2017 AIME I 第 11 题

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11.

考虑把 99 个数 1,2,3,,91, 2, 3, \ldots, 9 排列在一个 3×33 \times 3 方阵中。对每种排列, 令 a1a_1a2a_2, 和 a3a_3 分别为第 1122, 和 33, 行中三个数的中位数, 再令 mm 为集合 {a1,a2,a3}\{a_1, a_2, a_3\} 的中位数。设 QQ 为满足 m=5m = 5 的排列数。 求 QQ 除以 10001000 的余数。

Consider arrangements of the 99 numbers 1,2,3,,91, 2, 3, \ldots, 9 in a 3×33 \times 3 array. For each such arrangement, let a1,a_1, a2,a_2, and a3a_3 be the medians of the numbers in rows 1,1, 2,2, and 3,3, respectively, and then let mm be the median of {a1,a2,a3}.\{a_1, a_2, a_3\}. Let QQ be the number of arrangements for which m=5.m = 5. Find the remainder when QQ is divided by 1000.1000.

答案:360
知识点:中位数(数据)分类讨论排列
难度评级:2990
解答:

1,2,3,41, 2, 3, 4 各记为 L,将 6,7,8,96, 7, 8, 9 各记为 G。如果 55 不是某一行的中位数,那么没有一行的中位数等于 55,所以 m5m \neq 5。因此 55 所在行必须含有一个 L 和一个 G(按某种顺序为 L5G),而另外两行必须提供一个小于 55 的中位数和一个大于五的中位数。用剩下的三个 L 和三个 G,这两行要么是 LLL 与 GGG,要么是 LLG 与 LGG。

先数字母排列:这三种行类型可以分配给第 1,2,31, 2, 3 行,共 3!=63! = 6 种方法,而 L5G 这一行可以排列成 3!=63! = 6 种。第一种情况中,LLL 和 GGG 各只有 11 种排列, 得到 661=366 \cdot 6 \cdot 1 = 36 个模式;第二种情况中,LLG 和 LGG 各有 33 种排列, 得到 669=3246 \cdot 6 \cdot 9 = 324 个模式。总共有 360360 个字母模式。

最后四个 L 可由 1,2,3,41, 2, 3, 44!4! 种方式填入,四个 G 可由 6,7,8,96, 7, 8, 94!4! 种方式填入,所以 Q=360242=207360Q = 360 \cdot 24^2 = 207360,模 10001000 的余数为 360360

Rename each of 1,2,3,41, 2, 3, 4 as L and each of 6,7,8,96, 7, 8, 9 as G. If 55 is not a row median, then no row median equals 5,5, so m5.m \neq 5. Thus 55's row must contain one L and one G (reading L5G in some order), and the other two rows must supply one median below 55 and one above. With the remaining three L's and three G's, those rows are either LLL and GGG, or LLG and LGG.

Count arrangements of letters: the three row types can be assigned to rows 1,2,31, 2, 3 in 3!=63! = 6 ways, and the L5G row can be ordered in 3!=63! = 6 ways. In the first case LLL and GGG have 11 ordering each, giving 661=366 \cdot 6 \cdot 1 = 36 patterns; in the second, LLG and LGG each have 33 orderings, giving 669=3246 \cdot 6 \cdot 9 = 324 patterns. That is 360360 letter patterns in all.

Finally the four L's can be filled with 1,2,3,41, 2, 3, 4 in 4!4! ways and the four G's with 6,7,8,96, 7, 8, 9 in 4!4! ways, so Q=360242=207360,Q = 360 \cdot 24^2 = 207360, whose remainder mod 10001000 is 360.360.

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