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Armstrong Number in Java

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Java Armstrong Number

If the given number is equal to the sum of the power of n for each digit present in that integer then, that number can be Armstrong Number in Java.

For example 153 is Armstrong Number because Number of individual digits in 153 = 3 and

153 = 1³ + 5³ + 3³ ==> 1 + 125 + 27 ==> 153

Below steps will show you the common approach to check for the Armstrong Number in Java

Steps:

  1. Enter any number
  2. Divide the given number into individual digits (For Example, Divide 153 into 1, 5 and 3) and count them.
  3. Calculate the power of n for each individual and add those numbers
  4. Compare original value with Sum value.
  5. If they exactly matched then it is Armstrong number else it is not Armstrong

In this article we will show you, How to Write Java Armstrong Number program using While Loop, For Loop, Functions and Recursion. We will also show you, Armstrong Numbers between 1 to n.

Java Armstrong Number using While Loop

This Armstrong Number in Java program allows the user to enter any positive integer and, this program will check whether a number is Armstrong Number or Not using the Java While Loop

JAVA CODE

// Armstrong Number in Java Programming

package FrequentPrograms;

import java.util.Scanner;

public class ArmstrongNumberUsingWhile {
	private static Scanner sc;
	
	public static void main(String[] args) {
		int Number, Temp, Reminder, Times = 0;
		double Sum = 0;
		sc = new Scanner(System.in);		
		System.out.println("\n Please Enter number to Check for Armstrong: ");
		Number = sc.nextInt();

		//Helps to prevent altering the original value
		Temp = Number;
		while (Temp != 0) {
			Times = Times + 1;
			Temp = Temp / 10;
		   }
		   
		Temp = Number;
		while( Temp > 0)  {
			Reminder = Temp %10;
		    Sum = Sum + Math.pow(Reminder, Times);
		    Temp = Temp /10;
		   }
		System.out.format("\n Sum of entered number is = %.2f", Sum);
		
		if (Sum == Number) {
			System.out.format("\n% d is a Armstrong Number", Number);
		}
		else {
			System.out.format("\n% d is NOT a Armstrong Number", Number);
		}
	}
}

OUTPUT

Armstrong Number in Java using While Loop

ANALYSIS

In this Java Armstrong Number program, the following statement will ask the user to enter any positive integer and then, that number will be assigned to Number variable.

System.out.println("\n Please Enter number to Check for Armstrong: ");
Number = sc.nextInt();

Next, We assign the original value to the Temp variable. This will help us to preserve our original value and then do all the manipulation on Temp variable.

Below While loop will make sure that, the given number is greater than 0 and statements inside the while loop will split the numbers and counts number of individual digits inside the given number. If you don’t understand the logic, Please refer Java Program to Count Number Of Digits in a Number article.

while (Temp != 0) {
	Times = Times + 1;
	Temp = Temp / 10;
}

Second While loop will make sure that, the given number is greater than 0 and calculate the sum of the powers of given number.

while( Temp > 0)  {
    Reminder = Temp %10;
    Sum = Sum + Math.pow(Reminder, Times);
    Temp = Temp /10;
}

Let us see the working principle of this Java Armstrong Number while loop in iteration wise. Here, User Entered value: Number = 9474 and Sum = 0

Temp = Number
Temp = 9474

First Iteration
Reminder = Temp %10
Reminder = 9474 % 10 = 4

Sum = Sum + Math.pow (Reminder, Times)

For this example, Times = 4 because number of digits in 9474 = 4. So math.pow function will multiply the Reminder 4 times as shown below

Sum = Sum + (Reminder * Reminder * Reminder * Reminder)
Sum = 0 + (4 * 4 * 4 * 4)
Sum = 0 + 256
Sum = 256

Temp = Temp /10
Temp = 9474 /10
Temp = 947

NOTE: If the number of digits count is 5 then Reminder will be multiplied by 5 times.

Second Iteration
From the first Iteration the values of both Temp and Sum has been changed as: Temp = 947 and Sum = 256

Reminder = Temp %10
Reminder = 947 % 10 = 7

Sum = Sum + (Reminder * Reminder * Reminder * Reminder)
Sum = 256 + (7 * 7 * 7 * 7) => 256 + 2401
Sum = 2657

Temp = Temp /10 = 947 /10
Temp = 94

Third Iteration
From the Third Iteration the values of both Temp and Sum has been changed as: Temp = 94 and Sum = 2657

Reminder = Temp %10
Reminder = 94 % 10 = 4

Sum = 2657 + (4 * 4 * 4 * 4) = >2657 + 256
Sum = 2913

Temp = Temp /10 = 94 /10
Temp = 9

Fourth Iteration
From the Fourth Iteration the values of both Temp and Sum has been changed as: Temp = 9 and Sum = 2913

Reminder = Temp %10
Reminder = 9 % 10 = 9

Sum = 2913 + (9 * 9 * 9 * 9) => 2913 + 6561
Sum = 9474

Temp = Temp /10 = 9/10
Temp = 0

Here Number = 0 so, the while loop condition will fail

if ( Number == Sum ) – Condition will check whether the user enter number is exactly equal to Sum number or not. If this condition is True, then it is Armstrong else the given number is not Armstrong number in Java

