MathDB

A chocolate bar has five lengthwise dents and eight crosswise ones, which can be used to break up the bar into sections (one can get a total of

9 \times 6 = 54

cells). Two players play the following game with such a bar. At each move (the two players move alternatively) one player breaks off a section of width one from the bar along a single dent and eats it, the other player does the same with what’s left of the bar, and so on. When one of the players breaks up a section of width two into two strips of width one, he eats one of the strips and the other player eats the other strip. Prove that the player who has the first move can play so as to eat at least

6

cells more than his opponent (no matter how his opponent plays).
(R Fedorov)

Problem Statement