Problem:
Seven cubes, whose volumes are , and 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?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
The volume of each cube follows the pattern of , for is between and . 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 cube (which is just ). The sides areas can be measured as the sum , giving us . Structurally, if we examine the tower from the top, we see that it really just forms a square of area . Therefore, we can say that the total surface area is
Alternatively, for the area of the tops, we could have found the sum
giving us as well.
Note: The area on top and bottom are because the largest area is , and the other cubes are "inscribed" in it.
It can quickly be seen that the side lengths of the cubes are the integers from to inclusive.
First, we will calculate the total surface area of the cubes, ignoring overlap. This value is
Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to
Subtracting the overlapped surface area from the total surface area, we get
The problems on this page are the property of the MAA's American Mathematics Competitions