2005 AIME I Problem 9

Attempt Problem 9 of the 2005 AIME I 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 2005 AIME I solutions, or check the answer key.

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9.

Twenty-seven unit cubes are each painted orange on a set of four faces so that the two unpainted faces share an edge. The 2727 cubes are then randomly arranged to form a 3×3×33 \times 3 \times 3 cube. Given that the probability that the entire surface of the larger cube is orange is paqbrc,\frac{p^a}{q^b r^c}, where p,p, q,q, and rr are distinct primes and a,a, b,b, and cc are positive integers, find a+b+c+p+q+r.a + b + c + p + q + r.

Answer: 74
Concepts:basic probabilityindependent eventscube geometrycasework
Difficulty rating: 2920
Solution:

Each unit cube has one "bad edge": the edge shared by its two unpainted faces. The larger cube's surface is entirely orange exactly when every unit cube's bad edge touches no visible face. A uniformly random orientation places the bad edge uniformly among the cube's 1212 edge positions, so for each unit cube we count the edge positions both of whose faces are hidden.

A corner cube shows 33 faces meeting at a vertex; the safe edges are those of the 33 hidden faces meeting at the opposite vertex, so the probability is 312=14.\frac{3}{12} = \frac{1}{4}. An edge cube shows 22 adjacent faces, which touch 4+41=74 + 4 - 1 = 7 edges, leaving 55 safe: probability 512.\frac{5}{12}. A face-center cube shows 11 face touching 44 edges, leaving 88 safe: probability 812=23.\frac{8}{12} = \frac{2}{3}. The center cube is always fine.

With 88 corner, 1212 edge, and 66 face-center cubes, the probability is (14)8(512)12(23)6=512234318, \begin{aligned} &\left(\frac{1}{4}\right)^{8}\left(\frac{5}{12}\right)^{12}\left(\frac{2}{3}\right)^{6} \\ &= \frac{5^{12}}{2^{34} \cdot 3^{18}}, \end{aligned} so a+b+c+p+q+ra + b + c + p + q + r =12+34+18= 12 + 34 + 18 +5+2+3+ 5 + 2 + 3 =74.= 74.

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