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

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<math>\textbf{(A)} ~(6, 1, 1) \qquad\textbf{(B)} ~(6, 2, 1) \qquad\textbf{(C)} ~(6, 2, 2) \qquad\textbf{(D)} ~(6, 3, 1) \qquad\textbf{(E)} ~(6, 3, 2)</math> | <math>\textbf{(A)} ~(6, 1, 1) \qquad\textbf{(B)} ~(6, 2, 1) \qquad\textbf{(C)} ~(6, 2, 2) \qquad\textbf{(D)} ~(6, 3, 1) \qquad\textbf{(E)} ~(6, 3, 2)</math> | ||

− | == | + | == Solution == |

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

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-Note: In general, this game is very complicated. For example <math>(8, 7, 5, 3, 2)</math> is winning for the first player but good luck showing that. | -Note: In general, this game is very complicated. For example <math>(8, 7, 5, 3, 2)</math> is winning for the first player but good luck showing that. | ||

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+ | == Video Solution by OmegaLearn (Game Theory) == | ||

+ | https://youtu.be/zkSBMVAfYLo | ||

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+ | ~ pi_is_3.14 |

## Revision as of 00:20, 12 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?

## Solution

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

This proves that is losing for the first player.

-Note: In general, this game is very complicated. For example is winning for the first player but good luck showing that.

## Video Solution by OmegaLearn (Game Theory)

~ pi_is_3.14