2008 AMC 12B 第 1 题

先试着解答 2008 AMC 12B 第 1 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2008 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

1.

一名篮球运动员在一场比赛中投进了 55 个球。每次进球得 22 分或 33 分。该运动员总得分可能有多少种不同的数值?
A basketball player made 55 baskets during a game. Each basket was worth either 22 or 33 points. How many different numbers could represent the total points scored by the player?
22
33
44
55
66
答案:E
知识点:区间内整数计数
小提示:
总分最小时全是 22 分球,最大时全是 33 分球。
The total is smallest with all 22-point baskets and largest with all 33-point baskets
大提示:
5⋅25 \cdot 2 和 5⋅35 \cdot 3 之间的每个整数都能达到。
Every integer between 5⋅25 \cdot 2 and 5⋅35 \cdot 3 is attainable
解答:
总分从 5⋅2=105 \cdot 2 = 10(全是两分球)到 5⋅3=155 \cdot 3 = 15(全是三分球)。每把一个两分球换成三分球,总分正好增加 11,所以中间的每个整数都会出现。
可能的总分是 10,11,12,13,14,1510, 11, 12, 13, 14, 15,共有 66 个值。
因此,正确答案是 E。
The total ranges from 5⋅2=105 \cdot 2 = 10 (all two-pointers) to 5⋅3=155 \cdot 3 = 15 (all three-pointers). Swapping one two-pointer for a three-pointer raises the total by exactly 1,1, so every integer in between occurs.
The possible totals are 10,11,12,13,14,15,10, 11, 12, 13, 14, 15, which is 66 values.
Thus, the correct answer is E.
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