2003 AIME II Problems/Problem 12

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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?

Solution

Let $v_i$ be the number of votes candidate $i$ received, and let $s=v_1+\cdots+v_{27}$ be the total number of votes cast. Our goal is to determine the smallest possible $s$ .

Candidate $i$ got $\frac{v_i}s$ of the votes, hence the percentage of votes she received is $\frac{100v_i}s$ . The condition in the problem statement says that $\forall i: \frac{100v_i}s + 1 \leq v_i$ .

Obviously, if some $v_i$ would be $0$ or $1$ , the condition would be false. Thus $\forall i: v_i\geq 2$ . We can then rewrite the above inequality as $\forall i: s\geq\frac{100v_i}{v_i-1}$ .

If for some $i$ we have $v_i=2$ , then from the inequality we just derived we would have $s\geq 200$ . If for some $i$ we have $v_i=3$ , then $s\geq 150$ . And if for some $i$ we have $v_i=4$ , then $s\geq \frac{400}3 = 133\frac13$ , and hence $s\geq 134$ .

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

Now consider a situation where $26$ candidates got $5$ votes each, and one candidate got $4$ votes. In this situation, the total number of votes is exactly $134$ , and for each candidate the above inequality is satisfied. Hence the minimum number of committee members is $s=\boxed{134}$ .

Note: Each of the $26$ candidates received $\simeq 3.63\%$ votes, and the last candidate received $\simeq 2.985\%$ votes.

2003 AIME II (Problems • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2003 AIME II Problems/Problem 12

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Problem

Solution

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