Conference rusure::math

1860.0. "Crux Mathematicorum 1828" by RUSURE::EDP (Always mount a scratch monkey.) Fri Mar 18 1994 11:21

    Proposed by T. W. O. Richards, Cheddar Gorge, Great Britain.
    
    In the last century, the English mathematician Arthur Cayley introduced
    a permutation problem, loosely based on the card game Treize, which he
    called Mousetrap.  Suppose that the numbers 1, 2, ..., n are written on
    n cards, one on each card.  After shuffling (permuting) the cards,
    start counting the deck from the top card down.  If the number on the
    card does not equal the count, then put that card at the bottom of the
    deck and continue counting.  If the two are equal then put the card
    aside and start counting again from 1.
    
    Let's say the game is won if all the cards have been put aside.  In
    this case, form a new deck with the cards in the order in which they
    were set aside and play a new game with this deck.  For example, if we
    start with n=5 cards in the order 25143, we win:
    
    	25143 -> 3251 -> 3251 -> 513 -> 513 -> 51 -> 51 -> 1
           *              *               *          *     *
    
    and the new deck is 42351, which wins again:
    
    	42351 -> 3514 -> 351 -> 13 -> 3
         *          *     *     *     *
    
    but now our deck, 24513, puts aside no cards at all.  Is there an
    arrangement (using more cards, if necessary) which will give you three
    or more consecutive wins?
    
    [Solutions are in the current issue of _Crux Mathematicorum_:  1, 12,
    165342, 52173846, and eight arrangements of nine cards.  All were found
    by trial and error or computer.  Can we do better?  -- edp]