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 be the number of votes candidate received, and let be the total number of votes cast. Our goal is to determine the smallest possible .
Candidate got of the votes, hence the percentage of votes they received is . The condition in the problem statement says that . ( means "for all", so this means "For all , is true")
Obviously, if some would be or , the condition would be false. Thus . We can then rewrite the above inequality as .
If for some we have , then from the inequality we just derived we would have . If for some we have , then . And if for some we have , then , and hence .
Is it possible to have ? We just proved that to have such , all have to be at least . But then , which is a contradiction. Hence the smallest possible is at least .
Now consider a situation where candidates got votes each, and one candidate got votes. In this situation, the total number of votes is exactly , and for each candidate the above inequality is satisfied. Hence the minimum number of committee members is .
Note: Each of the candidates received votes, and the last candidate received votes.
Solution 2
Let there be members of the committee. Suppose candidate gets votes. Then as a percentage out of is . Setting up the inequality and simplifying, (the ceiling function is there because is an integer. Note that if we set all equal to we have . Clearly is the least such number that satisfies this inequality. Now we must show that we can find suitable . We can let 26 of them equal to and one of them equal to . Therefore, is the answer. - whatRthose
Solution 3
Let be the total number of people in the committee, and be the number of votes candidate gets where . The problem tells us that Therefore, and so . Trying , we get that a contradiction. Bashing out a few more, we find that works perfectly fine.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.