Armstrong Number Program in JAVA

Details

What is Armstrong Number ?

In an Armstrong Number, the sum of power of individual digits is equal to number itself.

In other words the following equation will hold true 

xy..z = xn + yn+.....+ zn

n is number of digites in number

For example this is a 3 digit Armstrong number 

370 = 33 + 73 + o3
 = 27 + 343 + 0
 = 370

Examples of Armstrong Numbers 

 0, 1, 4, 5, 9, 153, 371, 407, 8208, etc.

Let’s write this in a program:

Java Program to check whether a number is Armstrong Number

//ChecktempNumber is Armstrong or not using while loop
package com.guru99;
 
public class ArmstrongNumber {
 
	public static void main(String[] args) {
		
		 int inputArmstrongNumber = 153; //Input number to check armstrong  
		 int tempNumber, digit, digitCubeSum = 0;
 
	       tempNumber = inputArmstrongNumber;
	        while (tempNumber != 0)
	        {
	        	
	        	/* On each iteration, remainder is powered by thetempNumber of digits n
	        	 */
	            System.out.println("Current Number is "+tempNumber);
	            digit =tempNumber % 10;
				System.out.println("Current Digit is "+digit);
	            //sum of cubes of each digits is equal to thetempNumber itself
	            digitCubeSum = digitCubeSum + digit*digit*digit;
				System.out.println("Current digitCubeSum is "+digitCubeSum);
	            tempNumber /= 10;
	           
	        }
 
	        //check giventempNumber and digitCubeSum is equal to or not 
	        if(digitCubeSum == inputArmstrongNumber)
	            System.out.println(inputArmstrongNumber + " is an Armstrong Number");
	        else
	            System.out.println(inputArmstrongNumber + " is not an Armstrong Number");
 
	}
 
}
Output 
Current Number is 153
Current Digit is 3
Current digitCubeSum is 27
Current Number is 15
Current Digit is 5
Current digitCubeSum is 152
Current Number is 1
Current Digit is 1
Current digitCubeSum is 153
153 is an Armstrong Number

Java Program to Print Armstrong numbers from 0 to 999 

//ChecktempNumber is Armstrong or not using while loop
package com.guru99;

public class ArmstrongNumber {

    public static void main(String[] args) {
        int tempNumber, digit, digitCubeSum;

        for (int inputArmstrongNumber = 0; inputArmstrongNumber < 1000; inputArmstrongNumber++) {
            tempNumber = inputArmstrongNumber;
            digitCubeSum = 0;
            while (tempNumber != 0) {

                /* On each iteration, remainder is powered by thetempNumber of digits n
                 */

                digit = tempNumber % 10;

                //sum of cubes of each digits is equal to thetempNumber itself
                digitCubeSum = digitCubeSum + digit * digit * digit;

                tempNumber /= 10;

            }

            //check giventempNumber and digitCubeSum is equal to or not 
            if (digitCubeSum == inputArmstrongNumber)
                System.out.println(inputArmstrongNumber + " is an Armstrong Number");

        }

    }

}
Output 
0 is an Armstrong Number
1 is an Armstrong Number
153 is an Armstrong Number
370 is an Armstrong Number
371 is an Armstrong Number
407 is an Armstrong Number