1999 AMC 8 第 15 题

先试着解答 1999 AMC 8 第 15 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 1999 AMC 8 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。

15.

平原镇的自行车牌照每个含有三个字母。第一个字母从集合 {C,H,L,P,R}\{C,H,L,P,R\} 中选,第二个从 {A,I,O}\{A,I,O\} 中选,第三个从 {D,M,N,T}\{D,M,N,T\} 中选。
当平原镇需要更多牌照时,他们增加了两个新字母。这两个新字母可以都加入同一个集合,也可以分别加入两个不同集合。加入两个字母后,最多能多制作多少个额外牌照?
Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R},\{C,H,L,P,R\}, the second from {A,I,O},\{A,I,O\}, and the third from {D,M,N,T}.\{D,M,N,T\}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of additional license plates that can be made by adding two letters?
2424
3030
3636
4040
6060
答案:D
知识点:乘法原理最优化分类讨论
小提示:
原来的牌照数是 5⋅3⋅45\cdot3\cdot4
The old number of plates is 5⋅3⋅45\cdot3\cdot4
大提示:
要让乘积最大,应让三个因子尽量接近平衡
To maximize a product, make the three factors as balanced as possible
解答:
目前一共可以制作 5⋅3⋅4=605 \cdot 3 \cdot 4 = 60 个牌照。
如果两个字母都加入第一个集合,那么可以制作 7⋅3⋅4=84 7 \cdot 3 \cdot 4 = 84 个牌照。
如果两个字母都加入第二个集合,那么有 5⋅5⋅4=100 5 \cdot 5 \cdot 4 = 100 个牌照。
如果两个字母都加入第三个集合,那么有 5⋅3⋅6=90 5 \cdot 3 \cdot 6 = 90 种选择。
如果一个字母加入第一个集合,另一个加入第二个集合,那么有 6⋅4⋅4=96 6 \cdot 4 \cdot 4 = 96 个牌照。
如果另一个字母改为加入第三个集合,那么可以得到 6⋅3⋅5=90 6 \cdot 3 \cdot 5 = 90 个牌照。
最后,如果两个字母分别加入第二个和第三个集合,那么有 5⋅4⋅5=100 5 \cdot 4 \cdot 5 = 100 个牌照。
最大可以达到 100100 个牌照,因此额外增加 100−60=40100 - 60 = 40 个。
所以正确答案是 D。
There are currently 5⋅3⋅4=605 \cdot 3 \cdot 4 = 60 license plates that can be made.
If both letters are added to the first set, then there are 7⋅3⋅4=84 7 \cdot 3 \cdot 4 = 84 possible plates.
If they are both added to the second, there are 5⋅5⋅4=100 5 \cdot 5 \cdot 4 = 100 plates.
If they are added to the third, there are 5⋅3⋅6=90 5 \cdot 3 \cdot 6 = 90 choices.
If one is added to the first set and the other to the second set, there are 6⋅4⋅4=96 6 \cdot 4 \cdot 4 = 96 plates.
If the other is added to the third set, we get 6⋅3⋅5=90 6 \cdot 3 \cdot 5 = 90 possible plates.
Finally, if the letters are added to the second and third sets, there are 5⋅4⋅5=100 5 \cdot 4 \cdot 5 = 100 plates.
We see that 100100 is the greatest number of plates that we can achieve. This is an additional 100−60=40100 - 60 = 40 plates.
Thus, D is the correct answer.
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