2009 AIME II Problems/Problem 4
Contents
Problem
A group of children held a grape-eating contest. When the contest was over, the winner had eaten grapes, and the child in -th place had eaten grapes. The total number of grapes eaten in the contest was . Find the smallest possible value of .
Solution
Resolving the ambiguity
The problem statement is confusing, as it can be interpreted in two ways: Either as "there is a such that the child in -th place had eaten grapes", or "for all , the child in -th place had eaten grapes".
The second meaning was apparrently the intended one. Hence we will restate the problem statement in this way:
A group of children held a grape-eating contest. When the contest was over, the following was true: There was a such that for each between and , inclusive, the child in -th place had eaten exactly grapes. The total number of grapes eaten in the contest was . Find the smallest possible value of .
Solving this problem
The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term (the number of grapes eaten by the child in -st place), difference , and number of terms . We can easily compute that this sum is equal to .
Hence we have the equation , and we are looking for a solution , where both and are positive integers, , and is maximized. (The condition states that even the last child had to eat a non-negative number of grapes.)
The prime factorization of is . Hence there are ways how to factor into two positive terms and :
- , , then is a solution.
- , , then is a solution.
- , , then is a solution.
- In each of the other three cases, will obviously be less than , hence these are not valid solutions.
The smallest valid solution is therefore , .
Another Solution
Let us go in order of place, from to . The first kid eats grapes, the second kid eats grapes, and so on. These must sum up to 2009. Hence , or . Factoring, we get . We are only looking for integer solutions, so we can prime factorize 2009 and look for factor pairs. . So and , and , or (k + 1) = 7(2009, 0)(293, 6)(89, 48)n=\boxed{089}$.
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |