2001 AMC 10 Problem 12

Any three consecutive integers include a multiple of $3$ and at least one even number, so their product is a multiple of $6$. We are told $7$ divides $n$ as well, and $\gcd(6, 7) = 1$, so $42$ divides $n$. That guarantees choices (A), (B), (C) and (E), all of which divide $42$.

Only (D) is left, and it is genuinely not forced: $n = 5 \cdot 6 \cdot 7 = 210$ is divisible by $7$ but $210 = 28 \cdot 7 + 14$, so $28$ does not divide it. The product has only one factor of $2$ here because $6$ is the only even number among $5, 6, 7$.

The answer is $\boxed{\textbf{(D)}\ 28}$.

"Not necessarily a divisor" means one legal example that fails is enough. First use structure (consecutive integers force factors of $2$ and $3$) to certify the choices that always work; then hunt for a counterexample against what remains. Choosing the triple whose even member is $2 \pmod 4$ keeps the extra factor of $2$ out.

• Assuming the product must contain 4 because 'consecutive integers include even numbers'; only one even number is guaranteed among three.
• Testing only $6 \cdot 7 \cdot 8 = 336$, which happens to be divisible by $28$, and wrongly concluding that every choice divides $n$.