1985 AMC 12 Problem 20

Attempt Problem 20 of the 1985 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 1985 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

A wooden cube with edge length nn units (where nn is an integer >2\gt2) is painted black all over. By slices parallel to its faces, the cube is cut into n3n^3 smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is n?n?
55
66
77
88
none of these
Answer: D
Concepts:algebraic manipulationbasic countingcube geometry
Small hint:
Count the unpainted cubes in the interior cube of side length n−2n-2
Big hint:
On each face, exactly (n−2)2(n-2)^2 cubes avoid the edges
Solution:
There are (n−2)3(n-2)^3 completely unpainted cubes. Each of the six faces contributes (n−2)2(n-2)^2 cubes with exactly one painted face, so equality gives 6(n−2)2=(n−2)3. 6(n-2)^2=(n-2)^3. Since n>2,n\gt2, division by (n−2)2(n-2)^2 gives n−2=6,n-2=6, hence n=8.n=8.
Therefore the correct answer is D.
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