2020 AMC 10A Problem 10

Problem 10 of 25EasierAlgebra Geometry

Seven cubes, whose volumes are

1,

8,

27,

64,

125,

216,

and

343

cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

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Solution

The cube side lengths are

1,2,3,4,5,6,7

, stacked from largest on bottom to smallest on top. The sum of the surface areas of the separate cubes is

6(1^2+2^2+\cdots+7^2)

=6\cdot140=840

. Each contact hides two square faces, with areas

1^2,2^2,\ldots,6^2

. Subtracting these hidden faces gives

840

-2(1^2+2^2+\cdots+6^2)

=840-182=658

. Thus, B is the correct answer.

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