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# Putnam 2013

No it means that more than half of the people who take it often get zero. Not everybody. If the mean were zero

posted by Justinian at 10:11 PM on April 27 [10 favorites]

No, it means that often, more than half of the scores are zero.

The median of a finite set of numbers is calculated by rank-ordering the numbers and taking either the single number farthest from both ends or taking the arithmetic mean of the two numbers farthest from the ends.

Example #1: The median of {4, 1, 11, 56, 12, 8, 14} is 11, since in the ordered set <1, 4, 8, 11, 12, 14, 56>, 11 is right in the middle.

Example #2: The median of {5, 18, 7, 6, 23, 2} is 6.5, since in the ordered set <2, 5, 6, 7, 18, 23> there are two numbers in the middle -- 6 and 7 -- whose mean is (6 + 7) / 2 = 6.5.

In order to get a median equal to zero, then, it must be that if you rank-order all of the scores, then the exact middle score is zero in case the total number of scores is odd or both of the middle scores are zero in case the total number of scores is even.

So, minimally, it often happens that more than half the test takers score zero.

posted by Jonathan Livengood at 10:12 PM on April 27 [8 favorites]

One of the wonders of the modern Information Age is that it presents me, each and every day, with a new way to feel inferior. What a time to be alive.

posted by Snarl Furillo at 10:34 PM on April 27 [4 favorites]

Yeah, the problems I've looked at seem really well crafted. You go, "hey, that doesn't seem so bad, you just...wait. Hmm. Well, if you....nope. Huh. Maybe if I...wait, for

posted by jcreigh at 10:36 PM on April 27 [4 favorites]

Metafilter makes me feel like an idiot almost every day. So don't feel so bad.

posted by mrhappy at 10:57 PM on April 27 [2 favorites]

Best line in the whole article.

posted by cthuljew at 10:58 PM on April 27 [3 favorites]

I think I know how to solve that problem, though I'm a physicist so my proof may not be rigourus enough and I may be falling a trap.

First of all you set up the geometry of shape you're making. I'm fairly sure that the best way of doing this is to place a cone on the top and the bottom of the cylinder and give all of the circular faces an equal radius, so it's shaped like (=) but with the cones and cylinder having the same width. The arument I can give as to why that's the optimal arrangement is that every surface which can be on the interior of the shape is on the interior: the flat circles are completely covering each other, and the convex curves which cannot be covered by flat or convex surfaces are on the exterior.

In order to prove that this arrangement is optimal, you could draw the shape with an arbitary height, and set up an equation for the total volume and surface area with the radius of the cylinder as R and the radius of the cone as R + r (i.e different by some amount). By taking the ratio of volume to area, it looks like the height factors would cancel. Then you could set R = 1, and find the value of r which would maximise the ratio (which I strongly suspect would be zero). I think it's ok not to consider any cases in which the flat surfaces aren't touching at all, since they are obviously less efficient than the cases in which the flat surfaces are touching.

Having found r in terms of R (and I guess r = 0*R), the final thing to do would be to find the optimum ratio of the radius to the height. To do this I'd set the height equal to 1, and write an expression for the volume / surface area as a function of R. Then, finding the minimum of this expression would give R as a proportion of the height.

posted by Ned G at 4:50 AM on April 28

I'm not sure what you have in mind for that can buoy question, but the contest was a lot different in its early days — nowadays it's all about ingenuity, but many of the 1938 contest questions are routine. (Like, they wouldn't be out of place on an exam in a first- or second-year calculus or linear algebra course.)

posted by stebulus at 6:20 AM on April 28

I always justify reinventing the wheel by saying that I probably understand the idea better for having reinvented it instead of looking it up somewhere. However that may just be a scheme to reduce cognitive dissonance.

posted by madcaptenor at 8:34 AM on April 28 [1 favorite]

My husband is a math professor at a small college, and the Putnam is one of the big events of the year in their department. They have a year-long seminar class devoted to learning how to work on competition problems of this kind; the students really get into the spirit of the thing and their team usually does much better than the size/prestige of the college might suggest. (He says "this year's MIT team [which this article is about] was a

To answer your question, every year, he idly works on the exam while the students are taking it, so they can talk over their approaches during lunch and afterward. He says that giving it half-attention during the exam, he can always solve at least four problems, with varying degrees of progress on the rest. Then over the subsequent week or so, he can usually solve around nine; can't always get more than that, but it varies how much time he has to devote to it.

