Question

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.

Answer 1

Note:As U didnt mentioned In which language you want me to develop ,I developed in java..Plz tell me if u need this in c++ or c.

Could you plz go through this code and let me know if u need any changes in this.Thank You
_________________

// numbers.in

153
6
0

_________________________

package org.students;

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

public class CheckArmstrongOrPerfectNumber {

   public static void main(String[] args) {
   int num;
       try {
           Scanner sc=new Scanner(new File("numbers.in"));
           while(sc.hasNext())
           {
               num=sc.nextInt();
               if(num!=0)
               {
                   boolean b1=isArmstrongNum(num);
                   boolean b2=isPerfectNum(num);
                   if(b1)
                   {
                       System.out.print(num+" is an Armstrong number");
                   }
                   else
                   {
                       System.out.print(num+" is not an Armstrong number");
                   }
                  
                   if(b2)
                   {
                       System.out.println(" and it is a perfect number.");
                   }
                   else
                   {
                       System.out.println(" but it is not a perfect number.");
                   }
               }
               else
               {
                   break;
               }
           }
       } catch (FileNotFoundException e) {
   System.out.println("** File Not Found **");
       }
      

   }

   private static boolean isPerfectNum(int num) {
   //Declaring the variable
   int sum=0;
  
  
   for(int i=1;i<num-1;i++)
   {
   if(num%i==0)
   sum+=i;
   }
  
   /* if the user entered number is equal to the
   * sum value the number is perfect number
   */
   if(sum==num)
   return true;
   else
   return false;
   }

   private static boolean isArmstrongNum(int number) {
   // Declaring variables
   int rem = 0, tot = 0;
   int value = number;
   /* This while loop will get executes
   * until the number is greater than zero
   */
   while (number > 0)
   {
   rem = number % 10;
   tot += Math.pow(rem, 3);
   number = number / 10;
   }
   /* If the sum of cubes of individual digits is equal
   * to the number which was passed as parameter
   * then that number is Armstrong number.So it returns true.
   * else it returns false
   */
   if (tot == value)
   {
   return true;
   }
   else
   {
   return false;
   }

   }

}

_______________________________

Output:

153 is an Armstrong number but it is not a perfect number.
6 is not an Armstrong number and it is a perfect number.

_______________Could you plz rate me well.Thank You