1961 AMC 12 第 24 题

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：

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