2003 AIME II Problem 12

Below is the professionally curated solution for Problem 12 of the 2003 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AIME II solutions, or check the answer key.

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Concepts:percentageinequalityextremal argument

Difficulty rating: 2920

12.

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?

Solution:

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.

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