AMC10 2020 A

AMC10 2020 A · Q10

AMC10 2020 A · Q10. It mainly tests Exponents & radicals, 3D geometry (surface area).

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?

七个立方体，体积分别为1、8、27、64、125、216和343立方单位，沿垂直方向堆叠成塔，体积从底部到顶部递减。除了最底部的立方体，每个立方体的底面完全位于下方立方体的顶面上。塔的总表面积（包括底部）有多少平方单位？

(A) 644 644

(B) 658 658

(D) 720 720

(E) 749 749

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because the volumes of the cubes are given to be $1^3, 2^3, \ldots, 7^3$, the edge lengths are $1, 2, \ldots, 7$. If the cubes were not stacked, the total surface area would be $6\cdot(1^2+2^2+\cdots+7^2)$. When two cubes meet along a surface, two of the smaller face areas are lost, for a total area of $2\cdot(1^2+2^2+\cdots+6^2)$ that needs to be subtracted from the surface area of the unstacked cubes to get the total surface area of the tower. The result is $$ 4\cdot(1^2+2^2+\cdots+6^2)+6\cdot 7^2 =4\cdot 91+6\cdot 49 =658. $$

答案（B）：因为这些立方体的体积给定为 $1^3, 2^3, \ldots, 7^3$，所以它们的棱长分别为 $1, 2, \ldots, 7$。如果这些立方体不堆叠，总表面积为 $6\cdot(1^2+2^2+\cdots+7^2)$。当两个立方体沿一个面接触时，会损失两个较小的面面积，总共损失 $2\cdot(1^2+2^2+\cdots+6^2)$，需要从未堆叠时的表面积中减去，得到塔的总表面积。结果为 $$ 4\cdot(1^2+2^2+\cdots+6^2)+6\cdot 7^2 =4\cdot 91+6\cdot 49 =658. $$

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