**A problem posed in Nordic Mathematical Contest 1987:**

**Question:**

Nine journalists, each from a different country, participate in a press conference. None of them can speak more than three languages, and each two journalists have at least one common language. Prove that at least five of the journalists can speak the same language.

**Solution1:**

Assume the journalists are . Assume that no five of them have a common language. Assume the languages speaks are and . Group according to the language they speak with . No group can have more than three members. So, either there are three groups of three members each, or two groups with three members and one with two. Consider the first alternative. We may assume that speaks with , , and , with , , and , and with , and . Now, speaks with , , and , with , , and . must speak a fourth language, , with , , and . But, now speaks both and with , , and . So, has to use his third language with and . This contradicts the assumption we made.

So, we now may assume that speaks only with and . As is not special, we conclude that for each journalist , the remaining eight are divided into three mutually exclusive language groups, one of which has only two members. Now, uses with three others, and there has to be another language he also speaks with three others. If this were or , a group of five would arise (including ). So, speaks with three among . Either two of these three are among , , and , or among . Both alternatives lead to a contradiction to the already proved fact that no pair of journalists speaks two languages together. Hence, the proof. QED.

**Reference:**

**Nordic Mathematical Contest, 1987-2009.**

**Amazon India link:**

https://www.amazon.in/Nordic-Mathematical-Contest-1987-2009-Todev/dp/1450519830/ref=sr_1_1?s=books&ie=UTF8&qid=1515913392&sr=1-1&keywords=Nordic+mathematical+contest

Cheers,

Nalin Pithwa.

### Like this:

Like Loading...

*Related*