Maths Olympiad Prep

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, 2011

Combinatorics Difficulty 8.5 Shortlist Prove it Baltic Way

Two persons play the following game with positive integers. The initial number is 201120112011^{2011}. Each move consists of subtraction by an integer between 11 and 20102010 inclusive, or division by 20112011, rounding down when necessary. The player who obtains a non-positive integer wins. Who will win this game: the first player or the second?

Solution

Though the problem is taken from the recent article (A. Guo. Winning strategies for aperiodic subtraction games // arXiv: 1108.1239v2), it could be known for the smaller numbers, say, for 22 instead of 20112011.

The initial numbers NN for which the second player has a winning strategy are those ones that have odd numbers of trailing 00's in base 20112011 (i.e. if the biggest power of 20112011 that divides NN is odd). The main difficulty of the problem is to invent this answer. The proof is trivial: each move of the first player makes this biggest power to be even, and after that the second player can make this power odd by a suitable move.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.