# Difference between revisions of "2021 AMC 10B Problems/Problem 24"

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== Another Solution == | == Another Solution == | ||

− | First we note that symmetrical positions are losing for the player to move. Then we start checking small positions. (n) is always winning for the first player. Furthermore, (3, 2, 1) is losing and so is (4, 1). We look at all the positions created from (6, 2, 1). There are several different positions that can | + | First we note that symmetrical positions are losing for the player to move. Then we start checking small positions. (n) is always winning for the first player. Furthermore, (3, 2, 1) is losing and so is (4, 1). We look at all the positions created from (6, 2, 1), as (6, 1, 1) is obviously winning by playing (2, 2, 1, 1). There are several different positions that can be played by the first player from (6, 2, 1). They are (2, 2, 2, 1), (1, 3, 2, 1), (4, 2, 1), (6, 1), (5, 2, 1), (4, 1, 2, 1), (3, 2, 2, 1). Now we list refutations for each of these moves: |

## Revision as of 23:12, 11 February 2021

## Problem

Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes and can be changed into any of the following by one move: or . Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?

## Video Solution by OmegaLearn (Game Theory)

~ pi_is_3.14

## Another Solution

First we note that symmetrical positions are losing for the player to move. Then we start checking small positions. (n) is always winning for the first player. Furthermore, (3, 2, 1) is losing and so is (4, 1). We look at all the positions created from (6, 2, 1), as (6, 1, 1) is obviously winning by playing (2, 2, 1, 1). There are several different positions that can be played by the first player from (6, 2, 1). They are (2, 2, 2, 1), (1, 3, 2, 1), (4, 2, 1), (6, 1), (5, 2, 1), (4, 1, 2, 1), (3, 2, 2, 1). Now we list refutations for each of these moves:

(2, 2, 2, 1) - (2, 1, 2, 1)

(1, 3, 2, 1) - (3, 2, 1)

(4, 2, 1) - (4, 1)

(6, 1) - (4, 1)

(5, 2, 1) - (3, 2, 1)

(4, 1, 2, 1) - (2, 1, 2, 1)

(3, 2, 2, 1) - (1, 2, 2, 1)

This proves that (6, 2, 1) is losing for the first player.