France Olympiad Team Selection Test 2005 In an international meeting of $n ≥ 3$ participants, $14$ languages are spoken. We know that: 
- Any $3$ participants speak a common language. 
- No language is spoken by more than half of the participants.
What is the least value of $n$?

Set up the standard incidence matrix. 
Let the row be the $n$ participants and the 14 columns be the languages. Fill in the entry of 1 if the participant can speak the language.
We wish to count the number of column triples $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$. Since every 3 participants speak a common language, this number is at least $ n \choose 3$. Since every language is spoken by half or less (my interpretation), there is at most $15 \times { \frac{n}{2} \choose 3}$ such triples.
Solving ${ n \choose 3} \leq 14{ \frac{n}{2} \choose 3}$, which is equivalent to $(n-1)(n-8) \geq 0 $, we get $ n \leq 1, n \geq 8 $. Hence, the least possible value of $n$ is 8.
It remains to show that $n=8 $ is possible. This is quite easy to do so, if you remember that equality must hold throughout. Hence, no 3 participants speak 2 common languages, is a necessary and sufficient condition. There is a natural choice, which works.

For the first 7 languages, use 
$ \begin{array} { | l | l | l | l | l | l | l |}
\hline
1&1&1&1&0&0&0              \\ \hline
1&1&0&0&1&1&0              \\ \hline
1&0&1&0&1&0&1              \\ \hline
1&0&0&1&0&1&1              \\ \hline
0&1&1&1&0&0&0              \\ \hline
0&1&0&0&1&1&0              \\ \hline
0&0&1&0&1&0&1              \\ \hline
0&0&0&1&0&1&1              \\ \hline
\end{array} $
For the next 7 languages, use 1 minus entries in the above table.
