2018 AMC 12A Problem 15

Below is the professionally curated solution for Problem 15 of the 2018 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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Concepts:symmetrybasic counting

Difficulty rating: 1800

15.

A scanning code consists of a 7×77 \times 7 grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of 4949 squares. A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of 9090^\circ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?

510510

10221022

81908190

81928192

65,53465{,}534

Solution:

Under the symmetry group of the square, the 4949 cells break into orbits, and every cell in an orbit must have the same color. Classifying cells by distance from the center yields exactly 1010 orbits that can be colored independently. Each orbit is black or white, giving 2102^{10} colorings, but the all-black and all-white grids are excluded. So there are 2102=10222^{10} - 2 = 1022 symmetric scanning codes.

Thus, the correct answer is B.

Problem 15 in Other Years