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

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<math>100\frac{a_n}{N}</math>. Setting up the inequality  
 
<math>100\frac{a_n}{N}</math>. Setting up the inequality  
 
<math>a_n \geq 1 + 100\frac{a_n}{N}</math> and simplifying,
 
<math>a_n \geq 1 + 100\frac{a_n}{N}</math> and simplifying,
−
<math>a_n \geq \ceil(\frac{N}{N - 100})</math> (the ceiling function is there because <math>a_n</math> is an integer.
+
<math>a_n \geq \lceil(\frac{N}{N - 100})\rceil</math> (the ceiling function is there because <math>a_n</math> is an integer.
 
Note that if we set all <math>a_i</math> equal to <math>\ceil(\frac{N}{100 - N})</math> we have <math>N \geq 27\ceil(\frac{N}{100 - N})</math>. Clearly <math>N = 134</math> is the least such number that satisfies this inequality. Now we must show that we can find suitable <math>a_i</math>. We can let 26 of them equal to <math>5</math> and one of them equal to <math>4</math>. Therefore, <math>N = \boxed{134}</math> is the answer.
 
Note that if we set all <math>a_i</math> equal to <math>\ceil(\frac{N}{100 - N})</math> we have <math>N \geq 27\ceil(\frac{N}{100 - N})</math>. Clearly <math>N = 134</math> is the least such number that satisfies this inequality. Now we must show that we can find suitable <math>a_i</math>. We can let 26 of them equal to <math>5</math> and one of them equal to <math>4</math>. Therefore, <math>N = \boxed{134}</math> is the answer.
  

Revision as of 15:41, 8 July 2019

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.

Solution 2

Let there be $N$ members of the committee. Suppose candidate $n$ gets $a_n$ votes. Then $a_n$ as a percentage out of $N$ is $100\frac{a_n}{N}$. Setting up the inequality $a_n \geq 1 + 100\frac{a_n}{N}$ and simplifying, $a_n \geq \lceil(\frac{N}{N - 100})\rceil$ (the ceiling function is there because $a_n$ is an integer. Note that if we set all $a_i$ equal to $\ceil(\frac{N}{100 - N})$ (Error compiling LaTeX. Unknown error_msg) we have $N \geq 27\ceil(\frac{N}{100 - N})$ (Error compiling LaTeX. Unknown error_msg). Clearly $N = 134$ is the least such number that satisfies this inequality. Now we must show that we can find suitable $a_i$. We can let 26 of them equal to $5$ and one of them equal to $4$. Therefore, $N = \boxed{134}$ is the answer.

See also

2003 AIME II (Problems • Answer Key • Resources)
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

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