Difference between revisions of "2021 AMC 12B Problems/Problem 22"
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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== | + | == Video Solution by OmegaLearn (Game Theory) == |
| − | + | https://youtu.be/zkSBMVAfYLo | |
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=21|num-a=23}} | {{AMC12 box|year=2021|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:08, 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 
![[asy] unitsize(4mm); real[] boxes = {0,1,2,3,5,6,13,14,15,17,18,21,22,24,26,27,30,31,32,33}; for(real i:boxes){ draw(box((i,0),(i+1,3))); } draw((8,1.5)--(12,1.5),Arrow()); defaultpen(fontsize(20pt)); label(",",(20,0)); label(",",(29,0)); label(",...",(35.5,0)); [/asy]](http://latex.artofproblemsolving.com/1/a/5/1a5938c90db67cc867d798bb9693e149b9438071.png)
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
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
