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 .
Solution 1
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 minimized. (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 , .
Solution 2
If the first child ate grapes, then the maximum number of grapes eaten by all the children together is . Similarly, if the first child ate grapes, the maximum total number of grapes eaten is .
For the value is less than . Hence must be at least . For , the maximum possible sum is . And we can easily see that , hence grapes can indeed be achieved for by dropping the last four children.
Hence we found a solution with and kids, and we also showed that no smaller solution exists. Therefore the answer is .
Solution 3 (similar to solution 1)
If the winner ate n grapes, then 2nd place ate grapes, 3rd place ate grapes, 4th place ate grapes, and so on. Our sum can be written as . If there are x places, we can express this sum as , as there are occurrences of n, and is equal to . This can be factored as . Our factor pairs are (1,2009), (7,287), and (41,49). To minimize n we take (41,49). If , then and . (Note we would have come upon the same result had we used .) ~MC413551
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 |
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