Festivus/Christmas Adam Post (Yule Blog Challenge #4)

Table of Contents

1. Introduction

2. Yule Blog Prompt #6: Something Unexpected That I Faced/Learned

3. 2022 Putnam A5

4. Extra Credit and My Old Algebra I Class (as a Youngster)

Introduction

Today is Seinfeld Day -- a Festivus for the rest of us. It is also known as Christmas Adam -- the day before Christmas Eve. On this blog, I often use the word "Adam" to refer to a day before an "Eve" -- in particular, if a holiday is on Sunday, then the Adam is a Friday, the last day of school before the holiday (and hence the day on which the holiday is celebrated in the classroom).

This year, that applies to Christmas Adam as well -- in New York, students regularly attend school on December 23rd, and even here in Southern California, some districts are open on December 23rd when Christmas falls on a Sunday. This includes the district where I subbed from 2018-2020. (On the other hand, no school is open on Christmas Eve, despite what you see in Frosty the Snowman.)

My current school, meanwhile, has been closed for a full week -- the first of a three-week break. In fact, this marks my fourth post of winter break. I've written extensively about my Math I classes so far, so let me devote today's post to Math III.

Yule Blog Prompt #6: Something Unexpected That I Faced/Learned

Every year on the first Saturday in December, the Putnam exam is given at colleges across the country. It's the world's hardest math test -- there are twelve questions and six hours to complete the test. When I took the test as a young college student, my best score was 22 out of 120 -- 22 might not sound that great, but the most common score is usually zero. The top students earn scholarship money -- I was never close to getting in the money, though.

Ever since I got into teaching, I like to present a Putnam problem to my highest-level class -- this year that means Math III -- a few days after the exam is given. Since my students took their Chapter 5 Test from December 5th-7th, it meant that the 8th was Putnam day. I do it in order to inspire my students -- they might think their Chapter 5 Test or final exam to be hard, but they're easy compared to the Putnam.

The "unexpected" part of the Yule Blog prompt here refers to which question I showed my students. I'd expected to show them Question A1, as this should be the easiest question on the Putnam. Here's a link to this question, courtesy the Art of Problem Solving website:

https://artofproblemsolving.com/community/c7h2971706_2022_putnam_a1

Determine all ordered pairs of real numbers  such that the line  intersects the curve  in exactly one point.

On the surface, this question doesn't appear too difficult. It includes an ln function -- my Math III students had just learned about logs, though not e or ln. This caused several problems with the test -- first of all, on the Chapter 5 DeltaMath Test, DeltaMath kept including ln for some students, so I was forced to change the question to extra credit. And on the paper portion of the test, I was able to catch a cheater when he used ln to find the inverse of an exponential function -- since I never taught ln, I know he must have obtained the answer from an unauthorized online source. (This fits today's prompt -- my troubles with ln in Chapter 5 were "unexpected.")

But now look at the answer to this year's Putnam A1, as given by SpecialBeing2017:

- When , every  works.

- When , then there exists  s.t. .
Then any  with  or  will work

- When , only  works

- When , then there exists  s.t. .
Then any  with  will work

In his answer here, the expression 2t^2/(t^2 + 1) is easily recognizable as a derivative -- if f (x) is the original function ln(1 + x^2), then f '(t) = 2t^2/(t^2 + 1). This suggests that Calculus is required to solve this problem -- and thus it wasn't possible to present this question in my Math III class. But what does it say about the Putnam that even the so-called easiest Question A1 requires Calculus?

But even as I was trying to work out A1 on my own (before SpecialBeing posted the answer), I took a peek at the other questions, and saw Question A5 with the names "Alice and Bob." I remembered that there had been an "Alice and Bob" question two years earlier -- the names refer to two players taking turns in a game. Since "Alice and Bob" questions don't require Calculus, I realized that A5 was the problem that I'd like to show my class, not A1.

2022 Putnam A5

https://artofproblemsolving.com/community/c7h2971711_2022_putnam_a5

Alice and Bob play a game on a board consisting of one row of 2022 consecutive squares. They take turns placing tiles that cover two adjacent squares, with Alice going first. By rule, a tile must not cover a square that is already covered by another tile. The game ends when no tile can be placed according to this rule. Alice's goal is to maximize the number of uncovered squares when the game ends; Bob's goal is to minimize it. What is the greatest number of uncovered squares that Alice can ensure at the end of the game, no matter how Bob plays?

