C Program to Find Neon Number

Flow Control
➤ If-else Statement in C
➤ Programs on if-else
➤ Switch Case in C
➤ Switch case Programs
➤ Conditional Operator
➤ While loop in C
➤ Do-while loop in C
➤ While vs do-while
➤ For loop in C
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➤ Largest in 3 Numbers
➤ Find Grade of student
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➤ Leap Year Program
➤ Simple calculator in C
➤ Check Odd or Even
➤ Roots of Quadratic Equation
➤ Find Reverse of Number
➤ Factors of a number in C
➤ Generate Multiplication table
➤ Find Power of a Number
➤ Find GCD and LCM
➤ Find factorial of Number
➤ Count Number of Digits
➤ Sum of digits in Number
➤ Sum of N Natural Numbers
➤ Sum of Squares of Natural No.
➤ Find Sum of Odd Numbers
➤ Find the Sum of Series
➤ Find Fibonacci series in C
➤ Sum of the Fibonacci series
➤ Sum until enters +ve numbers
➤ Sum of max 10 no. & Skip -ve
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➤ Fahrenheit to Celsius
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➤ Octal ↔ Binary in C
☕️ Number Programs in C
➤ Prime Number in C
➤ Strong Number in C
➤ Krishnamurthy Number
➤ Neon Number in C
➤ Palindrome number
➤ Perfect Number in C
➤ Armstrong Number
☕️ Pattern Programs in C
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➤ Print Diamond Pattern
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➤ Diamond Pattern of Numbers

We will write a C program to find Neon number. For loop or while loop can be used. After that, we will write another C program to find the neon number within a range. A Prerequisite example is a C program to find the sum of digits. First of all, we should know that, what is the Neon number?

Neon Number

If the sum of digits of the square of the number is equal to the same number, then the number is called Neon number.

Example:- 9
Square of 9 = 9²= 81
Sum of the digits of the square = 8+1= 9
So, 9 is a Neon number.

Another Example:- 1
Square of 1= 1²= 1
Sum of the digits of the square = 1
S0, 1 is a Neon number.

Another Example :- 5
Square of 5 = 5² = 25
Sum of the digits of the square = 2+5 = 7
Here, 5 is not equal to 7.
So, 5 is not a Neon number.

C Program to check Neon Number using While Loop

#include<stdio.h>
int main()
{

    int n, sqr, rem, sum=0;

    printf("Enter Number: ");
    scanf("%d",&n);

    sqr = n*n; //we can also use pow()

    while(sqr!=0)
    {
        rem = sqr%10;
        sum += rem; //sum = sum + rem
        sqr /= 10;  //sqr = sqr / 10
    }

    if(sum==n)
        printf("%d is a neon number.\n",n);
    else
        printf("%d is NOT a neon number.\n",n);

    return 0;
}

Output:-

Enter Number: 1
1 is a neon number.

Enter Number: 5
5 is NOT a neon number.

The variables in this program are n, sqr, rem, and sum. The variable sum is initialized with 0, variable n, sqr, and rem will hold the value of the inputted number, the square of the number, and remainder respectively. First, we take a number as an input from the user, then calculate its square value, which will store in the variable sqr.

Using the while loop, we find every digit in the sqr and after finding every digit, the sum will be calculated. the sum is the value of the addition of every digit of sqr. While loop ends when sqr becomes 0. Now, using the if-else conditional statement we check that is the sum equal to the number inputted from the user? If yes then the number is a Neon number else number is not a Neon number.

Using For Loop

#include<stdio.h>
int main()
{
    int n, i, sqr, sum=0;

    printf("Enter Number: ");
    scanf("%d",&n);

    sqr = n*n;

    for(i=sqr;i>0;i/=10)
    {
        sum += (i % 10);
    }

    if(sum==n)
        printf("%d is a neon number.\n",n);
    else
        printf("%d is NOT a neon number.\n",n);

    return 0;
}

The same as the previous program, this program also finds that the number is neon number or not, nut using for loop. Logic is the same as while loop.