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3x3 Grid Digit Divisibility

1550AdeptCombinatoricsNumber TheoryProbability

AMC 8 · 2023 · Problem 23

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ gray and white tiles shown below on the right. [asy]size(5.663333333cm); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); fill((6,.33)--(7,.33)--(7,1.33)--cycle,mediumgray); draw((6,.33)--(7,.33)--(7,1.33)--(6,1.33)--cycle,gray); fill((6,1.67)--(7,2.67)--(6,2.67)--cycle,mediumgray); draw((6,1.67)--(7,1.67)--(7,2.67)--(6,2.67)--cycle,gray); fill((7.33,.33)--(8.33,.33)--(7.33,1.33)--cycle,mediumgray); draw((7.33,.33)--(8.33,.33)--(8.33,1.33)--(7.33,1.33)--cycle,gray); fill((8.33,1.67)--(8.33,2.67)--(7.33,2.67)--cycle,mediumgray); draw((7.33,1.67)--(8.33,1.67)--(8.33,2.67)--(7.33,2.67)--cycle,gray);[/asy] What is the probability that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids? Below is an example of such tiling. [asy]size(2cm); fill((1,0)--(0,1)--(0,2)--(1,1)--cycle,mediumgray); fill((2,0)--(3,1)--(2,2)--(1,1)--cycle,mediumgray); fill((1,2)--(1,3)--(0,3)--cycle,mediumgray); fill((1,2)--(2,2)--(2,3)--cycle,mediumgray); fill((3,2)--(3,3)--(2,3)--cycle,mediumgray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray);[/asy]

Answer choices

A
\frac{1}{1024}
B
\frac{1}{256}
C
\frac{1}{64}
D
\frac{1}{16}
E
\frac{1}{4}
0 students attempted0% solvedRating 1550

Related practice paths

AMC 8 PracticeMiddle school contest practice with hints and review.What Is a Good AMC 8 Score?Understand AMC 8 results and set the next practice target.AIME Practice StrategyHow to improve accuracy on high-difficulty problems.

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