C Program to Find Neon Number with Output and Explanation

A neon number is a number where the sum of digits of its square equals the number itself; in C, you can find it by squaring the number, summing the digits of the square, and comparing it to the original number using code like if (sum == num).

Examples

Input9

Output9 is a neon number

Input5

Output5 is not a neon number

Input1

Output1 is a neon number

How to Think About It

To find if a number is neon, first square the number. Then, add all digits of this squared value. If the sum equals the original number, it is a neon number; otherwise, it is not.

Algorithm

1

Get input number from the user

2

Calculate the square of the number

3

Initialize sum to zero

4

Extract each digit of the square and add to sum

5

Compare sum with the original number

6

Print if the number is neon or not

Code

c

#include <stdio.h>

int main() {
    int num, square, sum = 0, temp;
    printf("Enter a number: ");
    scanf("%d", &num);
    square = num * num;
    temp = square;
    while (temp > 0) {
        sum += temp % 10;
        temp /= 10;
    }
    if (sum == num)
        printf("%d is a neon number\n", num);
    else
        printf("%d is not a neon number\n", num);
    return 0;
}

Output

Enter a number: 9
9 is a neon number

Dry Run

Let's trace the input 9 through the code

1

Input number

num = 9

2

Calculate square

square = 9 * 9 = 81

3

Sum digits of square

sum = 8 + 1 = 9

4

Compare sum with original

sum (9) == num (9) → true

5

Print result

"9 is a neon number"

Iteration	temp	sum
1	81	0
2	8	1
3	0	9

Why This Works

Step 1: Square the number

We calculate square = num * num to get the number to analyze.

Step 2: Sum digits of the square

We extract each digit of square using modulus and division, adding them to sum.

Step 3: Compare sum with original number

If sum == num, the number is neon; otherwise, it is not.

Alternative Approaches

Using a function to sum digits

c

#include <stdio.h>

int sumDigits(int n) {
    int s = 0;
    while (n > 0) {
        s += n % 10;
        n /= 10;
    }
    return s;
}

int main() {
    int num;
    printf("Enter a number: ");
    scanf("%d", &num);
    if (sumDigits(num * num) == num)
        printf("%d is a neon number\n", num);
    else
        printf("%d is not a neon number\n", num);
    return 0;
}

This approach improves readability by separating digit summation into a reusable function.

Using recursion to sum digits

c

#include <stdio.h>

int sumDigits(int n) {
    if (n == 0) return 0;
    return (n % 10) + sumDigits(n / 10);
}

int main() {
    int num;
    printf("Enter a number: ");
    scanf("%d", &num);
    if (sumDigits(num * num) == num)
        printf("%d is a neon number\n", num);
    else
        printf("%d is not a neon number\n", num);
    return 0;
}

This recursive method is elegant but may be less efficient for very large numbers.

Complexity: O(log n) time, O(1) space

Time Complexity

The time depends on the number of digits in the square, which is proportional to log n. Summing digits requires iterating over each digit.

Space Complexity

Only a few integer variables are used, so space complexity is constant O(1).

Which Approach is Fastest?

The iterative approach is fastest and simplest; recursive methods add overhead but improve readability.

Approach	Time	Space	Best For
Iterative digit sum	O(log n)	O(1)	Simple and efficient
Function abstraction	O(log n)	O(1)	Readable and reusable code
Recursive digit sum	O(log n)	O(log n)	Elegant but uses call stack