Solutions Page for Logic Puzzle #4: Professor Inscrutable (February
2009)
The winner of the fourth logic puzzle contest is:
Alexa Ashworth '09!
The Puzzler was elated to receive so many correct submissions!
Click to go to:
Other correct answers with excellent explanations were submitted by:
Tom Irvin '09, Tongxin Lu '11, and Chris Smith '11
and
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
and
Dick Bedient (Mathematics), John Hind (Physical Education), Seth Hussey
(Physical Education), Pat Marino (Residential Life), and Jim Schreve
(Physics)
The next puzzle will appear on March 6.
Puzzle #4: Professor Inscrutable (February
2009)
Professor Inscrutable had three students in his contemporary meta-metaphysics
course: Amie, Matthew, and Sven. Each student had to write the same
number of papers for the course. Professor Inscrutable graded on a curve,
such that the best of the three papers got p points, the second-best
got q points, and the third-best got r points, where p, q, and r are
distinct integers such that p > q > r > 0. There were no ties,
and all students handed in all the required papers.
At the end of the term, Amie had earned 22 points, and Matthew and Sven
each had 9 points. Matthew had the best grade on the first paper.
Questions
1. How many papers were there?
2. What are the values of p, q, and r?
3. Who had the second-best grade on the second paper?
Solutions must contain explanations!
The Puzzler's solution:
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.
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