The first thing you might notice is that there are 2022 squares -- and the current year is 2022. It's a tradition to have at least one question mention the year (similar to Rebecca Rapoport making the answer equal the date on her calendars). Indeed, I recommend always knowing the prime factorization of the current year before entering the Putnam room, though the factors of 2022 don't help us here. (In case you're wondering for next year, 2023 = 7 * 17^2.)

When I discussed this question in my classes, I simplified the language a little. Instead of "maximize" and "minimize," I told my students that Bob must pay Alice $1 for each uncovered square. Then the final question becomes "What is the most amount of money Alice can ensure at the end of the game, no matter how Bob plays?"

How should we approach this problem? Let's simplify the problem with much fewer squares than 2022:

1. If there is just one square, then no one can play a tile. So Alice wins $1.
2. If there are two squares, then Alice must cover them both with her first tile. So she wins $0.
3. If there are three squares, then Alice must cover two squares with her first tile, and Bob can't play any tile. So Alice wins $1.
4. Suppose there are four squares. It quickly becomes apparently that Alice's best strategy is to place her tile on the two middle squares. If we number the squares 1-4, then Alice will place her tile on Squares 2 and 3. Then Bob can't place any title on Squares 1 or 4, so Alice wins $2.
5. If there are five squares, then no matter where Alice places a title, Bob can place one of his. So Alice wins $1.
6. Suppose there are six squares. Alice knows where to place her first title -- on Squares 2-3, which guarantees that Bob can't place a tile on Square 1. Then Bob can cover two of Squares 4-6, leaving a second square for Alice. So she wins $2.
7. Suppose there are seven squares. Alice places her first tile on Squares 2-3. If Bob were to place his tile on Squares 5-6, that would leave Squares 4 and 7 uncovered for Alice -- and that would be terrible strategy for Bob. Instead, he covers either Squares 4-5 or 6-7 -- whichever pair he covers, Alice must cover the other pair. So she wins $1.
8. Suppose there are eight squares. Alice places her first tile on Squares 2-3 as usual. Then on the remaining five squares 4-8, Bob and Alice place exactly one more tile. So Alice wins $2.
9. Suppose there are nine squares. Alice places her first title on Squares 2-3 as usual. At this point, Bob recognizes a new strategy -- he places his tile on Squares 6-7, leaving two squares, 4-5, empty -- for another tile, eventually. Indeed, Alice must cover either 4-5 or 8-9 -- whichever pair she covers, Bob covers the other pair. So Alice wins $1.
10. Suppose there are ten squares. Both players follow their established strategies -- Alice skips a single square and covers 2-3, while Bob skips two squares and covers 6-7. Once again, there's room for a tile at 4-5 and another to cover two of 8-10. So Alice wins $2.
11. At eleven squares, something different finally happens. Alice places her first tile on Squares 2-3 as usual, and Bob places his first tile on Squares 6-7 as usual. But now there are still four squares from 8-11 uncovered -- Alice sees this and so she covers 9-10. Now Bob can't cover Squares 8 or 11 -- instead, he can only cover Squares 4-5. and that's the last legal move. So Alice wins $3.

At this point, we look for patterns. We see that Alice can earn $2 with four squares, but it takes an extra seven squares for her to win her third dollar. And we can see where the seven squares come from -- one square for Alice to skip, two for her tile, two for Bob to skip, and two for his tile.

This pattern continues. So in order for Alice to win at least $4, there must be eighteen squares, and for her to guarantee a fin, there must be 25 squares.

Now we can finally solve the original problem. We use long division to divide 2022 by seven (no calculators allowed on the Putnam) and we get 288, remainder 6. So that's $2 for the first four squares and $288 for the next 2016 squares. The last two squares are easily covered by Bob. Therefore the total amount of money Alice can guarantee herself is $290 -- and that is the correct answer.

In my classroom, I obviously didn't have a board with 2022 squares, so I used 22 instead (as 22 is still the year, '22). I cut an index card into tiles and checked that each title covered two squares. Then I challenged my top students in each class to take the role of Alice, while I played Bob.

For 22 squares, the maximum Alice can guarantee is $4 (since she needs 25 squares for $5). In first period, my first two Alices won only $2, and only the third Alice won $4. But in fifth period, the first Alice figured out the trick and was able to win $4 right off the bat. After someone wins $4, I then go and down the rows and have each student try out one of the cases from one to eleven squares, until we see the general pattern.

This is a great game for the MTBoS. If you wish to play this game in your own class, it would be great to use dominoes to cover the squares -- but you'll have to create rows of squares of the correct size so that a domino covers exactly two squares. If our mini-version has 22 squares, then that would require eleven dominoes for each pair of students -- so that's nearly 200 for a full-sized class.

