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 A Mad tea party (Posted on 2002-04-30)
The Mad Hatter, March Hare and Dormouse are sitting down for a tea party. They sit at a table with twelve chairs, and twelve cups of tea.

Each day at six o'clock, everyone moves over two seats to the left or to the right (if any of those seats are free), then if there is tea in the cup at their seat, each one drinks it so the cup becomes empty.

After this, Alice comes and fills one of the empty cups on the table with tea again.

Prove that Alice can make sure that there are at least six full teacups on the table every day just before six.

 Submitted by levik Rating: 2.2000 (5 votes) Solution: (Hide) Each of the tea-drinking characters has a "parity" that they must stick to: if they start out with an even numbered seat, they must always stay even, or if they start out odd, they must stay odd. This is because they must always move 2 seats over, and the total number of seats is even. An "odd" character will never get to an "even" seat, and vise versa. There are two possibilities here: All three characters are the same parity. This means that that all three cannot get to the cups of the other parity, and those six cups will always stay filled. Problem solved for Alice. Two of the characters are of one parity and the third is of another. Then that character will be the only one drinking from a set of six cups, and it will be easy for Alice to simply fill up the cup that the character just drank from. (Of course, the other two will end up not getting any tea after 3 days of the party, but that's not Alice's problem.)

Comments: ( You must be logged in to post comments.)
 Subject Author Date my try ubergeek 2002-12-25 22:15:13 If Alice had disposable friends... Carey 2002-05-31 02:49:40

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