2004 AMC 12B 第 8 题

Question

一位杂货商摆放罐头，最上面一排有一个罐头，每下面一排都比上一排多两个罐头。若这个陈列共有

100

个罐头，它有多少排？

A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains

100

cans, how many rows does it contain?

5

8

9

10

11

各排罐头数为

1, 3, 5, \ldots

，即奇数。

The row counts are

1, 3, 5, \ldots,

the odd numbers

前

n

个奇数之和为

n^2

。

The sum of the first

n

odd numbers is

n^2

各排罐头数为

1, 3, 5, \ldots, (2n - 1)

，前

n

个奇数之和为

n^2

。令

n^2 = 100

，得

n = 10

。

因此正确答案是 D。

The rows contain

1, 3, 5, \ldots, (2n - 1)

cans, and the sum of the first

n

odd numbers is

n^2.

Setting

n^2 = 100

gives

n = 10.

Thus, the correct answer is D.

其他年份的第 8 题