Of course, none of us will see the inside of a classroom again until 2023. So we might consider using 23 as the size of our game. But 23 is a tough case -- Alice can't guarantee $5 until 25 squares, and since an odd number of squares are uncovered at the end of an odd-sized game, the most she can win is $3.

This is also an interesting game to write as a computer program. We might have the user play Alice and the computer represent Bob. The tricky thing is that a computer Bob should play perfect strategy. It's easy to figure out and code the correct strategy for Bob if we assume that Alice is perfect as well -- if Alice starts 2-3, Bob goes 6-7, the computer follows an Alice 9-10 with a Bob 13-14, and so on.

But what if Alice goes backwards -- she opens a 22-square game with 20-21, for example? Then a computer Bob should figure out to go backwards as well and cover 16-17. And if Alice follows a suboptimal strategy -- say by covering the middle squares 11-12 -- then Bob should be able to take advantage and figure out how to pay Alice less than the $4 maximum. But it's difficult to figure out how to code these wrong strategies in.

By the way, 2020 Putnam B2 was the old Alice and Bob problem. It also involves a row of squares (or holes), except that this time each peg covers only a single hole, and instead of placing the pegs, each player takes turns moving them, with the last player to make a legal move as the winner. I wrote about this game in my February 2021 posts (as the Putnam was delayed two months due to COVID).

Extra Credit and My Old Algebra I Class (as a Youngster)

As I wrote in my Thanksgiving break posts, I was expecting several students to ask for extra credit as the semester came to a close. I tried to run it the same way I did at my old school last year -- each student must tell me how he or she learned math, and then I set up some sort of assignment in order to prove it.

That worked well at my tiny magnet school, but it was tricky in full-sized classes. Most assignments are on DeltaMath, so it's logical for the extra credit to be on that website as well. But DeltaMath doesn't allow me to assign work to a single student -- instead, I must assign it to an entire period. Then other students in the class started doing the extra credit, and I quickly lost track of who was doing it.

I like to think back to the Algebra I class that I took as a young seventh grade student. My teacher (yes, the same one who sent me independent work in second grade) had an interesting grading system. I wish to describe it here, including her extra credit system.

First of all, each chapter had a quiz and a test. We covered about three chapters per quarter (as our school didn't have semester grades, only quarter grades), so that's six assessments. The grades for each assessment would be averaged, and that would be our quarter grade.

Let's try an example here. Suppose these are the six assessment grades:

B, B, B-, B+, C+, B-

To average these scores, each B counts as 85 (the midpoint of the B range), the B- as 80.5 (the midpoint of the B range, which went from 80-81 for this teacher), the B+ as 89.5 (midpoint of the 89-90 range), and the C+ as 79.5. Then we calculate:

(85 + 85 + 80.5 + 89.5 + 79.5 + 80.5)/6 = 500/6 = 83.333

which is in the B range. But, as I explained earlier, homework must be factored in -- if at least 75% of the homework is complete, then it becomes a B+, otherwise it's a B-.

You might wonder, what if the assessments average to an A, and the kid does homework as well? Then the final grade is an A+ -- which is a valid grade at our school. For our teacher, the A+ was 97.5-100, so that the midpoint of the A+ range was 98.75 and the midpoint of the A range was 94.25:

B+, D+, D-, D, C, A

The average works out to be:

(89.5 + 69.5 + 60.5 + 65 + 75 + 94.25)/6 = 75.625

which is in the C range. If enough homework is done, it's a C+, and if not, it's a C-.

The fact that A+ started at 97.5 rather than 99 (like the B+ at 89 and the C+ at 79) had a slight effect on the grades. It meant that three A+'s and an A averaged to 97.625, an A+ (whereas three B+'s and a B averaged to 88.375, a B), so it's easier to get an A+. But it also meant that it's harder for an A (or even an A+) to raise a grade lower than A. An A and a B averaged to 89.625, a B+ (but a B and a C averaged to 80.0, a B-). Likewise, an A+ and a B- averaged to a B+ (but a B+ and a C- averaged to a B-).

D-, F, B-, A-, D, B+

Here's the most controversial part of the system -- any F counted as 55. So the average is:

(60.5 + 55 + 80.5 + 90.5 + 65 + 89.5)/6 = 73.5

which is in the C range. Again, it could be C+ or C- depending on the homework. On the other hand, suppose we counted F's as their actual score instead of 55. If that F were really 35%, then the average would be 70.167 (a C-), if it were 30% the average would be 69.333 (a D+), and if it were 5% the average would be 65.167 (a D).