The Putnam is

posted by LobsterMitten at 11:16 AM on April 28 [1 favorite]

This is, IMO, patently false. There was at least one question this year which was intractable without Fourier analysis.

posted by TypographicalError at 11:50 AM on April 28

I'm not sure how typical that is. At my school, yes, the ones who actually ran the Putnam could readily solve several problems on the day, but they had plenty of experience with this kind of math. (One of them was a former Putnam Fellow, actually.) I doubt that a generic math professor — with little or no experience in competitions, with a research specialty outside the Putnam's ambit — would do so well.

posted by stebulus at 6:45 PM on April 28 [1 favorite]

But orthogonalising a function space with respect to a countable collection of basis functions which respect the underlying symmetries of the space is something that would occur to any reasonably bright sophomore....

(Yeah, they do use a bit of domain knowledge in the Putnam, but I think not a whole lot.)

posted by kaibutsu at 6:42 AM on April 29 [1 favorite]

I can't tell if this is in jest or not, but given the standardish US math undergrad curriculum, said sophomore would probably but not certainly know the meanings of all the individual words in that sentence. Linear Algebra is often not seen at all until the sophomore year, and then typically in a very computationally intensive fashion that is light on concepts.

Now if you want to say any reasonably bright

posted by PMdixon at 2:18 PM on April 29

Post

# Putnam 2013

April 27, 2014 8:58 PM Subscribe

“I wanted to use the intermediate value theorem but it just wasn’t happening.” MIT undergrad Zach Wener-Fligner reports from this year's William Lowell Putnam Mathematical Competition, the nation's premier math contest for college students, a test so hard that the median score is often zero.

I took the Putnam twice, a long time ago, and also managed a nonzero score (both times, but the first time only just barely). I wasn't a walk-in, either, I was on the team the second time. That test is hard in a way you can't even explain to your average "I took Calculus in college" sort of people, but when you tell someone you took a 6 hour math test with 12 questions on it and a median score of zero, and then you tell them you did it again a year later, they look at you funny. You have to actually like doing math to subject yourself to that kind of thing.

posted by axiom at 9:11 PM on April 27 [1 favorite]

posted by axiom at 9:11 PM on April 27 [1 favorite]

So that must mean everybody gets zero. Right? (Often.)

posted by carping demon at 9:41 PM on April 27

posted by carping demon at 9:41 PM on April 27

Just as a point of reference, when I was in my math doctoral program (not at a top program by any means, of course), the other grad students and I would attempt these problems for some amount of fun. I don't know that we ever really solved more than 4-5 out of 12 in a year's packet, although we were mostly trying for kicks and none of us had competitive math backgrounds. It really is a whole 'nother kettle of fish.

posted by TypographicalError at 9:43 PM on April 27

posted by TypographicalError at 9:43 PM on April 27

Also, in case anyone wanted to see the full suite of problems from 2013, here it is. You can also click on the problem numbers to see a discussion thread for the AoPS forums where people discuss their solutions.

posted by TypographicalError at 9:47 PM on April 27 [5 favorites]

posted by TypographicalError at 9:47 PM on April 27 [5 favorites]

*So that must mean everybody gets zero. Right? (Often.)*

No it means that more than half of the people who take it often get zero. Not everybody. If the mean were zero

*then*it would mean that everybody gets zero. Median != mean.

posted by Justinian at 10:11 PM on April 27 [10 favorites]

Aww, Putnams were fun! (For a definition of fun that I guess only really makes sense to a math major.) Especially the frantic sessions at the blackboard immediately afterwards, where those with a solution would try to explain it and others would try to poke holes in it, and sometimes you would realize how close you had been to an approach that would've worked.

There's a more detailed archive for recent years, with problems and solutions, here.

posted by parudox at 10:11 PM on April 27 [4 favorites]

There's a more detailed archive for recent years, with problems and solutions, here.

posted by parudox at 10:11 PM on April 27 [4 favorites]

*So that must mean everybody gets zero. Right? (Often.)*

No, it means that often, more than half of the scores are zero.

The median of a finite set of numbers is calculated by rank-ordering the numbers and taking either the single number farthest from both ends or taking the arithmetic mean of the two numbers farthest from the ends.

