Neon Number,Perfect Number,Strong Number

1.Neon Number
sum of digits of its square equals the original number.

Example
Number =
Step 1 – Square the number
9 × 9 = 81
Step 2 – Add digits of the square
8 + 1 = 9
Step 3 – Compare with original number
9 = 9
So 9 is a Neon Number.

Steps
1.Take a number n.
2.Calculate square = n × n.
3.Convert the square into digits.
4.Add all digits of the square.
5.Compare the sum with the original number.
6.If equal → Neon Number
If not → Not Neon Number

Python

def neon_no(no):
    sqr = no * no
    res = sqr
    sum = 0

    while res > 0:
        sum = sum + res % 10
        res = res // 10

    if sum == no:
        print("Neon")
    else:
        print("Not Neon")


neon_no(9)

Output

2.Strong Number
Sum of factorial of digits = the number itself
Example

Number = 145
Factorials = 1! + 4! + 5!
= 1 + 24 + 120 = 145

145=1!+4!+5!
145=1+24+120
145=1+24+120
145=145
145=145
So, 145 is a Strong Number.

Steps
Step 1:Take a number from the user.

Step 2:Store the original number in another variable so we can compare later.

Step 3:Extract the last digit using:

digit = n % 10
Step 4:Find the factorial of that digit.

Step 5:Add the factorial to a sum variable.

Step 6:Remove the last digit using:

n = n // 10
Step 7:Repeat the process until the number becomes 0.

Step 8:Compare:

sum == original number
If equal → Strong Number
If not equal → Not Strong Number
Java

import java.util.Scanner;

public class StrongNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.print("Enter number: ");
        int n = sc.nextInt();

        int temp = n;
        int sum = 0;

        while (n > 0) {
            int digit = n % 10;

            int fact = 1;
            for (int i = 1; i <= digit; i++) {
                fact = fact * i;
            }

            sum = sum + fact;
            n = n / 10;
        }

        if (sum == temp)
            System.out.println("Strong Number");
        else
            System.out.println("Not Strong Number");
    }
}

JavaScript

let n = parseInt(prompt("Enter number:"));
let temp = n;
let sum = 0;

while (n > 0) {
    let digit = n % 10;

    let fact = 1;
    for (let i = 1; i <= digit; i++) {
        fact = fact * i;
    }

    sum = sum + fact;
    n = Math.floor(n / 10);
}

if (sum === temp)
    console.log("Strong Number");
else
    console.log("Not Strong Number");

Python

n = int(input("Enter number: "))
temp = n
sum = 0

while n > 0:
    digit = n % 10

    fact = 1
    for i in range(1, digit + 1):
        fact = fact * i

    sum = sum + fact
    n = n // 10

if sum == temp:
    print("Strong Number")
else:
    print("Not Strong Number")

Output

3.Perfect Number
sum of its proper divisors (factors except the number itself) equals the number.

Example
Factors of 6 (excluding 6 itself):1, 2, 3
Sum of factors:1 + 2 + 3 = 6
Since sum = number, 6 is a Perfect Number.

Steps
1.Take a number n.
2.Find all numbers from 1 to n-1.
3.Check if they divide n exactly.
4.If yes, add them to sum.
5.After the loop, compare:
6.If sum == n → Perfect Number
7.Else → Not Perfect Number

Java

import java.util.Scanner;

public class PerfectNumber {

    static void perfect(int n) {
        int sum = 0;

        for (int i = 1; i < n; i++) {
            if (n % i == 0) {
                sum += i;
            }
        }

        if (sum == n)
            System.out.println("Perfect Number");
        else
            System.out.println("Not a Perfect Number");
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        perfect(n);
    }
}

JavaScript

function perfect(n) {
    let sum = 0;

    for (let i = 1; i < n; i++) {
        if (n % i === 0) {
            sum += i;
        }
    }

    if (sum === n)
        console.log("Perfect Number");
    else
        console.log("Not a Perfect Number");
}

let n = parseInt(prompt("Enter number:"));
perfect(n);

Python

def perfect(n):
    sum = 0
    for i in range(1,n):
        if n % i == 0:
            sum += i

    if sum == n:
        print("Perfect Number")
    else:
        print("Not a Perfect")

n = int(input())
perfect(n)

or

def perfect(n):
    sum = 0
    for i in range(1, n):
        if n % i == 0:
            sum += i

    if sum == n:
        return "Perfect Number"
    else:
        return "Not Perfect Number"

n = int(input())
print(perfect(n))

output