The winner of the fourth logic puzzle contest is Alexa Ashworth '09!
Other correct answers with excellent explanations were submitted by:
Tom Irvin '09, Tongxin Lu '11, and Chris Smith '11
Peter Rabinowitz (Comparative Literature)
Still other correct answers were submitted by:
Ari Abrams-Kudan '10, David Brown '10, Linnaea Chapman '10, Brendan Conway '09, Alex Gross '11, Kyle Hung '10, Alex Kim '12, Michael Kranz '10, Katheryn Kroleski '09, George Niden '09, Alex Pardy '11, Lindsay Shankman '12, William Thoreson-Green '12, Ian Thresher '12, Christopher Weitzman '11, Erin West '11, and Tom Williams '11
Dick Bedient (Mathematics), John Hind (Physical Education), Seth Hussey (Physical Education), Pat Marino (Residential Life), and Jim Schreve (Physics)
The total number of points awarded was 40. So, n(p + q + r) = 40, where n is the number of papers. p + q + r has to be at least 1 + 2 + 3 = 6, so n must be 1, 2, 4, or 5.
Since Matthew had the best grade on the first paper, we know that p must be less than 9. If there were only one or two papers, then Amie would not be able to collect her 22 points. So, n must be 4 or 5.
Assume n is 4. Then, p + q + r = 10. Thus, p must be at most 7. Since Amie earned 22 points in four papers, p must be at least 6. If p were 7, Matthew could not get 9 points, even if he received the fewest points on each of the other three papers. Thus, p would have to be 6, r would have to be 1, and q would have to be 3. But then there would be no way for Amie to collect 22 points.
So, n must be 5 and p + q + r = 8. Since Amie earned 22 points on five papers, p must be at least 5, which entails that p=5, q=2, and r=1. Thus, Matthew must have gotten the lowest grade on all papers except the first. The only way for Amie to earn 22 points would be for her to get the most points on every paper other than the first, and the second-best grade on the first assignment. Thus, Sven must have had the second-best grade on all papers other than the first, including the second.