This past weekend was the first Saturday in December. It's been my tradition on this blog to write about the annual William Lowell Putnam exam -- the world's hardest math test. This is what I wrote last year about the Putnam:
The first Saturday in December is the day of the annual William Lowell Putnam competition. It is a math contest for college students. Twelve years ago, a poster named John called it the "hardest math test in the world."
Every year, I like to inspire the students by showing them the first -- and usually easiest -- question, namely A1, from this year's Putnam exam. Just like last year, I will post the easiest Putnam question in my Tuesday post [which, of course, is today's post].
So let's look at this year's A1, which as usual I get from the Art of Problem Solving Website:
A1. Find the smallest positive integer j such that for every polynomial p(x) with integer coefficients and for every integer k, the jth derivative of p(x) at k is divisible by 2016.
At first, this question sounded a bit weird to me. Why should the derivative of every polynomial be divisible by 2016? (Oh, and that's another Putnam tradition -- to include the year in a question, just as Theoni Pappas gives the date as every answer.)
But the clue is that these are all integers. Here's something I realized -- the second derivative of every polynomial is always even. Here's why: we all know the power rule, where the derivative of the polynomial x^i is ix^(i - 1), so the second derivative is i(i - 1)x^(i - 2). Now no matter what integer i is, either i or i - 1 is even. So i(i - 1) must be even as well. This goes for the whole polynomial -- in the second derivative, every term is being multiplied by something even. Therefore the second derivative of any polynomial is even.
Likewise, we can show that the third derivative is always a multiple of 3 -- indeed, it must always be a multiple of 2 as well (hence a multiple of 6). So it's likely that if we take enough derivatives, we can even make the derivative a multiple of 2016.
I factor 2016 as 2^5 * 3^2 * 7. We need seven derivatives to guarantee a multiple of seven, as one of i, i - 1, ..., i - 6 must be a multiple of seven. If we're lucky, there are enough factors of 2 and 3 around to give us 2016. Let's see -- i to i - 6 must yield at least two multiples of three, so all we need now are the five factors of two. The worst-case scenario is if the middle number i - 3 is a multiple of 4 (but not of 8). That's only two factors of two, and i - 1 and i - 5 provide us two more factors of two, which leaves us short by one factor of two. That's so close, yet so far -- the seventh derivative must be divisible by 1008, but not necessarily 2016.
So we need one more derivative. The correct answer to A1 is eight. The above link gives a complete proof by Dr. Kent Merryfield, a mathematician from right here in Southern California.
I don't know how picky the graders will be with this; I'm going for overkill in the writeup.
The problem is that it's not enough to show that 8 works, we must also prove that no number lower than 8 works. The Putnam graders are notoriously tough -- an almost-correct answer is seemingly just as likely to earn 1 (out of 10) as a 9.
This question may seem difficult, but it's actually easy -- by Putnam standards, that is. In fact, here's what Merryfield has to say about the overall test:
I think that the median won’t be zero this year; I think that because of question A1. (Although, even with A1, there are questions about just how picky the graders are going to be whether they’re going to require people to say certain things.)
And some readers may be wondering here, in what world is a Calculus problem (derivatives) an "easy" problem? Then again, it's really a Number Theory problem masquerading as Calculus.
Anyway, today in class I tell my eighth graders about the Putnam. I decide not to give them A1 as an example, since it mentions derivatives. Instead, Question A4 sounds more accessible:
This one requires a visual, so I won't restate the problem here -- you'll have to click on the link to see the shapes anyway. Basically, we have a (2m - 1) x (2n - 1) grid and we have to fill it with tiles of two different shapes, and we want to know the minimum number of tiles requires.
The question stipulates that m and n are each at least four, so the smallest grid is 7 x 7. In class, I only mention the 7 x 7 case in order to avoid variables. Actually, as it turns out, the general case can be proved using induction with the recursive step being almost trivial. Most of the work is in proving the base case of 7 x 7 anyway!
The correct answer, by the way, is mn. For the base case, this means 16 tiles. This is counterintuitive since the two tiles have three and four squares each, so one would think that to get the minimum, we'd want as many of the larger tiles as possible. But the solution uses only one of the four-square tiles -- all the rest are the three-square tiles.
