AMC12 2021 B

AMC12 2021 B · Q19

AMC12 2021 B · Q19. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Two fair dice, each with at least $6$ faces are rolled. On each face of each die is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum of $7$ is $\frac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\frac{1}{12}$. What is the least possible number of faces on the two dice combined?

掷两个公平的骰子，每个骰子至少有 $6$ 个面。每个骰子的每个面上印有从 $1$ 到该骰子面数的不同整数。掷出和为 $7$ 的概率是掷出和为 $10$ 的概率的 $\frac34$，掷出和为 $12$ 的概率是 $\frac{1}{12}$。两骰子面数总和的最小可能值为多少？

(A) 16 16

(B) 17 17

(D) 19 19

(E) 20 20

Answer

Correct choice: (B)

正确答案：（B）

Solution

Suppose the dice have $a$ and $b$ faces, and WLOG $a\geq{b}$. Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$. As a result, there must be $\tfrac{4}{3}\cdot6=8$ ways to sum to $10$. There are at most nine distinct ways to get a sum of $10$, which are possible whenever $a,b\geq{9}$. To achieve exactly eight ways, $b$ must have $8$ faces, and $a\geq9$. Let $n$ be the number of ways to obtain a sum of $12$, then $\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a$. Since $b=8$, $n\leq8\implies a\leq{12}$. In addition to $3\mid{a}$, we only have to test $a=9,12$, of which both work. Taking the smaller one, our answer becomes $a+b=9+8=\boxed{\textbf{(B)}\ 17}$.

假设骰子有 $a$ 和 $b$ 个面，且不失一般性 $a\geq b$。由于每个骰子至少 $6$ 个面，和为 $7$ 总有 $6$ 种方式。因此，和为 $10$ 必须有 $\tfrac{4}{3}\cdot6=8$ 种方式。和为 $10$ 最多有九种不同方式，当 $a,b\geq{9}$ 时可能。要正好八种，$b$ 必须有 $8$ 个面，且 $a\geq9$。设 $n$ 为和为 $12$ 的方式数，则 $\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a$。由于 $b=8$，$n\leq8\implies a\leq{12}$。此外 $3\mid a$，只需测试 $a=9,12$，两者都可。取较小者，答案为 $a+b=9+8=\boxed{\textbf{(B)}\ 17}$。

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