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.

If the sum of the values obtained by raising the digits of a number to the power of their position is equal to the number itself, then the number is called Disarium 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

Enter a number
135
Entered number is a Disarium number

Enter a number
125
Entered number is not a Disarium number

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 andÂ  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.
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