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Geometry Difficulty 3.2 AMC 10/12 Find the answer

Seven cubes, whose volumes are 11, 88, 2727, 6464, 125125, 216216, and 343343 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?

Pick one

Solution

The volume of each cube follows the pattern of n3n^3, for nn is between 11 and 77.
We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the 7×7×77\times 7\times 7 cube (which is just 7×7=497 \times 7 = 49). The sides areas can be measured as the sum 4n=17n24\sum_{n=1}^{7} n^2, giving us 560560. Structurally, if we examine the tower from the top, we see that it really just forms a 7×77\times 7 square of area 4949. Therefore, we can say that the total surface area is 560+49+49=(B) 658560 + 49 + 49 = \boxed{\textbf{(B) }658}.
Alternatively, for the area of the tops, we could have found the sum n=27((n)2(n1)2)\sum_{n=2}^{7}((n)^{2}-(n-1)^{2}), giving us 4949 as well.
~ciceronii
Note: The area on top and bottom are 49 because the largest area is 49, and the other cubes are "inscribed" in it.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.