Java Program to Check a Number is Armstrong

Java Program to Check Armstrong Number

In this example, we will learn to check whether a given number is armstrong number or not.

A positive integer is called an Armstrong number of order n if

abcd... = aⁿ + bⁿ + cⁿ + dⁿ + ...

In case of an Armstrong number of 3 digits, the sum of cubes of each digits is equal to the number itself.

For example:

153 = 1*1*1 + 5*5*5 + 3*3*3 // 153 is an Armstrong number.

Example 1:

public class Armstrong {

public static void main(String[] args) {

int number = 171, originalNumber, remainder, result = 0;

originalNumber = number;

while (originalNumber != 0)
{
remainder = originalNumber % 10;
result += Math.pow(remainder, 3);
originalNumber /= 10;
}

if(result == number)
System.out.println(number + " is an Armstrong number.");
else
System.out.println(number + " is not an Armstrong number.");
}
}

Output :

171 is not an Armstrong number

Example 2: Check Armstrong number for n digits

public class Armstrong {

public static void main(String[] args) {

int number = 153, myNumber, remainder, result = 0, n = 0;

myNumber = number;

for (;myNumber != 0; myNumber /= 10, ++n);

myNumber = number;

for (;myNumber != 0; myNumber /= 10)
{
remainder = myNumber % 10;
result += Math.pow(remainder, n);
}

if(result == number)
System.out.println(number + " is an Armstrong number.");
else
System.out.println(number + " is not an Armstrong number.");
}
}

Output :

153 is an Armstrong number.