Question

三个立方体都由图中展开图折成。然后把它们一个叠一个放在桌上，使 $13$ 个可见数字之和尽可能大。这个和是多少？

Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum?

$154$

$159$

$164$

$167$

$189$

答案：C

知识点：正方体最优化

难度评级：1560

解答：

每个立方体六面数字和为 $1 + 2 + 4 + 8 + 16 + 32 = 63$ 。由展开图，相对面分别是 $1$ 与 $32$ ， $2$ 与 $16$ ， $4$ 与 $8$ 。

下面两个立方体各隐藏一对上下相对面，应隐藏和最小的 $4 + 8 = 12$ 。最上面立方体只隐藏底面，应隐藏 $1$ 。

最大可见和为 $\begin{aligned} &3 \cdot 63 - 2 \cdot 12 - 1 \\ &= 189 - 24 - 1 \\ &= 164. \end{aligned}$

所以正确答案是 C。

The six faces of each cube sum to $1 + 2 + 4 + 8 + 16 + 32 = 63.$ From the pattern, the pairs of opposite faces are $1$ & $32,$ $2$ & $16,$ and $4$ & $8.$

Each of the two lower cubes hides a pair of opposite faces (top and bottom); hiding the pair $4 + 8 = 12$ is best. The top cube hides only its bottom face, so hide the $1.$

The greatest sum is $\begin{aligned} &3 \cdot 63 - 2 \cdot 12 - 1 \\ &= 189 - 24 - 1 \\ &= 164. \end{aligned}$

Thus, C is the correct answer.

2008 AMC 12A 第 11 题

答案：C

难度评级：1560

解答：

其他年份的第 11 题