Difference between revisions of "2023 AMC 12B Problems/Problem 5"

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==Solution==
 
==Solution==
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First, we notice that there are a total of <math>12 \text{ } 2\times1</math> rectangles in a <math>3\times3</math> grid. Next, if we choose one of the corners, and the corner is not covered by a <math>2 \times 1</math> rectangle, we can eliminate a maximum of 2 rectangles.
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If we choose one of the side squares, we can eliminate a maximum of 3 rectangles.
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Finally, if we choose the center square, we can eliminate a maximum of 4 rectangles, but doing so means that if we choose a side square, we only eliminate 2 rectangles.

Revision as of 20:12, 15 November 2023

Problem

You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?

$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$

Solution

First, we notice that there are a total of $12 \text{ } 2\times1$ rectangles in a $3\times3$ grid. Next, if we choose one of the corners, and the corner is not covered by a $2 \times 1$ rectangle, we can eliminate a maximum of 2 rectangles.

If we choose one of the side squares, we can eliminate a maximum of 3 rectangles.

Finally, if we choose the center square, we can eliminate a maximum of 4 rectangles, but doing so means that if we choose a side square, we only eliminate 2 rectangles.