2018 AMC 12B Problem 7

The factors split into two telescoping chains. The odd-position factors form $$\begin{gathered} \log_3 7\cdot\log_7 11 \\ {}\cdot\log_{11} 15\cdots\log_{23} 27 \\ =\log_3 27=3, \end{gathered}$$ and the even-position factors form $$\begin{gathered} \log_5 9\cdot\log_9 13\cdots\log_{21} 25 \\ =\log_5 25=2. \end{gathered}$$ The product is $3\cdot2=6.$ Thus, the correct answer is C.