Challenge 3: Parametrized FizzBuzz Test — Possible Solution
====================================================================

import pytest

def fizzbuzz(n):
    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)

@pytest.mark.parametrize("n, expected", [
    (15, "FizzBuzz"),
    (9, "Fizz"),
    (10, "Buzz"),
    (7, "7"),
])
def test_fizzbuzz(n, expected):
    assert fizzbuzz(n) == expected

Running: pytest test_fizzbuzz.py

Output:
test_fizzbuzz.py::test_fizzbuzz[15-FizzBuzz] PASSED
test_fizzbuzz.py::test_fizzbuzz[9-Fizz] PASSED
test_fizzbuzz.py::test_fizzbuzz[10-Buzz] PASSED
test_fizzbuzz.py::test_fizzbuzz[7-7] PASSED

WHY THIS WORKS AS AN ANSWER
------------------------------
fizzbuzz(n) reuses the exact if/elif/else divisibility logic from
Course 1, Chapter 3's own FizzBuzz-style challenge, with one change
the challenge specifically asks for: return str(n) instead of
print(number), since a testable function needs to hand back a value
rather than just print one — str(n) reuses the int-to-string
conversion covered back in Course 1 to keep every branch's return
type consistently a string.

@pytest.mark.parametrize("n, expected", [...]) reuses this chapter's
own parametrize example structure exactly — a list of tuples, each
one becoming a separate, independently pass/fail test run. The four
chosen cases deliberately exercise every branch of fizzbuzz's
if/elif/else chain: 15 hits the combined "divisible by both" case, 9
hits Fizz-only, 10 hits Buzz-only, and 7 hits neither, falling through
to the plain-number case — together they cover the function's full
behavior in one test function instead of four separate ones.

def test_fizzbuzz(n, expected): receives each tuple's two values as
its own n and expected parameters, and assert fizzbuzz(n) == expected
reuses the same plain-assert-equality pattern from every earlier test
in this chapter, just run four times automatically instead of once.