Example #1: The median of {4, 1, 11, 56, 12, 8, 14} is 11, since in the ordered set <1, 4, 8, 11, 12, 14, 56>, 11 is right in the middle.

Example #2: The median of {5, 18, 7, 6, 23, 2} is 6.5, since in the ordered set <2, 5, 6, 7, 18, 23> there are two numbers in the middle -- 6 and 7 -- whose mean is (6 + 7) / 2 = 6.5.

In order to get a median equal to zero, then, it must be that if you rank-order all of the scores, then the exact middle score is zero in case the total number of scores is odd or both of the middle scores are zero in case the total number of scores is even.

So, minimally, it often happens that more than half the test takers score zero.

posted by Jonathan Livengood at 10:12 PM on April 27 [8 favorites]

Do you get points subtracted if you don't fill in all the little circles?

posted by Joe in Australia at 10:12 PM on April 27

posted by Joe in Australia at 10:12 PM on April 27

Cool, Jonathan Livengood, I never have understood that ordered set thing 'till now, though I heard it go by. Thank you.

posted by carping demon at 10:20 PM on April 27

posted by carping demon at 10:20 PM on April 27

It is mean as hell to not give the answers in that pdf paper. I am curious about the answer to 1938's #2, because that seems like it's obvious and there's no way it's obvious.

posted by kafziel at 10:24 PM on April 27

posted by kafziel at 10:24 PM on April 27

*“He’s a legendary competitor,” says Ben Gunby.“I feel like I would have to be an encyclopedia to remember all of his accomplishments.” Gunby is no slouch himself. He was on the United States IMO team with O’Dorney in Kazakhstan and in the Netherlands the following year, when Gunby placed 14th in the world. He would have likely competed the next year, too, but he left high school after his junior year to attend MIT.*

There is Mitchell Lee, Gunby’s college roommate last year, who was on the U.S. team in the Netherlands and the following year in Argentina–like Gunby, he left high school a year early after MIT accepted him, but spent a year doing math research rather than matriculating immediately.

And then there is Zipei Nei, another MIT junior, a Shanghai native who competed exactly once in the Olympiad, on the venerable Chinese team–“When China doesn’t win you figure there’s something fishy about the way the problems were written,” O’Dorney says. That one time was 2010 in Kazakhstan, the year that O’Dorney scored a remarkable 39 out of a possible 42 points for second in the world. Nei scored a perfect 42 out of 42.

There is Mitchell Lee, Gunby’s college roommate last year, who was on the U.S. team in the Netherlands and the following year in Argentina–like Gunby, he left high school a year early after MIT accepted him, but spent a year doing math research rather than matriculating immediately.

And then there is Zipei Nei, another MIT junior, a Shanghai native who competed exactly once in the Olympiad, on the venerable Chinese team–“When China doesn’t win you figure there’s something fishy about the way the problems were written,” O’Dorney says. That one time was 2010 in Kazakhstan, the year that O’Dorney scored a remarkable 39 out of a possible 42 points for second in the world. Nei scored a perfect 42 out of 42.

One of the wonders of the modern Information Age is that it presents me, each and every day, with a new way to feel inferior. What a time to be alive.

posted by Snarl Furillo at 10:34 PM on April 27 [4 favorites]

*that seems like it's obvious and there's no way it's obvious.*

Yeah, the problems I've looked at seem really well crafted. You go, "hey, that doesn't seem so bad, you just...wait. Hmm. Well, if you....nope. Huh. Maybe if I...wait, for

*every*integer? I dunno..."

posted by jcreigh at 10:36 PM on April 27 [4 favorites]

*One of the wonders of the modern Information Age is that it presents me, each and every day, with a new way to feel inferior. What a time to be alive.*

Metafilter makes me feel like an idiot almost every day. So don't feel so bad.

posted by mrhappy at 10:57 PM on April 27 [2 favorites]

I saw a little flyer on the bulletin board of the math department the week of the test during my freshman year and thought I might try it just for the experience; I don't remember having to sign up, but I saw my name on some list the day of the test, so maybe I did.

Only about 25 people took the test at my big state school. I had taken calculus in high school and completed the third semester of the calc sequence that fall; I had no idea the exam was any kind of big deal or I wouldn't have taken it.

