In this blog, we will develop the Armstrong number program in Java. First, we will develop a java program to check an Armstrong number, and then we will develop a java program for an Armstrong number between 1 to 1000.

A positive integer is called Armstrong number of order n if, abcd…. = aⁿ + bⁿ + cⁿ + dⁿ + ….

For Example :- 153
1³ + 5³ + 3³ = 1 + 125 + 27 = 153
So, 153 is an Armstrong number of order 3.

4150 = 4⁵ + 1⁵ + 5⁵ + 0⁵ = 1,024 + 1 + 3,125 + 0 = 4150
So, 4150 is an Armstrong number of order 5

Procedure to check Armstrong number of order N

1) Take a variable and take an order to check
2) Declare variables lastDigit, power, and sum Initialize sum with 0
3) Take a temporary variable n to store numbers
4) Find the last digit of n
5) Calculate the power of that lastDigit with order i.e. pow(lastDigit, order)
6) Add the result into the sum
7) Remove the last digit
8) Repeat steps 4 to 7 until the number becomes 0
9) Compare sum value and the actual number
==> If both are the same then it is the Armstrong number of the given order
==> Else it is not Armstrong number of the given order

Java method to check Armstrong number of order N

public static boolean isArmstrong(int number, int order){

   // declare variables
   int lastDigit = 0;
   int power = 0;
   int sum = 0;

   // temporary variable to store number
   int n = number;

   while(n!=0) {

      // find last digit
      lastDigit = n % 10;

      // find power of digit
      power = (int) Math.pow(lastDigit, order);

      // add power value into sum
      sum += power;

      // remove last digit
      n /= 10;
   }

   if(sum == number) return true;
   else return false;
}

In this method to find the power, we take the support of the pow(), which is defined in the Math class. The pow() is a static method so we can call it without creating an object of the Math class, and it returns double so we need to use type casting it to int type.

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Java program for Armstrong number between 1 to 1000

We can also find all Armstrong numbers of a given order in the given range. For this purpose, we need to take the minimum, and maximum values of the range and order to check the Armstrong number.

import java.util.Scanner;

public class ArmstrongNumberInRange {
   public static boolean isArmstrong(int number, int order){

      // declare variables
      int lastDigit = 0;
      int power = 0;
      int sum = 0;

      // temporary variable to store number
      int n = number;

      while(n!=0) {
         // find last digit
         lastDigit = n % 10;

         // find power of digit
         power = (int) Math.pow(lastDigit, order);

         // add power value into sum
         sum += power;

         // remove last digit
         n /= 10;
      }

      if(sum == number) return true;
      else return false;
  }

  public static void main(String[] args) {

      // declare variables
      int minRange , maxRange;
      int order = 0;

      // create Scanner class object
      Scanner scan = new Scanner(System.in);

      // read inputs
      System.out.print("Enter min & max "+
                       "Range value:: ");
      minRange = scan.nextInt();
      maxRange = scan.nextInt();

      System.out.print("Enter order to check::");
      order = scan.nextInt();

      // check in range
      System.out.println("Armstrong numbers"+
           " from "+minRange+" to "+maxRange+
           " of order " +order+" is:: ");

      for(int i = minRange; i<= maxRange; i++)
         if(isArmstrong(i, order)) 
            System.out.print( i + " ");

      // close Scanner class object
      scan.close();
  }
}

The output for different test-cases :-

Enter min and max Range value: 1 1000

Enter an order to check: 3

Armstrong numbers from 1 to 1000 of order 3 is:

1 152 370 371 407

Enter min and max Range value: 1 10000

Enter an order to check: 5

Armstrong numbers from 1 to 10000 of order 5 is:

1 4150 4151