Challenge 1: FizzBuzz-Style Check — Possible Solution
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number = 15

if number % 3 == 0 and number % 5 == 0:
    print("FizzBuzz")
elif number % 3 == 0:
    print("Fizz")
elif number % 5 == 0:
    print("Buzz")
else:
    print(number)

Output (for number = 15):
FizzBuzz

WHY THIS WORKS AS AN ANSWER
------------------------------
number % 3 == 0 checks divisibility by 3 using the modulo operator
from this chapter — a number is evenly divisible by 3 exactly when
the remainder of dividing by 3 is 0. The same logic applies to % 5 == 0
for divisibility by 5.

The FIRST condition combines both checks with and, matching this
chapter's logical-operators section — this MUST be checked before the
individual Fizz/Buzz checks, since a number divisible by both 3 and 5
(like 15) would otherwise incorrectly match the "only by 3" elif branch
first if the combined check came later in the chain.

The elif chain ensures exactly ONE branch runs per number — once
"FizzBuzz" or "Fizz" or "Buzz" matches, none of the later elif/else
branches are even evaluated, exactly this chapter's own if/elif/else
behavior. The final else only runs when none of the divisibility
checks matched at all, printing the plain number in that case.