Most of the problems talked about things I had heard of but couldn't have defined, such as groups, or hadn't heard of, and the ones where I at least knew what was being asked were much harder than I had expected, but they were somehow interesting and I kept plugging away for awhile before giving up. I scored in the low 20s, and was pretty embarrassed to have scored so low, and didn't realize until the TA who'd taught my calc 3 class told me in the hall that the high scorer at my school was in the upper 50s, but I had tied for 2nd and won $50. I bought the Feynman lectures with the money.

But the upshot was that my piddling score raised such expectations among my math profs, and caused them to pay such a ridiculous amount of attention to me and everything I did that I ended up dropping out of the math program at the end of my sophomore year. That was just what I'd come to a big state school in order to avoid. I even got contacts mainly in hopes that they and the physics people wouldn't recognize me without my thick, thick glasses, and they didn't. Whew.

Oh well, that was a long time (and a couple of serious brain injuries) ago.

posted by jamjam at 10:58 PM on April 27 [8 favorites]

Only about 25 people took the test at my big state school. I had taken calculus in high school and completed the third semester of the calc sequence that fall; I had no idea the exam was any kind of big deal or I wouldn't have taken it.

Most of the problems talked about things I had heard of but couldn't have defined, such as groups, or hadn't heard of, and the ones where I at least knew what was being asked were much harder than I had expected, but they were somehow interesting and I kept plugging away for awhile before giving up. I scored in the low 20s, and was pretty embarrassed to have scored so low, and didn't realize until the TA who'd taught my calc 3 class told me in the hall that the high scorer at my school was in the upper 50s, but I had tied for 2nd and won $50. I bought the Feynman lectures with the money.

But the upshot was that my piddling score raised such expectations among my math profs, and caused them to pay such a ridiculous amount of attention to me and everything I did that I ended up dropping out of the math program at the end of my sophomore year. That was just what I'd come to a big state school in order to avoid. I even got contacts mainly in hopes that they and the physics people wouldn't recognize me without my thick, thick glasses, and they didn't. Whew.

Oh well, that was a long time (and a couple of serious brain injuries) ago.

posted by jamjam at 10:58 PM on April 27 [8 favorites]

*“I like my way, because now I can say I used the Pigeonhole Principal on every problem except number two.”*

Best line in the whole article.

posted by cthuljew at 10:58 PM on April 27 [3 favorites]

Ah, the Putnam. I remember doing it my freshman year at UT and being super happy I got a 4. It was one of the higher scores at UT that year. The next year I got a big fat zero though. Oof.

posted by kmz at 11:24 PM on April 27

posted by kmz at 11:24 PM on April 27

What kills me is when I look at the solutions and each problem is dealt with in one neat paragraph.

posted by Quilford at 11:54 PM on April 27 [1 favorite]

posted by Quilford at 11:54 PM on April 27 [1 favorite]

If you gave tenured math profs the test, what would their typical score be?

posted by dontjumplarry at 4:34 AM on April 28

posted by dontjumplarry at 4:34 AM on April 28

I was told the scoring policy was that for a given problem, out of ten possible points, only 0, 1, 2, 8, 9, or 10 can be awarded. So you either just got it started, or you solved it, basically.

posted by PMdixon at 4:35 AM on April 28

posted by PMdixon at 4:35 AM on April 28

*It is mean as hell to not give the answers in that pdf paper. I am curious about the answer to 1938's #2, because that seems like it's obvious and there's no way it's obvious.*

I think I know how to solve that problem, though I'm a physicist so my proof may not be rigourus enough and I may be falling a trap.

First of all you set up the geometry of shape you're making. I'm fairly sure that the best way of doing this is to place a cone on the top and the bottom of the cylinder and give all of the circular faces an equal radius, so it's shaped like (=) but with the cones and cylinder having the same width. The arument I can give as to why that's the optimal arrangement is that every surface which can be on the interior of the shape is on the interior: the flat circles are completely covering each other, and the convex curves which cannot be covered by flat or convex surfaces are on the exterior.

In order to prove that this arrangement is optimal, you could draw the shape with an arbitary height, and set up an equation for the total volume and surface area with the radius of the cylinder as R and the radius of the cone as R + r (i.e different by some amount). By taking the ratio of volume to area, it looks like the height factors would cancel. Then you could set R = 1, and find the value of r which would maximise the ratio (which I strongly suspect would be zero). I think it's ok not to consider any cases in which the flat surfaces aren't touching at all, since they are obviously less efficient than the cases in which the flat surfaces are touching.

