C Program to Find Neon Number

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:

  1. Using For Loop
  2. 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.

Also Read:

Leave a Comment

Your email address will not be published.