neon,emirp,prime,armstrong,prime,sum,reverse -numbers

Sum of Number

package Afterfeb4;

public class sumofnumber {
    public static void main(String[] args) {
        int no = 96785;
        int sum = 0;
        while (no > 0) {
            sum = sum + no % 10;
            no = no / 10;
        }
        System.out.println("sum of number = " + sum);
    }

}

Output:sum of number = 35

ReverseNumber:

package Afterfeb4;

public class ReverseNumber {
    public static void main(String[] args) {
        int no = 12345;
        int reverse = 0;
        int count = 0;
        while (no > 0) {
            reverse = (reverse * 10) + no % 10;// 54321
            no = no / 10;// 1234 123 12 1
            count++;
        }
        System.out.println("reverse = " + reverse);
        System.out.println("count = " + count);
    }

}

Output:
reverse = 54321
count = 5

Palindrome reverse also same meaning ex;mam ex;414

package Afterfeb4;

public class Palindrome {
    public static void main(String[] args) {
        int no1 = 414;
        int no2 = no1;// no2=414
        int reverse = 0;
        while (no1 > 0) {
            reverse = (reverse * 10) + no1 % 10;
            no1 = no1 / 10;
        }
        if (reverse == no2)// if we give no1 it will give not palindrome because no1 is decease to zero
            System.out.println("palindrome");
        else
            System.out.println("NOT Palindrome");
    }

}

Output:
palindrome

ArmstrongNumber

An Armstrong number is a positive m-digit number that is equal to the sum of the mth powers of their digits. It is also known as pluperfect, or Plus Perfect, or Narcissistic number. Let’s understand it through an example.
Armstrong Number Example

1: 11 = 1

2: 21 = 2

3: 31 = 3

153: 13 + 53 + 33 = 1 + 125+ 27 = 153

125: 13 + 23 + 53 = 1 + 8 + 125 = 134 (Not an Armstrong Number)

1634: 14 + 64 + 34 + 44 = 1 + 1296 + 81 + 256 = 1643

Reference:https://www.javatpoint.com/armstrong-number-in-java

package Afterfeb4;

public class AmstrongNumber {
    public static void main(String[] args) {
        int no1 = 153;
        int no2 = no1;
        int armstrong = 0;
        while (no1 > 0) {
            int a = no1 % 10;
            armstrong = armstrong + (a * a * a);
            no1 = no1 / 10;

        }
        System.out.println(armstrong);
        if (armstrong == no2)
            System.out.println("armstrong");
        else
            System.out.println("not armstrong");
    }

}

Output:
153
armstrong

NeonNumber:
Neon Number

A positive integer whose sum of digits of its square is equal to the number itself is called a neon number.

Example of Neon Number

Let's take an example and check 9 and 45 are neon numbers or not.
Neon Number in Java

Steps to Find Neon Number

Read an integer from the user or initialize a number (n) to check.

Calculate the square of the given number (n) and store it in variable sq.

Find the sum of the digits of the square (sq) and store the sum in the variable (sum).

Compare the given number n with If both are equal, the given number is a neon number, else, not a neon number.

Referencehttps://www.javatpoint.com/neon-number-in-java

package Afterfeb4;

public class neonNumber {
    public static void main(String[] args) {
        int no1 = 9;
        int no2 = no1 * no1;// 9*9=*81=1+8=9 //81--> into loop then 8
        int neon = 0;

        while (no2 > 0) {

            neon = (neon) + no2 % 10;// 1 8
            no2 = no2 / 10;// 81---->8

        }
        System.out.println("Number = "+neon);
        if (no1 == neon)
            System.out.println("NeonNumber");
        else
            System.out.println("NOt NeonNumber");

    }

}

Output:
Number = 9
NeonNumber

emirpNumber:

A number is called an emirp number if we get another prime number on reversing the number itself. In other words, an emirp number is a number that is prime forwards or backward. It is also known as twisted prime numbers.
Note: Palindrome primes are excluded.

Emirp Number Example

Suppose, we want to check the number 79 is emirp or not.

We know that 79 is a prime number means that divisible by 1 and self only. On reversing the number, we get 97 which is another prime number. Therefore, 79 and 97 both are prime numbers. Hence, 79 is a prime number. Similarly, we can check other numbers also.

Some other emirp numbers are 13, 199, 107, 113, 1399, 1583, 1201, 3049, etc.

Steps to find Emirp Number

Read or initialize a number (n).

First, check the given number (n) is prime or not.
If not, break the execution and exit.
If prime, find the reverse (r) of the given number (n).

Check the reverse number (r) is prime or not.
If not, print number (n) is not emirp.
If prime, print the given number (n) as an emirp number.

Reference:https://www.javatpoint.com/emirp-number-in-java

package Afterfeb4;

public class emirpNumber {
    public static void main(String[] args) {
        int no =13;
        int emirp = 0;//13 --> 31reverse of no aslo a prime number
        while (no > 0) {
            emirp = (emirp * 10) + no % 10;
            no = no / 10;

        }
        System.out.println(emirp);
        if (emirp % 2 != 0 && emirp%3!=0 )
            System.out.println("emirpNumber");
        else
            System.out.println("NOt emirpNumber");
    }

}

Output:
71
emirpNumber

Primenumber:

package Afterfeb4;

public class primeNumber {
    public static void main(String[] args) {

            int no =5;
            if (no% 2 != 0 && no%3!=0  )
                System.out.println("primeNumber");
            else
                System.out.println("NOt primeNumber");
        }

    }

output:primeNumber