AMC12 2019 A

AMC12 2019 A · Q13

AMC12 2019 A · Q13. It mainly tests Basic counting (rules of product/sum), Divisibility & factors.

How many ways are there to paint each of the integers 2, 3, ..., 9 either red, green, or blue so that each number has a different color from each of its proper divisors?

有几种方法可以将整数2, 3, ..., 9每个涂成红色、绿色或蓝色，使得每个数与其每个真约数的颜色都不同？

(A) 144 144

(B) 216 216

(D) 384 384

(E) 432 432

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The restriction amounts to the following constraints: 2, 4, and 8 get different colors; 2 and 6 get different colors; 3 and 6 get different colors; and 3 and 9 get different colors. There are $3!$ ways to assign the colors for 2, 4, and 8. Once this is done, there are 2 ways to assign the color for 6, then 2 ways for 3, and 2 ways for 9. Finally, there are 3 ways to assign the color for each of 5 and 7. Hence the total number of paintings is $3!\cdot 2^3 \cdot 3^2 = 432$.

答案（E）：限制条件等价于以下约束：2、4、8 的颜色两两不同；2 与 6 颜色不同；3 与 6 颜色不同；以及 3 与 9 颜色不同。给 2、4、8 分配颜色共有 $3!$ 种方法。完成后，给 6 分配颜色有 2 种方法，然后给 3 有 2 种方法，给 9 也有 2 种方法。最后，给 5 和 7 各自分配颜色各有 3 种方法。因此，涂色总数为 $3!\cdot 2^3 \cdot 3^2 = 432$。

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