Conversing Computers

Any computer talks to $1957$ others. By the Pigeon-Hole Principle, this computer must talk to at least $327$ computers in the same language $L_1$ (say). If any two of these $327$ computers talk to each other in the same language $L_1$ , we have found our trio of computers. Otherwise we have found $327$ computers that only talk to each other in the other five languages ( $L_2$ ,..., $L_6$ ).

Any one of these $327$ computers talks to the other $326$ . By the Pigeon-Hole Principle, this computer must talk to at least $66$ of the other $327$ in the same language $L_2$ (say). If any two of these $66$ computers speak to each other in that language $L_2$ , we have found our trio of computers. Otherwise we have found $66$ computers that only talk to each other in four of the six languages $L_3$ ,..., $L_6$ .

Again, the Pigeon-Hole Principle tells us that any one of these $66$ computers talks to at least $17$ of the others in the same language $L_3$ (say). If any of these $17$ speak to each other in the same language $L_3$ , we are done. Otherwise we have $17$ computers which only use three languages $L_4$ , $L_5$ , $L_6$ to communicate with each other.

Once again, one of these $17$ computers must speak to $6$ of the others in the same language $L_4$ (say). If any two of these $6$ speak use the same language $L_4$ to speak to each other, we are done. Otherwise we have found $6$ computers which only use two of the languages ( $L_5$ , $L_6$ ) to speak to each other.

Finally, any one of these $6$ computers must speak to at least $3$ in the same language $L_5$ (say). If any two of these $3$ use the same language $L_5$ between each other, we are done. Otherwise we have a trio of computers which only use the last language $L_6$ , and we are done.

It is the same as making a closed figure with least number of line segments.. To make a closed figure, we need at least 3 lines, yielding a triangle. Same is the case here, as we can form a closed network of 3 computers communicating in the same language..