Question

Anh 读一本书。第一天她用 $t$ 分钟读了 $n$ 页，其中 $n$ 和 $t$ 都是正整数。第二天 Anh 用 $t + 1$ 分钟读了 $n + 1$ 页。此后每天 Anh 都比前一天多读一页，并且比前一天多用一分钟，直到她读完整本 $374$ 页的书。她读完这本书总共用了 $319$ 分钟。求 $n + t$ 。

Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t.$

答案

答案：53

知识点：等差数列最大公约数整除性

难度评级：2430

解答：

设 Anh 在第 $k$ 天读完。分别对每天页数和分钟数的等差数列求和， $\frac{k(2n + k - 1)}{2} = 374$ 且 $\frac{k(2t + k - 1)}{2} = 319,$ 所以 $k(2n + k - 1) = 748$ ，并且 $k(2t + k - 1) = 638$ 。

相减得 $2k(n - t) = 110$ ，所以 $k(n - t) = 55$ 。因此 $k$ 同时整除 $55$ 和 $\gcd(748, 638) = 22$ ，所以 $k \mid 11$ 。由于故事跨越不止一天， $k = 11$ 。

接着 $2n + 10 = \frac{748}{11} = 68$ ，得到 $n = 29$ ， $2t + 10 = \frac{638}{11} = 58$ ，得到 $t = 24$ 。因此 $n + t = 29 + 24 = 53$ 。

Say Anh finished on day $k.$ Summing the arithmetic progressions of pages and of minutes, $\frac{k(2n + k - 1)}{2} = 374$ and $\frac{k(2t + k - 1)}{2} = 319,$ so $k(2n + k - 1) = 748$ and $k(2t + k - 1) = 638.$

Subtracting, $2k(n - t) = 110,$ so $k(n - t) = 55.$ Thus $k$ divides both $55$ and $\gcd(748, 638) = 22,$ so $k \mid 11.$ Since the story spans more than one day, $k = 11.$

Then $2n + 10 = \frac{748}{11} = 68$ gives $n = 29,$ and $2t + 10 = \frac{638}{11} = 58$ gives $t = 24.$ Hence $n + t = 29 + 24 = 53.$

2016 AIME I 第 5 题

答案：53

难度评级：2430

解答：

其他年份的第 5 题