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AP Computer Science - The Armstrong Numbers

![APCSLogo.png](https://steemitimages.com/DQmXNmrb8ug59Z9Kck2Hq8wS9RJASeUFdrxruwWCNW8Fqri/APCSLogo.png) An Armstrong number is defined to be an integer where the sum each digit raised to the power of the number of digits in the number equals the number itself. Haha. Sound confusing? Well here is are two examples: ![](https://steemitimages.com/DQmWvoUGX4LTX5c53fWZ4d2D63bGSXAbARuF7TDW96yUHZQ/image.png) #### My program to find these numbers: Surprisingly this only took 53 lines of code... It can be done in less. ![Capture.PNG](https://steemitimages.com/DQmWiNHjLxNjviNViJUWeS3g6mTpwxM2XspAs5m9DfkZ7MF/Capture.PNG) #### The output: Here are all Armstrong numbers from 1 to 100,000 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748, 92727, 93084

If you want to try out the raw code yourself you can go to browxy.com or codiva.io and copy/paste this:

class Armstrong {
// This is the main class. It runs the program.
public static void main(String[] args) {
AllArmstrongNumbers(1,100000);
}
// This prints out all Armstrong numbers from 1st to 2nd argument.
public static void AllArmstrongNumbers(int min, int max) {
boolean ANum;
for (int i = min; i <= max; i++) {
ANum = isArmstrong(i);
if (ANum == true) {System.out.print(i + " ");}
}
}
// This counts the number of digits in a number.
public static int digitCount(int n) {
int count = 0;
while(n != 0) {
n = n / 10;
count++;
}
return count;
}
// This is an exponent calculator.
public static int power(int base, int exponent) {
int summ = base;
for(int i = 1; i < exponent; i++){
summ = summ*base;
}
return summ;
}
// This finds out if a number is Armstrong.
public static boolean isArmstrong(int num) {
int remainder;
int exp = digitCount(num);
int originalNumber = num;
boolean ArmstrongNumber;
int summ = 0;
for (int i = 1; i <= exp; i++) {
remainder = num % 10;
summ = summ + power(remainder,exp);
num = num / 10;
}
if (summ == originalNumber) {ArmstrongNumber = true;}
else {ArmstrongNumber = false;}
return ArmstrongNumber;
}
}

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Replies (23)

Hi, I took your great post as reason to implement this in python myself.

def getArmstongNumbers(start, end):
    armstongNumbers = []
    for nr in range(start, end):
        sum = 0
        for digit in str(nr):
            sum += pow(int(digit), len(str(nr)))
        if sum == nr:
            armstongNumbers += [nr]
    return armstongNumbers

print(getArmstongNumbers(1, 100000))

Maybe we will find more users who post their own implementation here.

[v for p,v in enumerate([sum([int(d)**len(str(i))for d in str(i)])for i in range(7**6)])if p==v]

The readability is lost, but it is even shorter.

I have just created a Java version:

import java.util.Vector;

public class ArmstrongNumbers {

	Vector<Integer> numbers = new Vector<Integer>();
	
	public ArmstrongNumbers(int from, int to) {
		for (int i=from; i<to; i++) {
			if (isArmstrongNumbers(i)) {
				numbers.add(i);
			}
		}
	}
	
	public static boolean isArmstrongNumbers(int number) {
		int len = (""+number).length();
		int workingNumber = number;
		int sum = 0;
		for (int i=0; i<len; i++) {
			sum += Math.pow(workingNumber%10, len);
			workingNumber /= 10;
		}
		return sum == number;
	}
	
	public String toString() {
		return numbers.toString();
	}
	
	public static void main(String[] args) {
		System.out.println(new ArmstrongNumbers(1,  10000000));
	}

}