2020 AMC 10A Problem 10

The volumes are $1^3, 2^3, \ldots, 7^3$, so the cubes have edges $1, 2, 3, 4, 5, 6, 7$. Split the outside of the tower into side faces and flat (top and bottom) faces, because they behave differently.

1. Side faces. Each cube sits on a bigger one, so nothing ever covers a side face. Every cube shows all $4$ sides. A cube with edge $s$ has sides of area $s^2$, so the sides give
$$4 \times (1 + 4 + 9 + 16 + 25 + 36 + 49) = 4 \times 140 = 560.$$
2. Top faces. Look straight down on the tower. Every spot of the big $7 \times 7$ square is covered by exactly one visible top surface: either part of the big cube's top, or the top of a smaller cube sitting on it. So the visible top area is $7 \times 7 = 49$.
3. Bottom face. Only the big cube's bottom shows: $49$.
4. Total: $560 + 49 + 49 = 658$.

The answer is $\boxed{\textbf{(B)}\ 658}$.

For a stack of boxes, the side faces never hide each other, and the flat faces you can see from above always add up to the footprint of the stack. Thinking about the view from above avoids subtracting overlaps one by one.

Start from the total surface of all seven cubes: $6 \times 140 = 840$. Where two cubes touch, the smaller cube's bottom and the same-sized patch on the larger cube's top are both hidden, so each contact area is lost twice. The contact areas are $1, 4, 9, 16, 25, 36$ (every cube except the biggest sits on another), which add to $91$. So the answer is
$$840 - 2 \times 91 = 840 - 182 = 658.$$