Having found r in terms of R (and I guess r = 0*R), the final thing to do would be to find the optimum ratio of the radius to the height. To do this I'd set the height equal to 1, and write an expression for the volume / surface area as a function of R. Then, finding the minimum of this expression would give R as a proportion of the height.

posted by Ned G at 4:50 AM on April 28

I got 1 out of 5 the year I took it for my college. And I got a nosebleed from the stress.

posted by Obscure Reference at 5:16 AM on April 28

posted by Obscure Reference at 5:16 AM on April 28

So far I've read 5 words of the 2013 test before realizing that my brain can't even say "icosahedron". It's not "isocahedron"...it says "icosahedron" dummy. I don't even know what either of those things are, and I had to double check what congruent meant, so this is not going well for me. Also, I would probably get kicked out when I folded my paper into a "isocahedron" and started rolling numbers to try and get the answer.

On the plus sign, I don't have a nosebleed.

posted by This_Will_Be_Good at 6:07 AM on April 28 [2 favorites]

On the plus sign, I don't have a nosebleed.

posted by This_Will_Be_Good at 6:07 AM on April 28 [2 favorites]

*It is mean as hell to not give the answers in that pdf paper. I am curious about the answer to 1938's #2, because that seems like it's obvious and there's no way it's obvious.*

I'm not sure what you have in mind for that can buoy question, but the contest was a lot different in its early days — nowadays it's all about ingenuity, but many of the 1938 contest questions are routine. (Like, they wouldn't be out of place on an exam in a first- or second-year calculus or linear algebra course.)

posted by stebulus at 6:20 AM on April 28

Sigh. I wish I understood math.

Also, math-teacher-but-really-phys-ed-teacher from 7th to 12th grade who I'm not naming simply because I'm not an asshole: you were a giant fucking asshole who utterly ruined math for all of us, year after year. Thanks a bunch.

posted by aramaic at 7:25 AM on April 28 [1 favorite]

Also, math-teacher-but-really-phys-ed-teacher from 7th to 12th grade who I'm not naming simply because I'm not an asshole: you were a giant fucking asshole who utterly ruined math for all of us, year after year. Thanks a bunch.

posted by aramaic at 7:25 AM on April 28 [1 favorite]

Background: I have a degree in mathematics.

When I was in school, the math department had a fun tradition of taking the Putnam test takers out to lunch in between sections. With students who were on the cusp of agreeing to do it, our professors always said, "Hey, you get a free lunch!"

Two years in a row I started the exam and became so despondent and miserable that I sneaked away after the first section and missed the free lunch after all. Because I knew that if I ate the lunch I would feel obligated to finish the exam, and the thought of facing the rest of the exam was so incredibly demoralizing.

Incidentally I would have scored a point my second year, but because I couldn't handle the psychic stress of writing the whole exam, it was never made official.

It's a damn hard test.

posted by telegraph at 7:28 AM on April 28 [1 favorite]

When I was in school, the math department had a fun tradition of taking the Putnam test takers out to lunch in between sections. With students who were on the cusp of agreeing to do it, our professors always said, "Hey, you get a free lunch!"

Two years in a row I started the exam and became so despondent and miserable that I sneaked away after the first section and missed the free lunch after all. Because I knew that if I ate the lunch I would feel obligated to finish the exam, and the thought of facing the rest of the exam was so incredibly demoralizing.

Incidentally I would have scored a point my second year, but because I couldn't handle the psychic stress of writing the whole exam, it was never made official.

It's a damn hard test.

posted by telegraph at 7:28 AM on April 28 [1 favorite]

I took the Putnam once, and got a bare handful of points.

The scoring contributes to the difficulty, and is different from most of the exams one takes in (say) calculus class. First, you have to have a complete correct answer to get any points. Then they start taking off points for mistakes. So there's no partial credit for 'mostly' getting it, or seeing the right idea: you have to get a full solution.

One of the difficulties with this kind of exam is that they need to be 'naively solvable' with a first-year math student's level of background knowledge. It may be that such a first-year student would need (like jamjam) to be extraordinarily clever to succeed with such a minimal background, but there you have it. As a result, the questions tend to have lots of pigeonhole principle, naive number theory, and symmetry arguments. Once you've done a enough 'hard' competition problems, you start to get a sense of how to take an arbitrary problem and cram it into one (or more) of these common problem types. The pigeonhole principle becomes a lens through which you view the world.

