2003 AIME II Problems/Problem 12

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.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS