AMC8 2007

AMC8 2007 · Q24

AMC8 2007 · Q24. It mainly tests Permutations, Digit properties (sum of digits, divisibility tests).

A bag contains four pieces of paper, each labeled with one of the digits 1, 2, 3 or 4, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

一个袋子里有四张纸，每张标有一个数字1、2、3或4，没有重复。从中不放回地抽三张纸，依次构造一个三位数。这个三位数是3的倍数的概率是多少？

(A) $\frac{1}{4}$ $\frac{1}{4}$

(B) $\frac{1}{3}$ $\frac{1}{3}$

(D) $\frac{2}{3}$ $\frac{2}{3}$

(E) $\frac{3}{4}$ $\frac{3}{4}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) A number is a multiple of three when the sum of its digits is a multiple of 3. If the number has three distinct digits drawn from the set $\{1,2,3,4\}$, then the sum of the digits will be a multiple of three when the digits are $\{1,2,3\}$ or $\{2,3,4\}$. That means the number formed is a multiple of three when, after the three draws, the number remaining in the bag is $1$ or $4$. The probability of this occurring is $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$.

（C）当一个数的各位数字之和是 $3$ 的倍数时，这个数就是 $3$ 的倍数。若该数由集合 $\{1,2,3,4\}$ 中抽取的三个互不相同的数字组成，则当数字为 $\{1,2,3\}$ 或 $\{2,3,4\}$ 时，数字和为 $3$ 的倍数。因此，当抽取三个数字后袋中剩下的数字是 $1$ 或 $4$ 时，所组成的数是 $3$ 的倍数。该事件发生的概率为 $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$。

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