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

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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>.
 
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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Candidate <math>i</math> got <math>\frac{v_i}s</math> of the votes, hence the percentage of votes they 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>. (<math>\forall</math> means "for all", so this means "For all <math>i</math>, <math>\frac{100v_i}s + 1 \leq v_i</math> is true")
  
 
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>.  
 
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>.  
  
−
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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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>.
  
 
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>.
 
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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<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,
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<math>a_n \geq \ceil(\frac{N}{N - 100})</math> (the ceiling function is there because <math>a_n</math> is an integer.
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<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>\lceil(\frac{N}{100 - N})\rceil</math> we have <math>N \geq 27\lceil(\frac{N}{100 - N})\rceil</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.
 +
- whatRthose
 +
 
 +
== Solution 3 ==
 +
Let <math>n</math> be the total number of people in the committee, and <math>a_i</math> be the number of votes candidate <math>i</math> gets where <math>1 \leq i \leq 27</math>.  The problem tells us that <cmath>\frac{100a_i}{n} \leq a_i - 1 \implies 100a_i \leq na_i - n \implies a_i \geq \frac{n}{n-100}.</cmath>Therefore, <cmath>\sum^{27}_{i=1} a_i = n \geq \sum^{27}_{i=1} \frac{n}{n-100} = \frac{27n}{n-100},</cmath>and so <math>n(n-127) \geq 0 \implies n \geq 127</math>.  Trying <math>n = 127</math>, we get that <cmath>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,</cmath>a contradiction.  Bashing out a few more, we find that <math>\boxed{n = 134}</math> works perfectly fine.
 +
 
 +
== Video Solution by Sal Khan ==
 +
https://www.youtube.com/watch?v=KD46pC_KFWk&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=10
 +
- AMBRIGGS
  
 
== 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}}
 +
 +
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:00, 11 December 2022

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 they received is $\frac{100v_i}s$. The condition in the problem statement says that $\forall i: \frac{100v_i}s + 1 \leq v_i$. ($\forall$ means "for all", so this means "For all $i$, $\frac{100v_i}s + 1 \leq v_i$ is true")

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 $\lceil(\frac{N}{100 - N})\rceil$ we have $N \geq 27\lceil(\frac{N}{100 - N})\rceil$. 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. - whatRthose

Solution 3

Let $n$ be the total number of people in the committee, and $a_i$ be the number of votes candidate $i$ gets where $1 \leq i \leq 27$. The problem tells us that \[\frac{100a_i}{n} \leq a_i - 1 \implies 100a_i \leq na_i - n \implies a_i \geq \frac{n}{n-100}.\]Therefore, \[\sum^{27}_{i=1} a_i = n \geq \sum^{27}_{i=1} \frac{n}{n-100} = \frac{27n}{n-100},\]and so $n(n-127) \geq 0 \implies n \geq 127$. Trying $n = 127$, we get that \[a_i \geq \frac{127}{27} \approx 4.7 \implies a_i \geq 5 \implies \sum^{27}_{a_i} a_i \geq 5 \cdot 27 = 135 \geq 127,\]a contradiction. Bashing out a few more, we find that $\boxed{n = 134}$ works perfectly fine.

Video Solution by Sal Khan

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

See also

2003 AIME II (Problems • Answer Key • Resources)
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Problem 11 		Followed by
Problem 13
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All AIME Problems and Solutions

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