pigeonhole principle - 17 mathematicians and 3 languages, prove that 3 communicate in the same language pairwise There are 17 mathematicians and 3 official languages. Every pair of mathematicians communicate in one of the official languages. Prove that there are 3 mathematicians communicating in the same language pairwise.

This is what I got:
At least one language is spoken by 6 people.

However I'm not really sure about the term pairwise, do we need 3 or 6 people speaking the same language? Does it mean 3 people can communicate with each other by forming 3 pairs, but not all pairs have to be speaking the same language? If so then I have proven by saying 6 people can speak the same language.
 A: This is also an old IMO problem. Posting a CW solution to get this off the list of unanswered questions.
Pick a mathematician, call them $X_0$. There are $16$ other mathematicians, and $16/3>5$, so $X_0$ is discussing with at least six others in the same language. Without loss of generality we can assume that they discuss in language $A$ with $X_1,X_2,\ldots,X_5$ and $X_6$. If any pair $X_i,X_j, 1\le i<j\le6$, also uses language $A$ in their mutual communication, then we have found our triple, $X_1,X_i,X_j$. So we can assume that every pair $\{X_i,X_j\}, 1\le i<j\le6$, uses either language $B$ or language $C$ in their communication.
We have thus simplified the problem to a set of six mathematicians and two languages. Let's repeat the dose. Pick a mathematician, say $X_1$. With the five peers $X_2,\ldots,X_6$ they are using the same language (only two choices) with at least three others. Again without loss of generality we can assume that $X_1$ communicates in $B$ with all of $X_2,X_3,X_4$. If any two of $X_2,X_3,X_4$ use language $B$ in their mutual correspondence, then we have again found our triple.
The remaining possibility is that in their pairwise communication all three of $X_2,X_3,X_4$ use language $C$. But then they form the required triple.