Counting any F as 55 reminds us of the 0=50 system that traditionalists complain about. If I recall correctly, my teacher had a score called "zero" that was separate from an F. So a 0 wouldn't turn into a 55, but getting just one question right (a score of 5%, if there were 20 questions) would be a 55. If we assume that the F above is really a zero, then the average becomes 64.333 (a D).

Also, I believe that since there was no F+ grade, a F assessment average with 75% homework would still be an F. Otherwise, a student could get 10% on every quiz, do 75% homework, and get a D- (while a kid with 60% on every quiz and 70% homework done would get an F).

Now let's get to the extra credit. The teacher offered two extra credit assignments per quarter. One of the assignments would raise one of the assessments by one letter grade, while the other one would let you change any assessment grade to an A+. Notice that it doesn't really matter which assessment you raise by one letter (though due to the 97.5 anomaly, you should choose one that's B- or lower), but of course you want to change your lowest grade to an A+. (Also, if I recall correctly, a zero can't be changed to an A+ 

Let's recalculate our grades for the three students from earlier, assuming that all do both extra credits:

Student 1: B, B, B-, B+, C+, B-

This student chooses to raise one of the B-'s to A- and the C+ to A+:

B, B, A-, B+, A+, B-

The average works out to be 88.208, which is still a B (since it's shy of 89). So unfortunately, the extra credit isn't enough to help this student, as their grades are already fairly high.

Student 2: B+, D+, D-, D, C, A

This student chooses to raise the D to C and the D- to A+:

B+, D+, A+, C, C, A

The average works out to be 83.667, which is now a B (originally a C before the extra credit).

Student 3: D-, F, B-, A-, D, B+

This student chooses to raise the D- to C- and the F (assuming it's not a zero) to A+:

C-, A+, B-, A-, D, B+

The average works out to be 82.458, which is now a B. We see that, unless a student already has fairly high grades, the extra credit is enough to raise their grade by as much as one letter.

Now suppose these were my own Math III students, and I decided to use the same system -- namely have two extra credits, one that raises an assessment by one letter and one that changes it to A+. How would that affect my students?

Let's say Student 3 had a grade of 73.5 before the extra credit. In my class, there were 1000 points (so 73.5% is 735 points), with each quiz/test worth 50 points. Then one letter grade would be 10% of this, or five points, so the one letter grade extra credit would raise the grade to 740 points or 74%.

As for the A+ extra credit, let's say that F was originally 27/50 (or 54%, close to the 55 we assumed earlier), and we defined A+ as 100% (or 50/50). Then that's an extra 23 points. Adding this to 740 points gives us 763 points, or 76.3% -- which is still a C. All that extra credit did nothing to the grade!

If Student 2 did this extra credit instead of 3, it might be enough to raise the grade to C+ (especially since C+ in our district starts at 77%, not 79%), but nowhere near the B it did at the school I attended.

The problem, of course, is obvious. At my old school, doing or not doing homework only affected the grade only by a plus or a minus, but at my current school, it's 20% of the grade. So doing homework at my old school was a C+ vs. C- matter, now it's a B vs. D matter. And the classwork was yet another 20% of the grade -- there was no classwork grade at my old school. Making homework and classwork be worth more meant that quizzes and tests are worth less -- and hence extra credit assignments applied to the quiz and test scores are also worth less.

In fact, there's only one test for which adding any extra credit is worth it -- the midterm. (The school I attended as a young student didn't even have finals, much less midterms.) The midterm and final are together worth 20% of the grade -- but because grades are weighted automatically in Aeries, the midterm by itself is worth 20% until the final is given.

And in Math III, our midterm was the district Benchmarks -- a test on which many students struggled, and so many grades dropped. This is why many students were demanding extra credit (and complaining that success on the Chapter 5 Test didn't raise their grades very much -- the 20% midterm was holding down their grades).

Thus this might be my solution to the extra credit problem -- have a single extra credit assignment on DeltaMath, and if they do well on it, raise the midterm grade to 100%. (The only worry is that a 100% midterm grade might artificially inflate their grades, and then they'll be disappointed when they score less than 100% on the final.)

Conclusion

This concludes my post. I hope you have a wonderful Christmas holiday -- or at least a semi-normal Christmas (after the last two were ruined by the pandemic, which is unfortunately still raging).