How to Write Java Armstrong Number program using While Loop, For Loop, Functions, and Recursion?. We will also show the Java Armstrong Numbers between 1 to n.
Java Armstrong Number
If the given number is equal to the sum of the cubes of each digit equals to the number, that can be Armstrong Number in Java. For example, 153 is Armstrong Number because of the Number of individual digits in 153 = 3 and
153 = 1³ + 5³ + 3³ ==> 1 + 125 + 27 ==> 153
The below steps will show the common approach to check for the Armstrong Number in Java
- Enter any number and divide that 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 the original value with Sum value.
- If they exactly matched, then it is Java Armstrong number else it is not Armstrong
Java Armstrong Number using While Loop
This program allows the user to enter any positive integer. This program will check whether a number is Armstrong or Not using the Java While Loop
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);
}
}
}
In this Java Armstrong Number program, the first two statement will ask the user to enter any positive integer and assign it to a variable.
Next, We assign the original value to the Temp variable. It will help us to preserve our original value and then do all the manipulation on the Temp variable.
The first 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 individual digits inside the given value. If you do not understand the logic, please refer to Java Program to Count Number Of Digits in a Number article.
The second Java while loop will keep the given number greater than 0 and calculates the sum of the powers.
Let us see the working principle of this Java Armstrong Number while loop in iteration wise. Here, the user Entered value are 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
Sum = Sum + (Reminder * Reminder * Reminder * Reminder)
Sum = 0 + (4 * 4 * 4 * 4) = 0 + 256 = 256
Temp = Temp /10
Temp = 9474 /10 = 947
If the digits count is 5, then Reminder will be multiplied by 5 times.
Armstrong number in Java Second Iteration: From the first Iteration, the values of both Temp and Sum changed as Temp = 947 and Sum = 256
Reminder = 947 % 10 = 7
Sum = 256 + (7 * 7 * 7 * 7) => 256 + 2401
Sum = 2657
Temp = 947 /10 = 94
Third Iteration: From the Third Iteration, the values of Temp = 94 and Sum = 2657
Reminder = 94 % 10 = 4
Sum = 2657 + (4 * 4 * 4 * 4) = >2657 + 256
Sum = 2913
Temp = 94 /10 = 9
Fourth Iteration: Temp = 9 and Sum = 2913
Reminder = 9 % 10 = 9
Sum = 2913 + (9 * 9 * 9 * 9) => 2913 + 6561
Sum = 9474
Temp = 9/10 = 0
Here Temp= 0. So, the while loop condition will fail
if ( Number== Sum ) – the condition will check whether the user enter one is exactly equal to Sum or not. If this condition is True, then it is Armstrong else it is not Armstrong number in Java
NOTE: If you are looking for the Armstrong number below 1000, then remove the while loop to count the of digits. 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 will check whether a number is Armstrong or Not using Java 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);
}
}
}
Please Enter number to Check for Armstrong:
1634
Sum of entered number is = 1634.00
1634 is a Armstrong NumberCheck Java Armstrong Number using Functions
This program for Armstrong Number in Java allows you to enter any positive integer. Then, this program will check whether a number is Armstrong or Not using Functions.
package FrequentPrograms;
import java.util.Scanner;
public class ArmstrongNumberUsingMethods {
private static Scanner sc;
public static void main(String[] args) {
int Num;
sc = new Scanner(System.in);
System.out.println("\n Please Enter integer to Check : ");
Num = sc.nextInt();
ArmstrongNumber(Num);
}
public static void ArmstrongNumber(int Num) {
int Temp, Reminder, Times = 0;
double Sum = 0;
//Helps to prevent altering the original value
Temp = Num;
while (Temp != 0) {
Times = Times + 1;
Temp = Temp / 10;
}
for(Temp = Num; Temp > 0; Temp = Temp /10 ) {
Reminder = Temp %10;
Sum = Sum + Math.pow(Reminder, Times);
}
System.out.format("\n Sum = %.2f", Sum);
if (Sum == Num) {
System.out.format("\n% d is a Armstrong", Num);
}
else {
System.out.format("\n% d is NOT a Armstrong", Num);
}
}
}
Please Enter integer to Check :
153
Sum = 153.00
153 is a ArmstrongWhen the Javac compiler reaches to ArmstrongNumber(Num) line in the main() program, then the compiler will immediately jump to below function:
public static void ArmstrongNumber(int Num) {
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 that holds a method to reverse an integer recursively.
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 and call the methods
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);
}
}
}
Please Enter number to Check for Armstrong:
371
Sum of entered number is = 371.00
371 is a Armstrong NumberArmstrongNumbers Class Analysis:
In this Java Armstrong Number program, we created one function to count the digits in it. 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 the 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 an 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 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 will check whether a number is Armstrong or Not using Recursion concept.
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
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);
}
}
}
Please Enter number to Check for Armstrong:
370
Sum of entered number is = 370.00
370 is a Armstrong NumberIn this Java Armstrong Number program, Within the function, Check_Armstrong(int Number, int Times), we are using the If statement to check whether the given number is greater than zero or not. It is very important, and if you forget this, then it ends up in an 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);
The above Statement will help to call the function Recursively with the updated value. If we miss this statement, then after completing the first line, it will terminate. For example, Number= 153 then the output will be 27.
Java Program to Find Armstrong Numbers between 1 to N
This Java Armstrong Number program allows the user to enter the minimum and maximum values. This program will find the Armstrong Numbers between the Minimum and Maximum values.
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);
}
}
}
}
Armstrong Numbers from 100 to 100000 output
Please Enter the Minimum Value:
100
Please Enter the Maximum Value:
100000
153 370 371 407 1634 8208 9474 54748 92727 93084 The For Loop helps the Javac compiler to iterate between Minimum and Maximum Variables. The iteration starts at the Minimum, and then it will not exceed the Maximum variable.
