The assignment said to make at least three comments.

gistfile1.java

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;


public final class DigitToolsApplication extends Application {
	public static void main(String[] args) {
		new DigitToolsApplication().run();
	}
	
	@Override
	protected void runCycle(Scanner scanner) {
		//Ask for input
		printInline("Please type a number: ");
		String nextLine = scanner.nextLine();
		//Try to convert the input (which is a string) to an Integer so that we can manipulate it
		Integer i;
		try {
			//Try to get the value of the string
			i = Integer.valueOf(nextLine);
		} catch (NumberFormatException e) {
			//But in the event that the string is invalid, we will throw an exception (error).
			throw new RuntimeException("Could not read passed digit", e);
		}
		//Now, we setup our variables to track the data we want
		Integer count, sum = 0, average, reverse = 0;
		//we want to also know if this is a perfect square
		boolean isPerfectSquare = false;
		//Along with some lists to store the digits and divisors
		List<Integer> divisors = new ArrayList<>();
		List<Integer> digits = new ArrayList<>();
		
		//Now, we create a temporary copy of "i" (which is the number we're manipulating), and start separating out the digits
		//This loop creates a copy of I, repeats as long as I > 0, and divides i by 10 after each loop (moving the decimal place)
		for (int workingI = i; workingI > 0; workingI = workingI/10) {
			//Now we get the remainder of division by 10
			//12 % 10 = 2 and the last digit in 12 is 2 thus we have the last digit
			digits.add(workingI % 10);
			//Now, we let the for loop cut off that last digit by dividing by 10 (see last statement in for loop above)
		}
		//Now that we have the digits in a list, we can get the size of that list as our digit count
		count = digits.size();
		//And we can iterate through all the digits to calculate the sum.
		for (Integer i1 : digits) {//For every digit, add it to sum
			sum += i1;
		}
		//Now that we have both the sum and the count, we can divide the two to get the average
		average = sum/count;
		//Next part, let's get the reverse
		//we go in reverse through the digits array list using a for loop starting at count-1
		//count-1 is the last index in the list of digits
		//Then, we use the N variable to keep track of how many times we've run the for loop
		//So, for each iteration, we add one to N (the number of times this for loop has run) and
		//remove one from index, so that we can traverse the array in reverse
		//this will continue until index < 0 (the first index)
		for (int index = count-1, n = 0; index >= 0; index--, n++) {
			//Now we add 10 to the power of N times the digit in question
			reverse += (int)Math.pow(10, n)*digits.get(index);
		}
		
		//Lastly, we need to calculate the divisors. We simply try every whole number between 1 and the square root of i
		//we use the square root because that is the "midpoint" where all divisors on either end will match up.
		//Imagine a number line going from 1 to i. In the middle of this number line is Math.sqrt(i)
		//Any divisor on the left side of that midpoint will have a partner on the right side that is equal to i/x
		//For example, all numbers have a divisor of 1. The "pair" for this divisor is i, and you can get that by doing i/1
		for (int x = 1, sqrt = (int) Math.ceil(Math.sqrt(i)); x <= sqrt ; x++) {
			//If the number evenly divides, it is a divisor.
			if (i % x == 0) {
				//If we have found a divisor, we add it to our divisor collection.
				divisors.add(x);
				//Then, to avoid adding i itself, we check if x == 1. If so, we do not add i/x (the matching divisor) to the array
				//otherwise, we add the matching divisor
				//We also want to avoid adding the same number twice. This is useful only for perfect squares, like 25, where x = 5; 25/5 = 5, thus 5 will be added twice
				int matchingPair = i/x;
				if (x != 1 && !divisors.contains(matchingPair)) divisors.add(matchingPair);
				//also, although we likely know this is true given that the above statement failed (only two cases in which if fails, when x == 1, or when i/x = x, or x*x = i ;)
				//we want to track if this is a perfect square.
				else if (x == sqrt) isPerfectSquare = true;
			}
		}
		//Lastly, since we added divisor matches next to their original divisors, the array will be out of order. We want to sort it
		//Java will automatically sort by least-->greatest in an array containing numbers.
		Collections.sort(divisors);
		//Now we need to check if the number is perfect, so we're going to calculate the sum of all the divisors.
		int divisorSum = 0;
		//For each divisor, add it to the sum.
		for (Integer divisor : divisors) {
			divisorSum += divisor;
		}
		//And now we can do the output
		print(">Number " + i, 
				"\t>Sum " + sum,
				"\t>Average " + average, 
				"\t>Reverse " + reverse, 
				"\t>Divisors " + divisors, 
				//This checks if the condition is true, and if so uses the first value. Otherwise, it will use the second.
				"\t>Is perfect " + (divisorSum == i ? "Yes" : "No"),
				"\t>Is perfect Square " + (isPerfectSquare ? "Yes" : "No"));
	}
}