ISC Practicals and Theory - Armstrong Number (OOP)

An Armstrong number is a number that equals the sum of its digits, each raised to the power of the number of digits. This version is written as a class Armstrong with data members and separate methods for accepting the number, counting digits, finding a power, checking and displaying.

Definition

A number $N$ with $k$ digits is an Armstrong number if the sum of the $k$-th powers of its digits is equal to $N$.

• $9474$ has 4 digits: $9^4 + 4^4 + 7^4 + 4^4 = 6561 + 256 + 2401 + 256 = 9474$. So 9474 is an Armstrong number.
• $100$ has 3 digits: $1^3 + 0^3 + 0^3 = 1$, which is not $100$. So 100 is not an Armstrong number.

Algorithm

1. In main, accept the number num using Scanner. If num <= 0, display an invalid input message and stop.
2. Create an object obj of class Armstrong; the constructor sets the data members n and digits to 0.
3. Call accept(num) to store the number in n.
4. Call display(), which calls isArmstrong() and prints the result.
5. isArmstrong() works as follows:
   • Call countDigits() and store the result in digits. It copies n to a local variable and counts how many times it can be divided by 10 before it becomes 0.
   • Copy n to temp and set sum = 0.
   • While temp > 0: extract d = temp % 10, add power(d, digits) to sum, then do temp = temp / 10.
   • Return true if sum == n, otherwise false.
6. power(b, e) multiplies p by b exactly e times with a for loop and returns it.

Program

The data members are n and digits. The methods are accept(), countDigits(), power(), isArmstrong() and display(); main only reads the input and creates the object.

import java.util.Scanner;
public class Armstrong
{
    int n, digits;
    Armstrong()
    {
        n = 0;
        digits = 0;
    }
    void accept(int num)
    {
        n = num;
    }
    int countDigits()
    {
        int temp = n, c = 0;
        while (temp > 0)
        {
            c++;
            temp = temp / 10;
        }
        return c;
    }
    int power(int b, int e)
    {
        int p = 1;
        for (int i = 1; i <= e; i++)
        {
            p = p * b;
        }
        return p;
    }
    boolean isArmstrong()
    {
        digits = countDigits();
        int temp = n, sum = 0;
        while (temp > 0)
        {
            int d = temp % 10;
            sum = sum + power(d, digits);
            temp = temp / 10;
        }
        return sum == n;
    }
    void display()
    {
        if (isArmstrong())
            System.out.println(n + " is an Armstrong number.");
        else
            System.out.println(n + " is not an Armstrong number.");
    }
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter a number: ");
        int num = sc.nextInt();
        if (num <= 0)
        {
            System.out.println("Invalid input. Enter a positive number.");
        }
        else
        {
            Armstrong obj = new Armstrong();
            obj.accept(num);
            obj.display();
        }
    }
}

Variable description table

Sample input and output

Enter a number: 9474
9474 is an Armstrong number.

Enter a number: 100
100 is not an Armstrong number.

Enter a number: 0
Invalid input. Enter a positive number.

Dry run

Take n = 9474. countDigits() returns 4, so digits = 4. Then isArmstrong() runs with temp = 9474 and sum = 0.

The loop stops as temp is 0. sum (9474) equals n (9474), so isArmstrong() returns true and display() prints that 9474 is an Armstrong number.

Variations

1. Armstrong numbers in a range "Accept two numbers m and n and display all the Armstrong numbers between m and n."
2. Three-digit Armstrong numbers "Display all the Armstrong numbers from 100 to 999."
3. Using a method that returns boolean "Write a method that accepts an integer and returns true if it is an Armstrong number, otherwise false."