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AIME 2003 II · 第 12 题

AIME 2003 II — Problem 12

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of the committee?

解析

Solution

Let viv_i be the number of votes candidate ii received, and let s=v1++v27s=v_1+\cdots+v_{27} be the total number of votes cast. Our goal is to determine the smallest possible ss.

Candidate ii got vis\frac{v_i}s of the votes, hence the percentage of votes they received is 100vis\frac{100v_i}s. The condition in the problem statement says that i:100vis+1vi\forall i: \frac{100v_i}s + 1 \leq v_i. (\forall means "for all", so this means "For all ii, 100vis+1vi\frac{100v_i}s + 1 \leq v_i is true")

Obviously, if some viv_i would be 00 or 11, the condition would be false. Thus i:vi2\forall i: v_i\geq 2. We can then rewrite the above inequality as i:s100vivi1\forall i: s\geq\frac{100v_i}{v_i-1}.

If for some ii we have vi=2v_i=2, then from the inequality we just derived we would have s200s\geq 200.

If for some ii we have vi=3v_i=3, then s150s\geq 150.

And if for some ii we have vi=4v_i=4, then s4003=13313s\geq \frac{400}3 = 133\frac13, and hence s134s\geq 134.

Is it possible to have s<134s<134? We just proved that to have such ss, all viv_i have to be at least 55. But then s=v1++v27275=135s=v_1+\cdots+v_{27}\geq 27\cdot 5 = 135, which is a contradiction. Hence the smallest possible ss is at least 134134.

Now consider a situation where 2626 candidates got 55 votes each, and one candidate got 44 votes. In this situation, the total number of votes is exactly 134134, and for each candidate the above inequality is satisfied. Hence the minimum number of committee members is s=134s=\boxed{134}.

Note: Each of the 2626 candidates received 3.63%\simeq 3.63\% votes, and the last candidate received 2.985%\simeq 2.985\% votes.

Solution 2

Let there be NN members of the committee. Suppose candidate nn gets ana_n votes. Then ana_n as a percentage out of NN is 100anN100\frac{a_n}{N}. Setting up the inequality an1+100anNa_n \geq 1 + 100\frac{a_n}{N} and simplifying, an(NN100)a_n \geq \lceil(\frac{N}{N - 100})\rceil (the ceiling function is there because ana_n is an integer. Note that if we set all aia_i equal to (N100N)\lceil(\frac{N}{100 - N})\rceil we have N27(N100N)N \geq 27\lceil(\frac{N}{100 - N})\rceil. Clearly N=134N = 134 is the least such number that satisfies this inequality. Now we must show that we can find suitable aia_i. We can let 26 of them equal to 55 and one of them equal to 44. Therefore, N=134N = \boxed{134} is the answer. - whatRthose

Solution 3

Let nn be the total number of people in the committee, and aia_i be the number of votes candidate ii gets where 1i271 \leq i \leq 27. The problem tells us that

100ainai1    100ainain    ainn100.\frac{100a_i}{n} \leq a_i - 1 \implies 100a_i \leq na_i - n \implies a_i \geq \frac{n}{n-100}. Therefore,

i=127ai=ni=127nn100=27nn100,\sum^{27}_{i=1} a_i = n \geq \sum^{27}_{i=1} \frac{n}{n-100} = \frac{27n}{n-100}, and so n(n127)0    n127n(n-127) \geq 0 \implies n \geq 127. Trying n=127n = 127, we get that

ai127274.7    ai5    ai27ai527=135127,a_i \geq \frac{127}{27} \approx 4.7 \implies a_i \geq 5 \implies \sum^{27}_{a_i} a_i \geq 5 \cdot 27 = 135 \geq 127, a contradiction. Bashing out a few more, we find that n=134\boxed{n = 134} works perfectly fine.

Video Solution by Sal Khan

https://www.youtube.com/watch?v=KD46pC_KFWk&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=10 - AMBRIGGS