I found my background with competition prep useful in early grad school: the analysis class was basically 'here's twenty theorems with some conceptual background; use them in some combination to solve the exam problems.' Competition had taught me to break things into small knowns and combine them in interesting ways on the fly.

This differs greatly from research mathematics, in that in research the tricks available in research are truly multitudinous, and you're not necessarily going to be sure which subset to focus on to build a solution to a problem. Indeed, a lot of my method for research involves inventing new methods, and then doing a bit of background search of the literature to figure out whose ideas I've just reinvented so I can cite them properly. (A most egregious example: A while ago I was prepping for tech job interviews, I realized that 'breadth first search' existed in the world before I started using it a few years ago. oops.) On some level, I'm probably just wasting lots of effort compared to just reading and knowing everything. But it's impossible to read everything, and I get some deeper understanding by reinventing things than I do from reading them right away. The best is to partially reinvent something and then go read someone who's done the actual work to flesh out and formalize the trick....

but I digress.

A really good book is 'The Art and Craft of Problem Solving,' by Paul Zeitz, who was the US coach for the IMO for a bunch of years. I loved it when I was in undergrad!

posted by kaibutsu at 7:58 AM on April 28 [9 favorites]

The scoring contributes to the difficulty, and is different from most of the exams one takes in (say) calculus class. First, you have to have a complete correct answer to get any points. Then they start taking off points for mistakes. So there's no partial credit for 'mostly' getting it, or seeing the right idea: you have to get a full solution.

One of the difficulties with this kind of exam is that they need to be 'naively solvable' with a first-year math student's level of background knowledge. It may be that such a first-year student would need (like jamjam) to be extraordinarily clever to succeed with such a minimal background, but there you have it. As a result, the questions tend to have lots of pigeonhole principle, naive number theory, and symmetry arguments. Once you've done a enough 'hard' competition problems, you start to get a sense of how to take an arbitrary problem and cram it into one (or more) of these common problem types. The pigeonhole principle becomes a lens through which you view the world.

I found my background with competition prep useful in early grad school: the analysis class was basically 'here's twenty theorems with some conceptual background; use them in some combination to solve the exam problems.' Competition had taught me to break things into small knowns and combine them in interesting ways on the fly.

This differs greatly from research mathematics, in that in research the tricks available in research are truly multitudinous, and you're not necessarily going to be sure which subset to focus on to build a solution to a problem. Indeed, a lot of my method for research involves inventing new methods, and then doing a bit of background search of the literature to figure out whose ideas I've just reinvented so I can cite them properly. (A most egregious example: A while ago I was prepping for tech job interviews, I realized that 'breadth first search' existed in the world before I started using it a few years ago. oops.) On some level, I'm probably just wasting lots of effort compared to just reading and knowing everything. But it's impossible to read everything, and I get some deeper understanding by reinventing things than I do from reading them right away. The best is to partially reinvent something and then go read someone who's done the actual work to flesh out and formalize the trick....

but I digress.

A really good book is 'The Art and Craft of Problem Solving,' by Paul Zeitz, who was the US coach for the IMO for a bunch of years. I loved it when I was in undergrad!

posted by kaibutsu at 7:58 AM on April 28 [9 favorites]

*Indeed, a lot of my method for research involves inventing new methods, and then doing a bit of background search of the literature to figure out whose ideas I've just reinvented so I can cite them properly.*

I always justify reinventing the wheel by saying that I probably understand the idea better for having reinvented it instead of looking it up somewhere. However that may just be a scheme to reduce cognitive dissonance.

posted by madcaptenor at 8:34 AM on April 28 [1 favorite]

*If you gave tenured math profs the test, what would their typical score be?*

posted by dontjumplarry at 7:34 AM on April 28 [edit] [x] [IP:whois|search] [user info] [+] [!]

posted by dontjumplarry at 7:34 AM on April 28 [edit] [x] [IP:whois|search] [user info] [+] [!]

My husband is a math professor at a small college, and the Putnam is one of the big events of the year in their department. They have a year-long seminar class devoted to learning how to work on competition problems of this kind; the students really get into the spirit of the thing and their team usually does much better than the size/prestige of the college might suggest. (He says "this year's MIT team [which this article is about] was a

*machine*.")

