2003 AMC 10B Problem 18

When $n$ is even, the factors are five consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by $3$ and exactly one by $5,$ so the product is always divisible by $15.$ To prove that no larger fixed divisor is forced, compare three cases: $$\begin{aligned} n=2 &: 3\cdot5\cdot7\cdot9\cdot11,\\ n=10 &: 11\cdot13\cdot15\cdot17\cdot19,\\ n=12 &: 13\cdot15\cdot17\cdot19\cdot21. \end{aligned}$$ The greatest common divisor of these three products is exactly $15,$ so a divisor common to every case cannot be any larger. Thus, the correct answer is D.