Question

三十一本书从左到右按价格递增排列。每本书与相邻书的价格相差 $\$2$ 。用最右端一本书的价格，顾客可以买到中间一本书和与它相邻的一本书。则：

Thirty-one books are arranged from left to right in order of increasing prices. The price of each book differs by $\$2$ from that of each adjacent book. For the price of the book at the extreme right a customer can buy the middle book and an adjacent one. Then:

所指的相邻书位于中间那本书的左侧

the adjacent book referred to is at the left of the middle book

中间那本书售价 $\$36$

the middle book sells for $\$36$

最便宜的书售价 $\$4$

the cheapest book sells for $\$4$

最贵的书售价 $\$64$

the most expensive book sells for $\$64$

以上都不正确

none of these are correct

答案：A

知识点：等差数列钱币分类讨论

难度评级：1530

小提示：

设最低价格为 $p$ ，写出第 $16$ 本和第 $31$ 本的价格

Let the cheapest price be $p$ ; identify the $16$ th and $31$ st prices

大提示：

分别检验中间书两侧的书，并排除使价格非正的情形

Test the two books adjacent to the middle book and reject any case forcing a nonpositive price

解答：

各书价格为 $p$ 、 $p+2$ 、 $\ldots$ 、 $p+60$ ，中间一本的价格为 $p+30$ 。若取右侧相邻书，则 $(p+30)+(p+32)=p+60\text{，}$ 得 $p=-2$ ，不可能是售价。取左侧相邻书时 $(p+30)+(p+28)=p+60\text{，}$ 所以 $p=2$ 。因此相邻书必须是左侧那一本。

因此，正确答案是 A。

The prices are $p,$ $p+2,$ $\ldots,$ $p+60,$ and the middle price is $p+30.$ If the right neighbor is used, then $(p+30)+(p+32)=p+60,$ which gives $p=-2,$ impossible for a selling price. The left neighbor gives $(p+30)+(p+28)=p+60,$ so $p=2.$ Thus the adjacent book must be the one to the left.

Therefore, the correct answer is A.

1961 AMC 12 第 24 题

答案：A

难度评级：1530

小提示：

大提示：

解答：

其他年份的第 24 题