Question

Steve 从左到右反复写数字 $1$ 、 $2$ 、 $3$ 、 $4$ 和 $5$ ，形成一个有 $10,000$ 位数字的列表，开头为 $123451234512\ldots$ 。然后他擦掉列表中每第三个数字，也就是从左数第 $3$ 、 $6$ 、 $9$ 、 $\ldots$ 个数字；接着在剩下的列表中擦掉每第四个数字，也就是剩余列表中第 $4$ 、 $8$ 、 $12$ 、 $\ldots$ 个数字；然后在此时剩下的列表中擦掉每第五个数字。此后位于第 $2019$ 、 $2020$ 、 $2021$ 位的三个数字之和是多少？

Steve wrote the digits $1,$ $2,$ $3,$ $4,$ and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$ th, $8$ th, $12$ th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in positions $2019,$ $2020,$ and $2021?$

$7$

$9$

$10$

$11$

$12$

答案：D

知识点：找规律过程模拟最小公倍数

难度评级：1660

解答：

初始循环块为 $12345$ 。删除每第三个数字时，每 $\operatorname{lcm}(3,5)=15$ 个原位置重复一次：所以新周期长度为 $10$ 。 $\begin{aligned} &123451234512345 \\ &\quad \longrightarrow 1245235134, \end{aligned}$

从这个周期中删除每第四个数字时，每 $\operatorname{lcm}(4,10)=20$ 个位置重复一次：所以新周期长度为 $15$ 。 $\begin{aligned} &12452351341245235134 \\ &\quad \longrightarrow 124235341452513, \end{aligned}$

再删除每第五个数字，得到其周期为 $12$ 。由于 $2019\equiv 3\pmod{12}$ ，第 $2019$ 、 $2020$ 、 $2021$ 位分别对应周期中的第三、四、五位，即 $4,2,5$ ，和为 $11$ 。 $\begin{aligned} &124235341452513 \\ &\quad \longrightarrow 124253415251, \end{aligned}$

所以正确答案是 D。

Start with the repeating block $12345.$ Deleting every third digit repeats over $\operatorname{lcm}(3,5)=15$ original positions: $\begin{aligned} &123451234512345 \\ &\quad \longrightarrow 1245235134, \end{aligned}$ so the new period has length $10.$

Deleting every fourth digit from this period repeats over $\operatorname{lcm}(4,10)=20$ positions: $\begin{aligned} &12452351341245235134 \\ &\quad \longrightarrow 124235341452513, \end{aligned}$ so the new period has length $15.$

Deleting every fifth digit from this period gives $\begin{aligned} &124235341452513 \\ &\quad \longrightarrow 124253415251, \end{aligned}$ which has period $12.$ Since $2019\equiv 3\pmod{12},$ the digits in positions $2019,$ $2020,$ and $2021$ are the 3rd, 4th, and 5th digits of this period: $4,2,5.$ Their sum is $11.$

Thus, the correct answer is D .

2020 AMC 10B 第 15 题

答案：D

难度评级：1660

解答：

其他年份的第 15 题