1994 AMC 12 第 19 题

先试着解答 1994 AMC 12 第 19 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 1994 AMC 12 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。

19.

在一张圆片上标“11”,两张圆片上标“22”,三张圆片上标“33”,…\ldots,五十张圆片上标“5050”。把这 1+2+3+⋯+50=12751+2+3+\cdots+50=1275 张带标号的圆片放进盒子里,再从盒中随机不放回地抽取圆片。至少必须抽取多少张圆片,才能保证抽到至少十张标号相同的圆片?
Label one disk “11,” two disks “22,” three disks “33,” …,\ldots, fifty disks “5050.” Put these 1+2+3+⋯+50=12751+2+3+\cdots+50=1275 labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is
1010
5151
415415
451451
501501
答案:C
知识点:抽屉原理extremal counting
小提示:
计算在每种标号最多抽到九张的条件下,最多可以抽取多少张圆片
Count the most disks that can be drawn while taking at most nine of each label
大提示:
可以抽出标号为 11 至 99 的所有圆片,但标号为 1010 至 5050 的每种最多只能抽九张
All disks labeled 11 through 99 may be drawn, but only nine of each label 1010 through 5050
解答:
为了避免出现十张标号相同的圆片,可以抽出 1+2+⋯+9=45 1+2+\cdots+9=45 张标号小于 1010 的所有圆片,并从每种标号中至多抽 99 张;这样的标号共有 4141 种,从 1010 到 5050。因此,在尚未被迫得到十张相同标号的圆片前,最多可抽 45+9(41)=41445+9(41)=414 张;再抽一张就能保证达到要求。最小数量为 415415。所以正确答案是 C。
To avoid ten equal labels, one may draw all 1+2+⋯+9=45 1+2+\cdots+9=45 disks with labels below 10,10, and at most 99 from each of the 4141 labels 1010 through 50.50. Thus 45+9(41)=41445+9(41)=414 disks can be drawn without forcing ten alike, and the next draw guarantees them. The minimum is 415.415. Thus the correct answer is C.
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