To answer your question, every year, he idly works on the exam while the students are taking it, so they can talk over their approaches during lunch and afterward. He says that giving it half-attention during the exam, he can always solve at least four problems, with varying degrees of progress on the rest. Then over the subsequent week or so, he can usually solve around nine; can't always get more than that, but it varies how much time he has to devote to it.

The Putnam is

*such*a cool institution; I wish more disciplines had things like this.

posted by LobsterMitten at 11:16 AM on April 28 [1 favorite]

*One of the difficulties with this kind of exam is that they need to be 'naively solvable' with a first-year math student's level of background knowledge. It may be that such a first-year student would need (like jamjam) to be extraordinarily clever to succeed with such a minimal background, but there you have it. As a result, the questions tend to have lots of pigeonhole principle, naive number theory, and symmetry arguments. Once you've done a enough 'hard' competition problems, you start to get a sense of how to take an arbitrary problem and cram it into one (or more) of these common problem types. The pigeonhole principle becomes a lens through which you view the world.*

This is, IMO, patently false. There was at least one question this year which was intractable without Fourier analysis.

posted by TypographicalError at 11:50 AM on April 28

Speaking of books, How to Solve It is a classic on how to go about solving problems (mathematical, but not only). It's a really good place to start.

posted by parudox at 11:54 AM on April 28 [2 favorites]

posted by parudox at 11:54 AM on April 28 [2 favorites]

If you gave tenured math profs the test, what would their typical score be?

Math proffing and Putnam taking are slightly separate skills. The two best Putnamers I've known were at the NSA and a software company the last I knew, and these guys were top-5 scorers.

[BTW, LobsterMitten, interesting look at the Mods' toolkit in your quotation of dontjumplarry. "[edit] [x] [IP:whois|search] [user info]" but no "[nuke from the sky]?"

posted by benito.strauss at 11:54 AM on April 28

Math proffing and Putnam taking are slightly separate skills. The two best Putnamers I've known were at the NSA and a software company the last I knew, and these guys were top-5 scorers.

[BTW, LobsterMitten, interesting look at the Mods' toolkit in your quotation of dontjumplarry. "[edit] [x] [IP:whois|search] [user info]" but no "[nuke from the sky]?"

posted by benito.strauss at 11:54 AM on April 28

Ha, oops.

posted by LobsterMitten at 12:02 PM on April 28 [1 favorite]

posted by LobsterMitten at 12:02 PM on April 28 [1 favorite]

We have to get two mods to turn separate keys simultaneously before the nuke from orbit widget will instantiate.

posted by cortex at 12:04 PM on April 28 [4 favorites]

posted by cortex at 12:04 PM on April 28 [4 favorites]

Soooo...these MIT people can help me explain Common Core to my kids?

posted by Chuffy at 5:02 PM on April 28 [1 favorite]

posted by Chuffy at 5:02 PM on April 28 [1 favorite]

*He says that giving it half-attention during the exam, he can always solve at least four problems, with varying degrees of progress on the rest.*

I'm not sure how typical that is. At my school, yes, the ones who actually ran the Putnam could readily solve several problems on the day, but they had plenty of experience with this kind of math. (One of them was a former Putnam Fellow, actually.) I doubt that a generic math professor — with little or no experience in competitions, with a research specialty outside the Putnam's ambit — would do so well.

posted by stebulus at 6:45 PM on April 28 [1 favorite]

*This is, IMO, patently false. There was at least one question this year which was intractable without Fourier analysis.*

But orthogonalising a function space with respect to a countable collection of basis functions which respect the underlying symmetries of the space is something that would occur to any reasonably bright sophomore....

(Yeah, they do use a bit of domain knowledge in the Putnam, but I think not a whole lot.)

posted by kaibutsu at 6:42 AM on April 29 [1 favorite]

*But orthogonalising a function space with respect to a countable collection of basis functions which respect the underlying symmetries of the space is something that would occur to any reasonably bright sophomore....*

I can't tell if this is in jest or not, but given the standardish US math undergrad curriculum, said sophomore would probably but not certainly know the meanings of all the individual words in that sentence. Linear Algebra is often not seen at all until the sophomore year, and then typically in a very computationally intensive fashion that is light on concepts.

Now if you want to say any reasonably bright

*MIT*sophomore, sure.

posted by PMdixon at 2:18 PM on April 29

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posted by leahwrenn at 9:06 PM on April 27 [7 favorites]