2022 AMC 10B Problem 19

Attempt Problem 19 of the 2022 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

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19.

Each square in a 5×55 \times 5 grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:

• Any filled square with two or three filled neighbors remains filled.

• Any empty square with exactly three filled neighbors becomes a filled square.

• All other squares remain empty or become empty.

A sample transformation is shown in the figure below.

Suppose the 5×55 \times 5 grid has a border of empty squares surrounding a 3×33 \times 3 subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)

 14 \ 14

 18 \ 18

 22 \ 22

 26 \ 26

 30 \ 30

Answer: C
Concepts:process simulationcaseworksymmetry
Difficulty rating: 2390
Solution:

First suppose the center is initially filled. It must have exactly 22 or 33 filled neighbors to survive. Every such neighbor already touches the center, so to disappear it cannot touch any other filled neighbor. Checking these pairwise nonadjacent positions, the only choices that do not also give some empty square exactly 33 filled neighbors are two opposite corners. There are 22 such configurations.

Now suppose the center is initially empty. Exactly 33 of its eight neighbors must be filled. Each of those three must disappear, so none may be adjacent to both of the others. Also, no empty square besides the center may be adjacent to all three. Applying these two tests gives the following four representative patterns:

Each of the first three patterns has 44 distinct rotations. The last has 44 rotations and their 44 reflected images, for 88 configurations. Thus the center-empty case contributes 4+4+4+8=20,4+4+4+8=20, and the total is 20+2=22.20+2=22.

Thus, the answer is C .

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