Problem 2016

Notice that $6$ is the first index $k$ such that $p_1p_2 \cdots p_{k-2} > p_{k-1}p_k$. Now, if $p_1p_2 \cdots p_{k-2} > p_{k-1}p_k$ for some index $k \ge 6$, then (by Bertrand-Tchebysheff) $p_1p_2 \cdots p_{k-1} > p_{k-1}^2p_k > 2p_{k-1} \cdot 2p_k > p_kp_{k+1}$, so $p_1p_2 \cdots p_{k-2} > p_{k-1}p_k$ for all indices $k \ge 6$.
Consequently, $m \le 5$, $r = p_m \le p_5 = 11$, $q \le p_4 = 7$, and $n < qr \le p_4p_5 = 7 \cdot 11 = 77$. Examination of the integers less than $77$ quickly yields the required numbers: $2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 18, 20, 24, 30, 42, 60$.