Challenge 1: Annotate a Function — Possible Solution
====================================================================

def fizzbuzz(n: int) -> str:
    if n % 3 == 0 and n % 5 == 0:
        return "FizzBuzz"
    elif n % 3 == 0:
        return "Fizz"
    elif n % 5 == 0:
        return "Buzz"
    else:
        return str(n)

print(fizzbuzz(15))

Output:
FizzBuzz

WHY THIS WORKS AS AN ANSWER
------------------------------
This is the exact fizzbuzz(n) function from Course 2, Chapter 8's own
testing challenge, with two additions matching this chapter's basic
type-hint syntax: n: int reuses the name: Type parameter-annotation
form from this chapter's add(a: int, b: int) example, documenting that
n is expected to be an integer.

-> str reuses this chapter's return-type annotation syntax, placed
after the parameter list and before the colon. This is accurate
because every branch of the if/elif/else chain returns a string
value — "FizzBuzz", "Fizz", "Buzz", or str(n) (which explicitly
converts n to a string in the fallback case) — so str is the correct,
honest return type for every possible path through the function, not
just the common case.

Nothing about the function's actual LOGIC changed at all — per this
chapter's own explanation, type hints are purely documentation as far
as the Python interpreter itself is concerned, so adding them here
doesn't change fizzbuzz(15)'s behavior or output in any way; it only
adds information a tool like mypy (or a code editor) could use to
catch a mismatched call before it ever runs.