Counting with inclusion–exclusion — How many integers between −11 and 11 (inclusive) are multiples of 2 or multiples of 3? Count 0 as a multiple of every integer.

Introduction / Context:
This problem checks your comfort with inclusion–exclusion and with symmetric intervals around 0. We count integers from −11 to 11 inclusive that are multiples of 2 or 3. Because 0 is a multiple of every integer, ensure it is correctly included once, not multiple times.

Given Data / Assumptions:

• Interval: from −11 to 11, inclusive (i.e., −11, −10, …, 10, 11).
• Criterion: divisible by 2 or by 3 (or by both).
• 0 qualifies (divisible by every integer).

Concept / Approach:
Use inclusion–exclusion: count multiples of 2, add multiples of 3, subtract multiples of 6 (since those were counted twice). Alternatively, list quickly near 0 by symmetry. Either approach should deliver the same number.

Step-by-Step Solution:
1) Multiples of 2 in [−11, 11]: −10, −8, −6, −4, −2, 0, 2, 4, 6, 8, 10 → 11 numbers.2) Multiples of 3 in [−11, 11]: −9, −6, −3, 0, 3, 6, 9 → 7 numbers.3) Multiples of 6 (common to both sets): −6, 0, 6 → 3 numbers.4) Inclusion–exclusion total = 11 + 7 − 3 = 15.

Verification / Alternative check:
A direct, quick listing centered at 0 confirms there are exactly 15 such integers. Many students find symmetry helps avoid misses on the negative side.

Why Other Options Are Wrong:
11 and 14 undercount by missing some 3-multiples or misusing inclusion–exclusion; 16 overcounts by double-counting common multiples; “None of these” is invalid because 15 is attainable.

Common Pitfalls:
Forgetting to include 0; forgetting to subtract multiples of 6; omitting − multiples on the negative side; treating the interval as exclusive of endpoints by mistake.

Final Answer:
15