Conclusion**

I hope after going through this post, you understand** how to find neon numbers using C **Programming language.

If you have any doubt regarding the program, then contact us in the comment section. We will be delighted to assist you.

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