C program to check if a number is Neon or not

C program to check if a number is Neon or not:

In this post, we will learn how to check if a number is a Neon number or not in C programming language. A number is called a Neon number if the sum of the digits of its square value is equal to the number.

For example, 9 is a Neon number because the square of 9 is 81 and 8 + 1 is equal to 9 that is the number.

This program will take a number as input from the user and print one message if it is a Neon Number or not.

Step to check if a number is Neon or not:

Take the number as input from the user
Find the square of the number
Find the sum of the digits in the square
Compare the sum with the number. If both are equal then it is a Neon number.

Now let’s write down the program:

Method 1: C program to check if a number is Neon or not:

#include <stdio.h>

int main()
{
    int num, square, sumDigits = 0;

    printf("Enter a number: ");
    scanf("%d", &num);

    square = num * num;

    while (square != 0)
    {
        sumDigits += square % 10;
        square /= 10;
    }

    if (num == sumDigits)
    {
        printf("It is a Neon number\n");
    }
    else
    {
        printf("It is not a Neon number\n");
    }

    return 0;
}

In this program,

num is an integer variable to hold the number
square is an integer variable to hold a square value
sumDigits is an integer variable to hold the sum of digits of the square. It is Initialised as 0
The program is asking the user to enter a number and it stores that number in the num variable
It finds the square of the number by multiplying the number with itself and stores that value in the square variable
It uses a while loop to find the sum of digits in the square :
- The while loop will run until the value of square is not equal to zero
- It gets the last digit of square and adds it to sumDigits
- It removes the last digit from the square variable.
After the loop, it checks if the number is equal to the sum of digits of the square value. If yes, it is a Neon number. Else, it is not.

If you run this program, it will give output as like below:

Enter a number: 9
It is a Neon number

Enter a number: 8
It is not a Neon number

Method 2: C program to check if a number is Neon or not by using a different method:

Let’s use a different method to check if a number is Neon or not in C:

#include <stdio.h>

int isNeon(int num)
{
    int square, sumDigits = 0;
    square = num * num;

    while (square != 0)
    {
        sumDigits += square % 10;
        square /= 10;
    }

    return num == sumDigits;
}

int main()
{
    int num;

    printf("Enter a number: ");
    scanf("%d", &num);

    if (isNeon(num))
    {
        printf("It is a Neon number\n");
    }
    else
    {
        printf("It is not a Neon number\n");
    }

    return 0;
}

Here,

We created a new method called isNeon. This method is used to check if a number is Neon or not.
- It takes one integer variable as input and returns an integer.
- It uses the same steps that we did in the last program.
- It returns 1 if the number is Neon, else it returns 0.
Inside main, we are calling isNeon in the if block. Based on its return value, we are printing a message.

Print all Neon numbers in a range:

We can use a loop and print all Neon numbers in a given range. Let’s find out all Neon numbers between 0 to 10000.

#include <stdio.h>

int isNeon(int num)
{
    int square, sumDigits = 0;
    square = num * num;

    while (square != 0)
    {
        sumDigits += square % 10;
        square /= 10;
    }

    return num == sumDigits;
}

int main()
{

    for (int i = 0; i <= 10000; i++)
    {
        if (isNeon(i))
        {
            printf("%d ", i);
        }
    }

    return 0;
}

Here,

We are using the same method isNeon as discussed in the previous example.
The for loop runs from 0 to 10000. For each value, it calls isNeon to check if it is a Neon number or not. Since we are using i with the loop, it passes i to isNeon.
If it is a Neon number, it prints that value.

If will print the below output:

0 1 9

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Updated on 27 Nov 2021

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