Question

一组文员被安排整理 $1775$ 份文件。每名文员以每小时 $30$ 份文件的恒定速度整理。第一小时结束时，其中一些文员被调去做另一项任务；第二小时结束时，剩余文员中同样数量的人也被调走；第三小时结束时也发生类似调配。这组人在 $3$ 小时 $10$ 分钟内完成整理。求整理开始后的前一个半小时内整理了多少份文件。

A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar reassignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.

答案

答案：945

知识点：速率丢番图方程模运算

难度评级：2310

解答：

设开始时有 $n$ 名文员，每小时结束时调走 $k$ 名。最后 $10$ 分钟每名剩余文员整理 $5$ 份文件，所以化简为 $95n - 105k = 1775$ ，即 $19n - 21k = 355$ 。 $\begin{aligned} &30n + 30(n - k) + 30(n - 2k) \\ &\quad {}+ 5(n - 3k) = 1775, \end{aligned}$

对 $19$ 取模，得到 $-2k \equiv 13$ ，所以 $2k \equiv 6$ ，且 $k \equiv 3 \pmod{19}$ 。取 $k = 3$ 得 $n = \frac{355 + 63}{19} = 22$ ；下一个候选 $k = 22$ 给出 $n = 43$ ，此时 $n - 3k \lt 0$ 。所以 $n = 22$ ， $k = 3$ 。

第一小时 $22$ 名文员整理 $30 \cdot 22 = 660$ 份文件，接下来的半小时 $19$ 名文员整理 $15 \cdot 19 = 285$ 份文件，总计 $660 + 285 = 945$ 。

Let $n$ clerks start and $k$ be reassigned at the end of each hour. In the final $10$ minutes each remaining clerk sorts $5$ files, so $\begin{aligned} &30n + 30(n - k) + 30(n - 2k) \\ &\quad {}+ 5(n - 3k) = 1775, \end{aligned}$ which simplifies to $95n - 105k = 1775,$ or $19n - 21k = 355.$

Modulo $19$ this reads $-2k \equiv 13,$ so $2k \equiv 6$ and $k \equiv 3 \pmod{19}.$ Taking $k = 3$ gives $n = \frac{355 + 63}{19} = 22;$ the next candidate, $k = 22,$ gives $n = 43,$ for which $n - 3k \lt 0.$ So $n = 22$ and $k = 3.$

In the first hour $22$ clerks sort $30 \cdot 22 = 660$ files, and in the next half hour $19$ clerks sort $15 \cdot 19 = 285,$ for a total of $660 + 285 = 945.$

2013 AIME II 第 7 题

答案：945

难度评级：2310

解答：

其他年份的第 7 题