Challenge 1: Catching ArithmeticException With a finally Block — Possible Solution
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DivideDemo.java:

public class DivideDemo {
    public static void main(String[] args) {
        try {
            int result = 10 / 0;
            System.out.println("Result: " + result);
        } catch (ArithmeticException e) {
            System.out.println("Caught: " + e.getMessage());
        } finally {
            System.out.println("This always runs.");
        }
    }
}

Output:

    Caught: / by zero
    This always runs.

Explanation:
10 / 0 with two ints throws ArithmeticException at runtime (integer
division by zero is not a compile-time error in Java). The catch
block matches it specifically and prints its message. The finally
block then runs unconditionally -- it would have run identically even
if no exception had been thrown at all, or if a completely different
exception type had propagated past this catch.

WHY THIS WORKS AS AN ANSWER
------------------------------
This catches a specific exception type (ArithmeticException) rather
than a broad Exception, per the chapter's own tip-box, and demonstrates
finally running regardless of the try block's outcome.