2003 AIME I 第 3 题

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3.

设集合 S={8,5,1,13,34,3,21,2}\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}。 Susan 按如下方式列出一个表: 对于 S\mathcal{S} 的每个二元子集,她在表上写下该子集中较大的元素。求表上所有数的和。

Let the set S={8,5,1,13,34,3,21,2}.\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}. Susan makes a list as follows: for each two-element subset of S,\mathcal{S}, she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.

答案:484
知识点:数对计数双重计数
难度评级:1840
解答:

元素 xx 与集合中每个比它小的元素组成二元子集时,都会成为较大的元素,因此 xx 对总和的贡献次数等于比它小的元素个数。将集合排序为 1,2,3,5,8,13,21,341, 2, 3, 5, 8, 13, 21, 34,表上所有数的和为 0(1)+1(2)+2(3)+3(5)+4(8)+5(13)+6(21)+7(34)=2+6+15+32+65+126+238=484. \begin{aligned} &0(1) + 1(2) + 2(3) + 3(5) \\ &\quad {}+ 4(8) + 5(13) + 6(21) \\ &\quad {}+ 7(34) \\ &= 2 + 6 + 15 + 32 \\ &\quad {}+ 65 + 126 + 238 = 484. \end{aligned}

An element xx is the greater element of a two-element subset exactly once for each smaller element of the set, so xx contributes to the sum once per element below it. Sorting the set as 1,2,3,5,8,13,21,34,1, 2, 3, 5, 8, 13, 21, 34, the sum of the list is 0(1)+1(2)+2(3)+3(5)+4(8)+5(13)+6(21)+7(34)=2+6+15+32+65+126+238=484. \begin{aligned} &0(1) + 1(2) + 2(3) + 3(5) \\ &\quad {}+ 4(8) + 5(13) + 6(21) \\ &\quad {}+ 7(34) \\ &= 2 + 6 + 15 + 32 \\ &\quad {}+ 65 + 126 + 238 = 484. \end{aligned}

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