Question

Problem 1:

More than 2500 years ago, mathematicians got interested in numbers.

Armstrong Numbers: The number 153 has the odd property that 13+53 + 33 = 1 + 125 + 27 = 153. Namely, 153 is equal to the sum of the cubes of its own digits.

Perfect Numbers: A number is said to be perfect if it is the sum of its own divisors (excluding itself). For example, 6 is perfect since 1, 2, and 3 divide evenly into 6 and 1+ 2 + 3 = 6.

Input File

The input is taken from a file named number.in and which a sequence of numbers, one per line, terminated by a line containing the number 0.

Output File

All the numbers read from the input file, are printed one per line followed by a sentence indicating whether the number is or is not Armstrong number and whether it is or is not a perfect number. Sample Input 153 6 0 Sample Output 153 is an Armstrong number but it is not a perfect number. 6 is not an Armstrong number but it is a perfect number.

Note: the program must be write in java

Answer 1

import java.io.File;
import java.io.PrintWriter;
import java.util.Scanner;

public class NumbersPro {
   public static void main(String[] args) throws Exception {
       Scanner sc = new Scanner(new File("numbers.in"));
       PrintWriter pw = new PrintWriter(new File("output.txt"));
       int num;
       String out = "";
       boolean isAmstrong, isPerfect;
       while (sc.hasNext()) {
           out = "";
           num = sc.nextInt();
           if (num == 0)
               break;
           isAmstrong = isAmstrong(num);
           if (isAmstrong)
               out = num + " is an Armstrong number";
           else
               out = num + " is not an Armstrong number";

           isPerfect = isPerfect(num);
           if (isPerfect)
               out += " and is an Perfect number";
           else
               out += " but not an perfect number";
           pw.println(out);

       }
       pw.close();
   }

   private static boolean isPerfect(int number) {
       int sum = 0, i;
       for (i = 1; i < number; i++) {
           if (number % i == 0) {
               sum = sum + i;
           }
       }
       return sum == number;
   }

   private static boolean isAmstrong(int number) {
       int originalNumber = number;
       int result = 0;
       while (originalNumber != 0) {
           int remainder = originalNumber % 10;
           result += Math.pow(remainder, 3);
           originalNumber /= 10;
       }

       return result == number;
   }
}

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