How to check for a Disarium number in java

What is Disarium Number?
Consider a number such as 135. Its digits are 1, 3 and 5 with 1 at position 1, 3 at position 2 and 5 at position 3 in the number.
Now, raise the digits to the power of their position and add the resultant values, which means 1¹ + 3² + 5³ = 1 + 9 + 125 = 135 which is the same as the number itself.
Such numbers whose sum of digits raised to the power of their position is equal to the number are called Disarium Numbers.
Note that the position is counted starting from 1 and from the left of number.

Java Program
Program to check for a Disarium number is given below

import java.util.Scanner;

public class DisariumNumberChecker {

   public static void main(String[] args) {
      // initialize reader to read user input
      Scanner reader = new Scanner(System.in);
      System.out.println("Enter a number");
      String numberStr = reader.nextLine();
      // close reader
      reader.close();
      // calculate length of entered number
      int length = numberStr.length();
      // convert entered string to integer
      int number = Integer.parseInt(numberStr);
      // initialize variable to hold sum
      int sum = 0;
      // make a copy of number so that original number can be compared with
      // the sum at the end
      int copy = number;
      while (copy > 0) {
         // get the rightmost digit
	 int digit = copy % 10;
	 // raise the digit to its position. Position will be the length of
	 // the number
	 int digitRaisedToPower = (int) Math.pow(digit, length);
	 // add it to sum
	 sum += digitRaisedToPower;
	 // remove rightmost digit
	 copy = copy / 10;
	 // decrease length by 1
	 length--;
      }
      // compare sum with number to check
      if (sum == number) {
	 System.out.println("Entered number is a Disarium number");
      } else {
	 System.out.println("Entered number is not a Disarium number");
      }
   }
}

Output

Logic
Number is read from the user and its length is calculated. Initially, length will be the number of digits. A variable sum is initialized to 0 to hold the sum of digits raised to the power of position.
Note that a copy of original number is made and all operations are performed on the copy so that the actual number is still with us for comparison with the sum later.
Now we have to get individual digits of the number so that power raised to their position can be calculated.

For getting the digits, modulus operator(%) is used and number % 10 is calculated. Number % 10 gives the rightmost digit of a number.
This digit is raised to the power of length since length of the number will be the position of rightmost digit. Power is calculated using pow method of java.lang.Math class and the result is added to the sum variable.
Since the rightmost digit is now processed, it is removed using the division operator(/). Number / 10 removes the rightmost digit and gives the remaining number. Example,

135 /10 gives 13. Length is also decreased by 1 since now there is one digit less in the number. Again, the rightmost digit is determined and  power of its position is calculated and added to the sum as above.
Since this operation has to be repeated for all the digits of the number, it is enclosed in a loop. The condition of the loop is number > 0.
Number will become 0 when there will be only 1 digit left in the number because then “number / 10” will return 0.
After the loop completes, the original number is compared with the resultant sum.
If they are same, the number is Disarium.

Let’s tweak in :

1. Modulus(%) operator gives the remainder of division. Thus 135 % 10 = 5.
2. Division(/) operator gives the quotient of division. Thus 135 / 10 = 13.