The audience chooses two of five cards, numbered from $1$ to $5$ respectively. The assistant of a magician chooses two of the remaining three cards, and asks a member of the audience to take them to the magician, who is in another room. The two cards are presented to the magician in arbitrary order. By an arrangement with the assistant beforehand, the magician is able to deduce which two cards the audience has chosen only from the two cards he receives. Explain how this may be done.
Problem
Source: Tournament of Towns 2007 - Fall - Junior O-Level - P5
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fishythefish
03.09.2011 22:51
Is the assistant aware of the audience's choice? Is anyone aware of what the last card is?
talkinaway
03.09.2011 23:21
There are $\binom {3}{2} = 3$ ways for the assistant to pick two cards from the remaining three cards. But knowing these two cards are NOT in the audience means there are also $\binom {3}{2} = 3$ possible combinations of cards available to the audience. Setting up table a bijection can easily be done. You probably could do it based on something like $\mod 3$ so it's easy to remember...however, there are only $\binom{5}{2, 2, 1} = 30$ possibilities, so memorizing a table wouldn't be hard. (After all, third graders memorize a table with $100$ entries.)
62861
08.08.2015 18:02
Sorry for revival, but here's an easy explicit solution. Put the numbers $1,2,3,4,5$ on the vertices of a regular pentagon in any order. Each pair of numbers corresponds to either a side or diagonal of the pentagon. If the audience chooses two numbers on a side (resp. diagonal), then the assistant chooses the two numbers on the parallel diagonal (resp. side). This works for any odd number of cards, with very small modifications.