One thing I tell my students about the Putnam is that all answers must be proved. Notice that it's possible to come up with an answer of mn by pure guessing -- since the larger tiles have four squares, we expect the answer to be about one-fourth the total number of squares, and (2m - 1)(2n - 1)/4 is approximately (2m)(2n)/4 or mn. But if I were to write that, Putnam would award me zero points.
My students wonder why anyone would choose to take the Putnam anyway. Well, I tell them that the top prize is a full scholarship to Harvard for grad school, and the next four students win cash, so that's always an incentive.
Meanwhile, on Friday, I received a comment on the blog. It was a post from last year -- the summer between my two years of subbing. The subject was spherical geometry. In particular, a college student was researching why AAS fails in that geometry, and came up on this blog, where she learned about the counterexample.
You may notice that so far, I've written all about Putnam and spherical geometry, and nothing about middle school math. OK, so I've been putting it off -- I've graded those tests on volumes of cylinders, cones, and spheres, and while there were a few bright spots (two 90's, with the new girl who had already learned this getting 100%), the results were mostly dismal. The students made several expected mistakes -- plugging in diameter for radius and forgetting to divide by three for cones. One student was completely mixed up -- not only did he fail to divide by three for cones, but he did divide by three for cylinders! And there was another student who switched the radius with the height -- but in her defense, she had been absent for most of the week.
So ironically, it's possible that I taught more people spherical geometry on Friday than ordinary Euclidean geometry! And it's definitely true that I taught more people about triangles on the sphere than the volume of a sphere. So many students wrote the wrong formula for a sphere that I believe I may have written down an incorrect formula on the board -- simply multiplying the formula for a cylinder by 4/3 instead of the correct formula. (Otherwise, where did the students get 4/3 from?)
I already wrote on Friday how I could have taught the lesson better (besides giving the correct sphere volume formula, that is). But there's a way I could have improved the test itself. You see, I avoided multiple choice on this test for two reasons. The first is that Illinois State did not provide multiple choice for any questions on this standard. The other is the CASIO calculator problem -- the students are taught to use 3.14 for pi, yet some (but not all) of the calculators cover this to an (improper) fraction -- and one of our calculators even gives a mixed number. So I'd have to write three correct answers -- a decimal, an improper fraction, an a mixed number!
Now consider the following problem:
Find the volume of a cylinder with radius 10 and height 1.
We work it out as follows:
V = pi r^2 h
= 314 cubic units
And notice that the answer shows up as 314 on all three calculators! Of course, this works because pi is approximated as 3.14 = 314/100, so the 10^2 term cancels the 100 in the denominator.
We could almost have a radius of 5 instead of 10. Notice that 314/100 = 157/50, and r^2 is 25, so we're only missing a factor of two. (Gee, it seems as if we're still working on Putnam A1!) So all we need is for the height to be even and the answer is a whole number on all calculators.
If we make the height 6, this allows us to change "cylinder" to "cone" and still have a whole number as the volume. Indeed, this allows us to write four multiple-choice questions:
1. Find the volume of a cylinder with radius 10 and height 6.
2. Find the volume of a cylinder with diameter 10 and height 6.
3. Find the volume of a cone with radius 10 and height 6.
4. Find the volume of a cone with diameter 10 and height 6.
And we can use the same choices for all four questions:
a. 157 cubic units
b. 471 cubic units
c. 628 cubic units
d. 1854 cubic units
For the sphere, we can't use a radius of 5 or 10. This is because of the 4/3 -- the extra 4 helps us in that we may have an odd radius, but then we can't cancel the extra 3. So 15 is the smallest sphere radius that we can use.
I'm angry at myself because both "eighth grade" and "geometry" are my focus -- especially as I reflect here on the blog. This should be one of my best lessons -- and I feel that I should have been clever enough to come up with these questions on my own. As soon as I realized that there was a problem with the CASIO calculators, I should have abandoned Illinois State and ask these problems.
In fact, I think I did a better job a month ago on the Laws of Exponents than on volume -- even though geometry, again, is my favorite topic to teach!