NOTE: If you are looking for the Java Armstrong number below 1000 then you can simply remove the while loop to count the number of digits in a number and then replace the below code

Sum = Sum + Math.pow(Reminder, Times);

With

Sum = Sum + (Reminder * Reminder * Reminder)

Check Armstrong Number in Java using For Loop

This Java Armstrong Number program allows the user to enter any positive integer and then, this program will check whether a number is Armstrong Number or Not using Java For Loop

// Check Armstrong Number in Java using For Loop

package FrequentPrograms;

import java.util.Scanner;

public class ArmstrongNumberUsingFor {
	private static Scanner sc;
	
	public static void main(String[] args) {
		int Number, Temp, Reminder, Times = 0;
		double Sum = 0;
		sc = new Scanner(System.in);		
		System.out.println("\n Please Enter number to Check for Armstrong: ");
		Number = sc.nextInt();

		//Helps to prevent altering the original value
		Temp = Number;
		while (Temp != 0) {
			Times = Times + 1;
			Temp = Temp / 10;
		   }
		   
		for(Temp = Number; Temp > 0; Temp =  Temp /10 )  {
			Reminder = Temp %10;
		    Sum = Sum + Math.pow(Reminder, Times);
		   }
		System.out.format("\n Sum of entered number is = %.2f", Sum);
		
		if (Sum == Number) {
			System.out.format("\n% d is a Armstrong Number", Number);
		}
		else {
			System.out.format("\n% d is NOT a Armstrong Number", Number);
		}
	}
}

OUTPUT

Armstrong Number in Java using For Loop

We just replaced the While loop in the above Java Armstrong Number example with the For loop.

Check Java Armstrong Number using Functions

This program for Armstrong Number in Java allows you to enter any positive integer and then, this program will check whether a number is Armstrong Number or Not using Functions.

// Check Armstrong Number in Java using FUnctions

package FrequentPrograms;

import java.util.Scanner;

public class ArmstrongNumberUsingMethods {
	private static Scanner sc;
	
	public static void main(String[] args) {
		int Number;
		sc = new Scanner(System.in);		
		System.out.println("\n Please Enter number to Check for Armstrong: ");
		Number = sc.nextInt();
		ArmstrongNumber(Number);
	}
	public static void ArmstrongNumber(int Number) {
		int Temp, Reminder, Times = 0;
		double Sum = 0;
		//Helps to prevent altering the original value
		Temp = Number;
		while (Temp != 0) {
			Times = Times + 1;
			Temp = Temp / 10;
		   }
		   
		for(Temp = Number; Temp > 0; Temp =  Temp /10 )  {
			Reminder = Temp %10;
		    Sum = Sum + Math.pow(Reminder, Times);
		   }
		System.out.format("\n Sum of entered number is = %.2f", Sum);
		
		if (Sum == Number) {
			System.out.format("\n% d is a Armstrong Number", Number);
		}
		else {
			System.out.format("\n% d is NOT a Armstrong Number", Number);
		}
	}
}

OUTPUT

Armstrong Number in Java using Functions

ANALYSIS

When the compiler reaches to ArmstrongNumber (Number) line in main() program of Java Armstrong Number then the compiler will immediately jump to below function:

public static void ArmstrongNumber(int Number) {

We already explained LOGIC above example.

NOTE: If we create a function with Void then there is no need to return any value but, if we declared a function with any data type (int, float etc) then we have return something out from the function.

Check Armstrong Number in Java using Oops

In this program to find Armstrong Number in Java, we are dividing the code using the Object Oriented Programming. To do this, First we will create a class which holds a method to reverse an integer recursively.

JAVA CODE

package FrequentPrograms;

public class ArmstrongNumbers {
	int Times, Reminder; 		

	public int Count_Numbers(int Number) {
		while (Number != 0) {
			Times = Times + 1;
			Number = Number / 10;
		}
		return Times;
	}
	
	public double Check_Armstrong(int Number) {
		double Sum = 0;
		Times = Count_Numbers(Number);
		while(Number > 0) {
			Reminder = Number %10;
			Sum = Sum + Math.pow(Reminder, Times);
			Number = Number /10;
		   }
		return Sum;
	}
	
}

Within the Main program, we will create an instance of the above specified class and call the methods

JAVA CODE

package FrequentPrograms;

import java.util.Scanner;

public class ArmstrongNumberUsingClass {
	private static Scanner sc;
	
	public static void main(String[] args) {
		int Number;
		double Sum = 0;
		
		sc = new Scanner(System.in);		
		System.out.println("\n Please Enter number to Check for Armstrong: ");
		Number = sc.nextInt();

		ArmstrongNumbers arms = new ArmstrongNumbers();
		Sum = arms.Check_Armstrong(Number);   
		
		System.out.format("\n Sum of entered number is = %.2f", Sum);
		
		if (Sum == Number) {
			System.out.format("\n% d is a Armstrong Number", Number);
		}
		else {
			System.out.format("\n% d is NOT a Armstrong Number", Number);
		}
	}
}

OUTPUT

Armstrong Number in Java using Class

ANALYSIS

ArmstrongNumbers Class Analysis:

In this Java Armstrong Number program, First we created one function to count number of digits in a given Number. Next, we declared one more function to calculate the sum of the power of n for each digit present in that integer. We already explained the Logic in above example.

