We have two piles with $2000$ and $2017$ coins respectively. Ann and Bob take alternate turns making the following moves: The player whose turn is to move picks a pile with at least two coins, removes from that pile $t$ coins for some $2\le t \le 4$, and adds to the other pile $1$ coin. The players can choose a different $t$ at each turn, and the player who cannot make a move loses. If Ann plays first determine which player has a winning strategy.
Problem
Source: JBMO Shortlist 2017 C3
Tags: combinatorics, game strategy, coins
ayan.nmath
25.07.2018 18:19
The main idea is to plot the points $(a, b) $ on the Cartesian plane where $a, b$ are number of coins in both the piles and bash out all the losing positions.
relay400
12.05.2020 22:46
Apologies for the bump, anyone know where the solution to this problem can be found? Not exactly sure where to find the shortlisted problems.
parmenides51
13.05.2020 07:03
JBMO shortlists may be found on the JBMO/ SHL section of this site the above problem, is solved at page 10 of the pdf here
sttsmet
25.05.2022 16:21
Please correct me if I am wrong
We will prove that $Bob$ haw a winning strategy. Let $S$ be the sum of the coins in both of the piles. $Bob$ must try to keep the value of $S \pmod3$ stable. When $Ann$ chooses $t=2$ $Bob$ chooses $t=3$ and vise versa, and when $Ann$ chooses $t=4$ then $Bob$ also chooses $t=4$. This guarantees that $S \equiv 0 \pmod3$ in every move of the game. Since the value of S will be always decreasing it will reach 3 and it will be Ann's turn. But we cannot have 0 coins at any pile thus one pile will have $1$ coin and the other $2$ coins so $Ann$ can't make a move.
Hoapham235
22.12.2022 05:05
parmenides51 wrote: JBMO shortlists may be found on the JBMO/ SHL section of this site the above problem, is solved at page 10 of the pdf here Excume, could you show me the link for the solution of JBMO Shortlist every year, please? Thank you so much.
parmenides51
22.12.2022 05:11
It has a drop downmenu on top of the site, path is: JBMO / SHL / SHL year
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