Difference between revisions of "2003 AIME II Problems/Problem 12"

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== Solution ==
 
== Solution ==
{{solution}}
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Let <math>v_i</math> be the number of votes candidate <math>i</math> received, and let <math>s=v_1+\cdots+v_{27}</math> be the total number of votes cast. Our goal is to determine the smallest possible <math>s</math>.
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Candidate <math>i</math> got <math>\frac{v_i}s</math> of the votes, hence the percentage of votes she received is <math>\frac{100v_i}s</math>. The condition in the problem statement says that <math>\forall i: \frac{100v_i}s + 1 \leq v_i</math>.
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Obviously, if some <math>v_i</math> would be <math>0</math> or <math>1</math>, the condition would be false. Thus <math>\forall i: v_i\geq 2</math>. We can then rewrite the above inequality as <math>\forall i: s\geq\frac{100v_i}{v_i-1}</math>.
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If for some <math>i</math> we have <math>v_i=2</math>, then from the inequality we just derived we would have <math>s\geq 200</math>. If for some <math>i</math> we have <math>v_i=3</math>, then <math>s\geq 150</math>. And if for some <math>i</math> we have <math>v_i=4</math>, then <math>s\geq \frac{400}3 = 133\frac13</math>, and hence <math>s\geq 134</math>.
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Is it possible to have <math>s<134</math>? We just proved that to have such <math>s</math>, all <math>v_i</math> have to be at least <math>5</math>. But then <math>s=v_1+\cdots+v_{27}\geq 27\cdot 5 = 135</math>, which is a contradiction. Hence the smallest possible <math>s</math> is at least <math>134</math>.
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Now consider a situation where <math>26</math> candidates got <math>5</math> votes each, and one candidate got <math>4</math> votes. In this situation, the total number of votes is exactly <math>134</math>, and for each candidate the above inequality is satisfied. Hence the minimum number of committee members is <math>s=\boxed{134}</math>.
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Note: Each of the <math>26</math> candidates received <math>\simeq 3.63\%</math> votes, and the last candidate received <math>\simeq 2.985\%</math> votes.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2003|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2003|n=II|num-b=11|num-a=13}}
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v_1+\cdots+v_{27}

Revision as of 22:58, 28 January 2009

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.

See also

2003 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

v_1+\cdots+v_{27}