NOTE: With the Check_Armstrong(int Number) method we are calling the Count_Number() function to get the Times value. If you want to pass the power value as argument then declare one more argument and remove this line.

Main Class Analysis:

First we created an instance / created an Object of the ArmstrongNumbers Class

ArmstrongNumbers arms = new ArmstrongNumbers();

Next, we are calling the Check_Armstrong(Number) method.

Sum = arms.Check_Armstrong(Number);

Lastly, System.out.println statement will print the output.

Check Armstrong Number in Java using Recursion

This Armstrong Number in Java program allows the user to enter any positive integer and then, this program will check whether a number is Armstrong Number or Not using Recursion concept.

JAVA CODE

package FrequentPrograms;

public class ArmstrongNumber {
	int Reminder; 		
	double Sum = 0;
	
	public double Check_Armstrong(int Number, int Times) {
		if (Number > 0) {
			Reminder = Number %10;
			Sum = Sum + Math.pow(Reminder, Times);
			Check_Armstrong(Number /10, Times);
		}
		return Sum;
	}
}

Within the Main program, we will create an instance of the above specified class and call the methods

JAVA CODE

// Armstrong Number in Java Programming

package FrequentPrograms;

import java.util.Scanner;

public class ArmstrongNumberUsingRecursion {

	private static Scanner sc;
	
	public static void main(String[] args) {
		int Number, Times;
		double Sum = 0;
		
		sc = new Scanner(System.in);		
		System.out.println("\n Please Enter number to Check for Armstrong: ");
		Number = sc.nextInt();

		ArmstrongNumbers arms = new ArmstrongNumbers();
		Times = arms.Count_Numbers(Number);   
		
		ArmstrongNumber arm = new ArmstrongNumber();
		Sum = arm.Check_Armstrong(Number, Times); 
		
		System.out.format("\n Sum of entered number is = %.2f", Sum);
		
		if (Sum == Number) {
			System.out.format("\n% d is a Armstrong Number", Number);
		}
		else {
			System.out.format("\n% d is NOT a Armstrong Number", Number);
		}
	}
}

OUTPUT

Armstrong Number in Java using Recursion

ANALYSIS

In this Java Armstrong Number program, Within the function, Check_Armstrong(int Number, int Times) we are using If statement to check whether the given number is greater than zero or not. This is very important and if you forget this then you will end up in infinite loop.

if (Number > 0) {

We already explained the logic in our previous examples. Within the If statement we used one 1 extra line.

Check_Armstrong(Number /10, Times);

Above Statement will help to call the function Recursively with updated value. If you miss this statement then, after completing the first line it will terminate. For example,

Number = 153

Then the output will be 27

Please be careful :)

Java Program to Find Armstrong Numbers between 1 to N

This Java Armstrong Number program allows the user to enter minimum and maximum values. This program will find the Armstrong Numbers between the Minimum and Maximum values.

JAVA CODE

// Java Program to Find Armstrong Numbers between 1 to n

package FrequentPrograms;

import java.util.Scanner;

public class ArmstrongNumbersbetween {
	private static Scanner sc;
	
	public static void main(String[] args) {
		int Temp, Number, Minimum, Maximum, Reminder;
		sc = new Scanner(System.in);	
		
		System.out.println("\n Please Enter the Minimum Value: ");
		Minimum = sc.nextInt();
		
		System.out.println("\n Please Enter the Maximum Value: ");
		Maximum = sc.nextInt();
		
		//ArmstrongNumbers arms = new ArmstrongNumbers();
		for(Number = Minimum; Number <= Maximum; Number++) {
			int sum = 0, Times = 0;
			Temp = Number;
			while (Temp > 0) {
				Times = Times + 1;
				Temp = Temp / 10;
			}
			Temp = Number;
			while( Temp > 0)  {
				Reminder = Temp %10;
				sum = sum + (int)Math.pow(Reminder, Times);
				Temp = Temp /10;
			}
			if(sum == Number) {
				System.out.format(" %d  ", Number);	
			}
		}	
	}
}

OUTPUT

Java Program Find Armstrong Numbers between 1 and N

ANALYSIS

This Java Armstrong Number program allows to enter minimum and maximum values and following For Loop helps compiler to iterate between Minimum and Maximum Variables, iteration starts at the Minimum and then it will not exceed Maximum variable.

for(Number = Minimum; Number <= Maximum; Number++) {

We already explained the logic in our first example so, please recheck the Find Armstrong Number in Java using while loop example.

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