1998 AMC 12 Problem 27

Attempt Problem 27 of the 1998 AMC 12 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 1998 AMC 12 solutions, or check the answer key.

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27.

A 9×9×99\times9\times9 cube is composed of twenty-seven 3×3×33\times3\times3 cubes. The big cube is “tunneled” as follows: First, the six 3×3×33\times3\times3 cubes which make up the center of each face as well as the center 3×3×33\times3\times3 cube are removed as shown. Second, each of the twenty remaining 3×3×33\times3\times3 cubes is diminished in the same way. That is, the center facial unit cubes as well as each center cube are removed. The surface area of the final figure is

384384

729729

864864

10241024

10561056

Answer: E
Concepts:surface areathree-dimensional counting
Difficulty rating: 2290
Small Hint:

After the first stage, classify the twenty remaining large subcubes as corner or edge cubes

Big Hint:

For each second-stage tunnel, subtract exposed center squares and add the newly exposed tunnel walls

Solution:

After the first stage, 88 corner subcubes contribute 2727 exposed units each, and 1212 edge subcubes contribute 3636 each. Tunneling a corner subcube removes 33 exposed unit squares and adds 2424 tunnel-wall squares; tunneling an edge subcube removes 44 and also adds 24.24. Hence the final area is 8(273+24)+12(364+24)=384+672=1056. \begin{aligned} &8(27-3+24)\\ &\qquad+12(36-4+24)\\ &=384+672=1056. \end{aligned} Thus E is correct.

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