Java Ice Breakers

How to check whether a number is Armstrong number or not

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by

anagendrakumar2020
in ,

This is the first solution to check the number for Armstrong. This solution uses String class, and a couple of Character functions to evaluate the sum of the cubes of each digit in a given number. The Armstrong number is say for a given number 153, evaluate the powers of cubes from any side as power (1 * 3) + power (5 * 3) + power (3 *3) , the sum of all powers must be equal to the given number 153.

public class ArmstrongNumbers {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		int sumCheck=0;
		System.out.println("The Armstrong numbers between 1 to 1000");
		for (int numValue=1;numValue<=1000;numValue++) {
			// Convert the number to String for easy processing and to avoid the second loop
			String numToStr=String.valueOf(numValue);
			int numLen=numToStr.length();

			// Calculate the sum of each digit's cube power value from any side
			for (;numLen>0;numLen--)
				if(Character.isDigit(numToStr.charAt(numLen-1))) 
					sumCheck = (int) (sumCheck + Math.pow(Character.getNumericValue(numToStr.charAt(numLen-1)),3));

			//Compare the sumCheck with actual number       
			if (sumCheck == numValue)
				System.out.println(numValue+" is Armstrong number");
			//Reset sumCheck to 0 for next iteration
			sumCheck=0;
		}
	}
}
Try it >>
Output:
The Armstrong numbers between 1 to 1000
1 is Armstrong number
153 is Armstrong number
370 is Armstrong number
371 is Armstrong number
407 is Armstrong number

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