Question

三个面上编号为 $1$ 到 $6$ 的骰子如图堆叠。十八个面中有七个面可见，剩下十一个面被隐藏（背面、底面、相接面）。从这个视角看不到的点数总和是

Three dice with faces numbered $1$ through $6$ are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is

$21$

$22$

$31$

$41$

$53$

答案：D

知识点：补集计数

难度评级：960

解答：

一个骰子各面的点数和为 $1 + 2 + 3 + 4 + 5 + 6 = 21.$ 因此 $3$ 个骰子的总点数是 $3 \cdot 21 = 63$ 。

可见点数之和为 $1 + 1 + 2 + 3 + 4 + 5 + 6 = 22.$ 所以不可见点数之和是 $63 - 22 = 41$ 。

所以正确答案是 D。

The sum of the numbers on one die is $1 + 2 + 3 + 4 + 5 + 6 = 21.$ Therefore, the sum of the numbers on all $3$ dice is $3 \cdot 21 = 63.$

The visible numbers add up to $1 + 1 + 2 + 3 + 4 + 5 + 6 = 22.$ This makes the sum of the unseen numbers $63 - 22 = 41.$

Thus, D is the correct answer.

2000 AMC 8 第 8 题

答案：D

难度评级：960

解答：

其他年份的第 8 题