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. 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
Steps:
- Enter any number
- Divide the given number into individual digits (For Example, Divide 153 into 1, 5 and 3) and count them.
- Calculate the power of n for each individual and add those numbers
- Compare original value with Sum value.
- 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 Program for Armstrong Number Using While Loop, For Loop, Functions and Recursion. We will also show you, Armstrong Numbers between 1 to n.
Find Armstrong Number in Java Using While Loop
This 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
ANALYSIS
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 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)
Sum = 256 + 2401
Sum = 2657
Temp = Temp /10
Temp = 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 = Sum + (Reminder * Reminder * Reminder * Reminder)
Sum = 2657 + (4 * 4 * 4 * 4)
Sum = 2657 + 256
Sum = 2913
Temp = Temp /10
Temp = 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 = Sum + (Reminder * Reminder * Reminder * Reminder)
Sum = 2913 + (9 * 9 * 9 * 9)
Sum = 2913 + 6561
Sum = 9474
Temp = Temp /10
Temp = 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
NOTE: If you are looking for the 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 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
JAVA CODE
// 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
We just replaced the While loop in the above example with the For loop.
Check Armstrong Number in Java using Functions
This 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 Functions
JAVA CODE
// 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
ANALYSIS
When the compiler reaches to ArmstrongNumber (Number) line in main() program 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 Java program, 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
ANALYSIS
ArmstrongNumbers Class Analysis:
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 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
ANALYSIS
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 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
ANALYSIS
This program allows the user 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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