# 2020 AMC 10A Problems/Problem 10

The following problem is from both the 2020 AMC 12A #7 and 2020 AMC 10A #10, so both problems redirect to this page.

## Problem

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?

$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$

## Solution 1

The volume of each cube follows the pattern of $n^3$ ascending, for $n$ is between $1$ and $7$.

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 $7\times 7\times 7$ cube (which is just $7 \times 7 = 49$). The sides areas can be measured as the sum $4\sum_{n=0}^{7} n^2$, giving us $560$. Structurally, if we examine the tower from the top, we see that it really just forms a $7\times 7$ square of area $49$. Therefore, we can say that the total surface area is $560 + 49 + 49 = \boxed{\textbf{(B) }658}$. Alternatively, for the area of the tops, we could have found the sum $\sum_{n=0}^{6}((n+1)^{2}-n^{2})$, giving us $49$ as well.

~ciceronii

## Solution 2

It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.

First, we will calculate the total surface area of the cubes, ignoring overlap. This value is $6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840$. 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 $2\sum_{n=1}^{6} n^2 = 182$. Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = \boxed{\textbf{(B) }658}$. ~emerald_block

## Solution 3 (a bit more tedious than other solutions)

It can be seen that the side lengths of the cubes using cube roots are all integers from $1$ to $7$, inclusive.

Only the cubes with side length $1$ and $7$ have $5$ faces in the surface area and the rest have $3$. Also, since the

cubes are stacked, we have to find the difference between each $n^2$ and $(n-1)^2$ side length as $n$ ranges from $7$ to

$2$.

We then come up with this: $5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)$.

We then add all of this and get $\boxed{\textbf{(B) }658}$.

~aryam

~IceMatrix

## Solution 5

Notice that the surface area of the top cube is $6s^2$ and the others are $4s^2$. Then we can directly compute. The edge length for the first cube is $7$ and has a surface area of $294$. The surface area of the next cube is $144$. The surface area of the next cube $100$. The surface area of the next cube is $64$. The surface area of the next cube is $36$. The surface area of the next cube is $16$. The surface area of the next cube is $4$. The surface area of the next cube is $4$. We then sum up $294+144+100+64+36+16+4$ to get $658$. -smartatmath

## See Also

 2020 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2020 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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