2013 AIME I Problem 4

Below is the professionally curated solution for Problem 4 of the 2013 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AIME I solutions, or check the answer key.

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Concepts:basic probabilitycombinationssymmetry

Difficulty rating: 2300

4.

In the array of 1313 squares shown below, 88 squares are colored red, and the remaining 55 squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated 9090^\circ around the central square is 1n,\frac{1}{n}, where nn is a positive integer. Find n.n.

Solution:

The rotation cycles the four L-shaped arms, so a symmetric coloring colors all four arms identically, and the 1212 outer squares contain 44 copies of whatever the arm shows. The number of red squares among the outer twelve is therefore a multiple of 4.4. Since there are 88 red squares in all, the center must be blue and each arm must contain exactly 22 red squares and 11 blue square.

The blue square within the arm can be chosen in 33 ways, so exactly 33 of the (135)=1287\binom{13}{5} = 1287 equally likely colorings are symmetric. The probability is 31287=1429,\frac{3}{1287} = \frac{1}{429}, so n=429.n = 429.

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