Problem: Let $n$ be a positive integer.
\begin{enumerate}
  \ii[(a)] Prove that there exists a set $S$ of $6n$ positive integers
  such that the least common multiple of any two is at most $32n^2$.
  \ii[(b)] Show that every set $T$ of $6n$ positive integers
  contains two elements with least common multiple exceeding $9n^2$.
\end{enumerate}

Solution: For part (a), let
\[ S = \{1, 2, \dots, 4n\} \cup \{4n+2, 4n+4, \dots, 8n\} \]
which works.

For (b),
the main idea is as follows.
Let $x_0 < x_1 < \dots < x_m$ be these integers.
Assume for contradiction the LCM's are less than $L$.
The main idea is the estimate
\[ L \ge \frac{x_i x_{i+1}}{\gcd(x_i, x_{i+1})} \ge \frac{x_i x_{i+1}}{x_{i+1} - x_i} \]
whence
\[ \frac{1}{L} \le \frac{1}{x_i} - \frac{1}{x_{i+1}}. \]
so taking the sum starting from any $k$ gives
\[ \frac{m+1-k}{L} \le \frac{1}{x_k} - \frac{1}{x_m} < \frac 1{x_k}. \]
For $(k,m) = (3n,6n)$ and $L = 9n^2$ this is a contradiction (since $x_k \ge 3n$).

\begin{remark*}
  China TST 2007/6 follows the same idea.
\end{remark*}