AMC12 2008 A

AMC12 2008 A · Q11

AMC12 2008 A · Q11. It mainly tests 3D geometry (volume), 3D geometry (surface area).

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?

三个立方体每个都由所示图案制成。然后它们被一个叠一个地堆放在桌子上，使得 $13$ 个可见数字的和尽可能大。这个和是多少？

(A) 154 154

(B) 159 159

(D) 167 167

(E) 189 189

Answer

Correct choice: (C)

正确答案：（C）

Solution

To maximize the sum of the $13$ faces that are showing, we can minimize the sum of the numbers of the $5$ faces that are not showing. The bottom $2$ cubes each have a pair of opposite faces that are covered up. When the cube is folded, $(1,32)$; $(2,16)$; and $(4,8)$ are opposite pairs. Clearly $4+8=12$ has the smallest sum. The top cube has 1 number that is not showing. The smallest number on a face is $1$. So, the minimum sum of the $5$ unexposed faces is $2\cdot12+1=25$. Since the sum of the numbers on all the cubes is $3(32+16+8+4+2+1)=189$, the maximum possible sum of $13$ visible numbers is $189-25=164 \Rightarrow C$.

为了使显示出来的 $13$ 个面的数字和最大，我们可以使未显示出来的 $5$ 个面的数字和最小。下面的 $2$ 个立方体各有一对相对的面被遮住。将立方体折叠后，$(1,32)$、$(2,16)$ 和 $(4,8)$ 是相对面。显然 $4+8=12$ 的和最小。最上面的立方体有 $1$ 个数字不显示。面上的最小数字是 $1$。因此，$5$ 个未露出的面的最小和为 $2\cdot12+1=25$。由于所有立方体上数字总和为 $3(32+16+8+4+2+1)=189$，所以 $13$ 个可见数字的最大可能和为 $189-25=164 \Rightarrow C$。

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