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heavybody
03-11-2003, 01:05 AM
Alice plays the following game of solitaire on a 20 × 20 chessboard. She begins by placing
100 pennies, 100 nickels, 100 dimes, and 100 quarters on the board so that each of the 400
squares contains exactly one coin. She then chooses 59 of these coins and removes them from
the board. After that, she removes coins, one at a time, subject to the following rules:
• A penny may be removed only if there are four squares of the board adjacent to its square
(up, down, left, and right) that are vacant (do not contain coins). Squares “off the board”
do not count towards this four: for example, a non-corner square bordering the edge of
the board has three adjacent squares, so a penny in such a square cannot be removed under
this rule, even if all three adjacent squares are vacant.
• A nickel may be removed only if there are at least three vacant squares adjacent to its
square. (And again, “off the board” squares do not count.)
• A dime may be removed only if there are at least two vacant squares adjacent to its square
(“off the board” squares do not count).
• A quarter may be removed only if there is at least one vacant square adjacent to its square
(“off the board” squares do not count).
Alice wins if she eventually succeeds in removing all the coins. Prove that it is impossible
for her to win.

irchans
03-11-2003, 08:34 AM
No Solution, but I really like the question!

cavalier
03-13-2003, 07:16 PM
I like it too. I'll be wasting time at work tomorrow trying to figure out some proof.

heavybody
03-16-2003, 02:30 AM
The solution to the problem is at this website
http://mathcircle.berkeley.edu/
Click: Bay Area Math Olympiad in Past Years then click BAMO 2000 exams and solutions