# Armstrong Number Program in Java

Armstrong Number Program in Java: A positive number is called an Armstrong number if the sum of the cube of each digit is equal to the number itself. There are only six Armstrong numbers present between 0 to 999. For example:

0 = 03 = 0

1 = 13 = 1

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

370 = 33 + 73 + 03 = 27 + 343 + 0 = 370

371 = 33 + 73 + 13 = 27 + 343 + 1 = 371

407 = 43 + 03 + 73 = 64 + 0 + 343 = 407

As we can see the given numbers are equal to the sum of cube of its digits. Hence, they are Armstrong number.

The following Java program demonstrates how to check a given number is an Armstrong number or not by using a while loop.

### Iterative Approach

Filename: ArmstrongNumberExample.java

```public class ArmstrongNumberExample
{
public static void main(String[] args)
{
int no = 407; // the number to be check if armstrong or not
int tempVar = no; // Creating a copy of the variable no
int sum = 0;
while(no > 0)
{
int dig = no % 10; //determining the modulo or remainder
sum = sum + (dig * dig * dig);//generating and adding cube of the remainder
no = no / 10; //removing the last digit
}
if(tempVar == sum)
System.out.print("The number " + tempVar +" is an armstrong number.");
else
System.out.print("The number " + tempVar +" is not an armstrong number.");
}
}```

Output:

`The number 407 is an Armstrong number.`

Explanation: The approach is quite straight-forward. In the first step, we assign the given number to a temporary variable named tempVar. It generates a copy of the variable no. After that, determine the remainder of the assigned no by using the modulo operator (no % 10) and the remainder store in the variable dig.

Now we find the cube of the remainder and resultant added to the variable sum. To remove the, we divide the no by 10 (no = no / 10). The same process is repeated till the given number is reduced to 0. Finally, we compare the copy of the given number with the value present in the sum variable. If the comparison between the copy (tempVar) and the sum variable yields true, then we can say the given number is an Armstrong number else not.

Let us discuss the recursive approach too.

Recursive Approach

Filename: ArmstrongNumberExample1.java

```public class ArmstrongNumberExample1
{
//function to check if the number is Armstrong or no
static boolean isArmstrong(int no, int tempVar, int sum)
{
if(no < 0)
return false;
if(no == 0) return sum == tempVar;
int dig = no % 10;
return isArmstrong(no/10, tempVar, sum + (dig * dig * dig));
}
public static void main(String[] args)
{
int a[] = {0, 1, 7, 79, 370, 407, 797, 152};
for(int i = 0; i < a.length; i++)
if(isArmstrong (a[i], a[i], 0))
System.out.println("The number " + a[i] +" is an Armstrong number.");
else
System.out.println("The number " + a[i] +" is not an Armstrong number.");
}
}```

Output:

```The number 0 is an Armstrong number.
The number 1 is an Armstrong number.
The number 7 is not an Armstrong number.
The number 79 is not an Armstrong number.
The number 370 is an Armstrong number.
The number 407 is an Armstrong number.
The number 797 is not an Armstrong number.
The number 152 is not an Armstrong number.```

Explanation: In the recursive approach also, we are doing the same thing. Firstly, we generate a copy of the given number. Then, reducing the number to 0 by removing its last digit and adding the cube of the last digit in sum variable. Finally, making a comparison to achieve the result.

Note: A negative number never become an Armstrong number.

### Armstrong Numbers Between the Two Numbers

We can also find the Armstrong number between the two given numbers. The following code illustrates the same.

Filename: ArmstrongNumberExample2.java

```public class ArmstrongNumberExample2
{
//function to check if the number is Armstrong or not
static boolean isArmstrong(int no, int tempVar, int sum)
{
if(no < 0) return false;
if(no == 0) return sum == tempVar;
int dig = no % 10;
return isArmstrong(no/10, tempVar, sum + (dig * dig * dig));
}
public static void main(String[] args)
{
int n1 = 0, n2 = 500 // Between n1 and n2, we will be checking Armstrong numbers.
System.out.println("Armstrong numbers between " + n1 + " and " + n2 + " are:" );
for(int i = n1; i <= n2; i++)
if(isArmstrong (i, i, 0))
System.out.print(i + " ");
}```

Output:

```Armstrong number between 0 and 500 are:
0 1 153 370 371 407```