2003 AIME II 第 12 题

先试着解答 2003 AIME II 第 12 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 2003 AIME II 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。

12.

一个杰出委员会的成员正在选主席,每位成员都给 2727 位候选人中的一位投一票。对每位候选人来说,其所得票数的准确百分比至少比其所得票数小 11。委员会成员人数的最小可能值是多少?

The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 2727 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 11 than the number of votes for that candidate. What is the smallest possible number of members of the committee?

答案:134
知识点:百分数不等式极端原理
难度评级:2920
解答:

设成员人数为 tt 得到 nn 票的候选人的得票百分比为 100nt\frac{100n}{t}, 因此条件为 100ntn1\frac{100n}{t} \le n - 1, 整理得 n(t100)tn(t - 100) \ge t。 这迫使 t>100t \gt 100,并且 ntt100.n \ge \frac{t}{t - 100}.

t133t \le 133, 则 tt10013333>4\frac{t}{t - 100} \ge \frac{133}{33} \gt 4, 所以每位候选人都至少需要 55 票,总票数至少为 275=135>t27 \cdot 5 = 135 \gt t,不可能。

t=134t = 134 时,每位候选人需要 n13434n \ge \frac{134}{34}, 即至少 44 票,这是可以做到的: 让 2626 位候选人各得 44 票,另一个候选人得 3030。 票。确实 4001342.993\frac{400}{134} \approx 2.99 \le 3300013422.429\frac{3000}{134} \approx 22.4 \le 29。 所以最小可能成员人数为 134134

Let tt be the number of members. A candidate with nn votes has percentage 100nt,\frac{100n}{t}, so the condition is 100ntn1,\frac{100n}{t} \le n - 1, which rearranges to n(t100)t.n(t - 100) \ge t. This forces t>100t \gt 100 and ntt100.n \ge \frac{t}{t - 100}.

If t133,t \le 133, then tt10013333>4,\frac{t}{t - 100} \ge \frac{133}{33} \gt 4, so every candidate needs at least 55 votes, and the total is at least 275=135>t27 \cdot 5 = 135 \gt t — impossible.

For t=134,t = 134, each candidate needs n13434,n \ge \frac{134}{34}, i.e. at least 44 votes, and this is achievable: let 2626 candidates receive 44 votes each and one receive 30.30. Indeed 4001342.993\frac{400}{134} \approx 2.99 \le 3 and 300013422.429.\frac{3000}{134} \approx 22.4 \le 29. So the smallest possible number of members is 134.134.

← 第 11 题#11
完整试卷

其他年份的第 12 题