Two people take turns cutting up a rectangular chocolate bar which is 6 x 8 squares in size. You are only allowed to cut the bar along a division between the squares, and your cut can only be a straight line. For example, you can turn the original bar into a 6 x 2 piece and a 6 x 6 piece, and this latter piece can be turned into a 1 x 6 piece and a 5 x 6 piece. The last player who can (legally) break the chocolate wins (and gets to eat the chocolate bar). Is there a winning strategy for the first or second player. What about the general case (when the starting bar is n x n)? |
 |
Solution by David Zhang, Ann Arbor Huron High School
In the general case of m x n squares, the total number of
breaks (or cuts) will be mn - 1. Therefore, if m and n
are both odd, the total number of breaks is even; so the second player will
win. Otherwise, the first player will win. In the given chocolate bar, the total
number of breaks is 47; so the first player will win.
Also solved by: Travis Rhynard, Noel Dominick, Brett Krasill,
Kim Curtiss, Alex Dawe (Shepherd H.S.).
Partial solutions were submitted by: Nicholas Timkovich
(East Grand Rapids H.S.); Dana Harrison
(Grand Rapids Forest Hills Northern H.S.); Laura Merritt, Jeff Beltinck
(Shepherd H.S.); Jeff Guo
(Grosse Pointe South H.S.).