Today we begin the next STEM project. Learning Module 7 of the Illinois State text is called "Shapes, Angles, and Structures." This project begins by having the students look at some pictures of skylines (that I found on the web), and then they go outside to hunt for various geometrical shapes in real life.
Meanwhile, the seventh graders start a project on cutting straws and forming triangles. In some ways, this reminds me of something mentioned by Dr. David Joyce, a critic of modern Geometry texts:
The text again shows contempt for logic in the section on triangle inequalities. In a silly "work together" students try to form triangles out of various length straws. Rather than try to figure out the relations between the sides of a triangle for themselves, they're led by the nose to "conjecture about the sum of the lengths of two sides of a triangle compared to the length of the third side. Think and discuss. Triangle Inequality Theorem. Your observations from the Work Together suggest the following theorem," and the statement of the theorem follows. Honesty out the window. Can any student armed with this book prove this theorem? Can a teacher? Can the authors?
A key difference is that in this project, students are to make other conjectures about the triangle, such as the angle sum -- not just Triangle Inequality.
The sixth graders begin a project on animal movements. This one is awkward, because Illinois State asks students to measure an actual animal -- where are they supposed to find one at school? (Though ironically, I see a dog walk on the parking lot after school today.) It ends up turning into a research project -- but even then, students can easily find the height and mass of a animal, but not the length of its legs or strides, as Illinois State asks.
Music break is interesting today. First, we found out that our Christmas concert is cancelled because the LAUSD elementary school with whom we share a campus has booked the auditorium every day until our students' last day (due to that school's own holiday program). So I decide to make it up to them by featuring parodies of Christmas songs during music break.
For Jingle Bells, I give "Jingle Algebra," and for O Christmas Tree, I provide "Oh, Number Pi." Here are links to both songs:
(Notice that the author of this website appears to be a Canadian elementary school teacher. This link is from six years ago -- so far she's made only two posts in 2016.)
But at the end of the day during IXL time, I have some extra time with my sixth graders. So I decide to go to YouTube and start playing some Square One TV songs. In addition to the theme song, I end up showing them the following videos:
2. Neighborhood Super Spy
3. Change Your Point of View
6. Triangle Song
7. Perpendicular Lines
9. Round It Off
10. Graph of Love
Here's a link again to Barry Carter's website:
Notice that only songs 1, 2, 4, and 7 have lyrics listed there. Carter does have a separate link for Song 5, which is "Timekeeper":
(I mentioned "Timekeeper" back in my October 5th post.)
The "Day in the Life" poster with a monthly posting day of the sixth is Dawneen, AKA Ms. Z:
Ms. Z, if you recall, is a South Carolina middle school teacher, so I definitely want to spend some time looking at her website. She hasn't made her December 6th post, so let's look at some of her recent posts, including her November 6th post:
November 6th was a Sunday, so she wrote mostly about Friday and Saturday. She begins:
First period went over distance between points and reflecting points over the x- or y-axis on a coordinate plane. Students continued using their graphs from this post activity. We also reviewed how to order rational numbers on a number line.
This sounds like an eighth grade lesson -- except it's actually for her honors sixth grade class!
Second, third, and sixth periods (all seventh grade math classes) were uneventful as we continued a lesson that was started on Thursday talking about simplifying algebraic expressions.
OK, this is logical for seventh grade. It appears on our seventh grade Benchmarks -- but it seems as if we can never reach the actual lesson!
After school, our boys cross country team had a pasta party in prep for the next day's state meet. They also watched the movie, "The 4-Minute Mile."
Indeed, Ms. Z writes:
This is my fifth year as a cross country coach along with my co-coach. Every year we get better. This year we finally had a runner place in the Top 15 for our division. He placed 7th overall with a time of 19:13. We were ranked 12th as a team last year; this year we ranked 7th as a team.
Let's end today's post on a high note. For Ms. Z's weekend reflection question, she replies:
My cross country athletes. As described above, I've known many of these young men since they were sixth and seventh graders. They are now seniors and juniors in high school. It's been an amazing journey to watch. For three of them, it was their last cross country meet with our school. One is a senior and will be out for track and field (he's our star 400-meter hurdler - placing third at state last year). But for the other two, they'll be going to a different school next year. And I've contemplated not coming back next year as well, so this just might be my last time training with them.