The Putnam was delightful. I didn’t realistically expect to solve eight questions, despite my apparent confidence beforehand. I still don’t know how I did it. So let’s see what happened:

A1: I did it without much trouble. I tried to construct a counterexample at first and then found the reason I was prevented from doing so, so I found a nice little proof instead.

A2: I missed a very elegant two-liner and came up with a rather ugly solution. Fortunately Putnam graders don’t give style points.

A3: I was confused for a while because I was trying to prove two things by induction at once. I tried to understand how that would work, but then I realized that the two things were actually equivalent and that I only had to prove one, and then it was easy from there.

A4: I didn’t have time to write this up brilliantly, so I wrote down a constructive approach that works. If the grader doesn’t question my ability to multiply out polynomials, I might get points, but I’m not really counting on more than one or two.

A5 and A6: I didn’t submit these.

B1: I had seen the idea before in the proof of the Liouville Approximation Theorem, so I didn’t have to think very much to solve this problem. I found it gets a bit ugly when the denominator of r is not a prime (and especially when all the prime factors are repeated). It was even more annoying that the argument was exactly the same, but one has to be very picky in writing up the solution. I messed it up the first time and had to get another B1 solution sheet.

B2: The left side remains unchanged when m increases and n decreases, whereas the right side can decrease if it is done right. Then I ended up with something about (1+1/x)^x being an increasing function, and I stopped there and claimed it was well-known.

B3: I found a solution I am quite proud of, even though it’s the same solution everyone else found.

B4: Complex numbers are so great. I knew a theorem that says it had to be a translation, but then I couldn’t find the image of any one point more easily than an arbitrary point, so I didn’t get a chance to use it. This problem would have been in the AoPS trig/complex numbers class if it had been on last year’s Putnam.

B5 and B6: I think B5 is supposed to be easy, but I was so drained by the time I got to it that I didn’t even think of taking logarithms despite the fact that it is by far the most obvious thing to do. Thus I left these two blank.

I’m expecting something like 70 points. Perhaps I’m being optimistic, but I’ll keep expecting 70ish until March. I’m still completely shocked.

Hi. I'm Simon Rubinstein-Salzedo. I'm a mathematics postdoc at Dartmouth College. I'm also a musician; I play piano and cello, and I also sometimes compose music and study musicology. I also like to play chess and write calligraphy. This blog is a catalogue of some of my thoughts. I write them down so that I understand them better. But sometimes other people find them interesting as well, so I happily share them with my small corner of the world.
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14 Responses to

1. aaronlehmann says:

Can you link to the test?

2. z9r4c3 says:

*claps*
i don’t remember how many you solved last year, but 8 definitely beats it!! congrats!

• Simon says:

Re: *claps*
Yes, 8>3.

• teratoma says:

Re: *claps*
8 <33
;-*
^_^

3. mathfanatic says:

Glad to hear you did well. Eight is indeed greater than three, at least in most textbooks.

• Anonymous says:

Not in Z/4

• Simon says:

But Z_4 is not a partially ordered set under anything remotely like the usual <= relation.

4. teratoma says:

awesome job dude

5. Fantabulous!

6. itsthomson says:

Hi! I read your post through a friend’s “Friends” page, and saw that you took the Putnam. I got 8 questions too! The only difference though, was that I didn’t do B3, but I did do B5, although the series expansions were a little ugly for me.
I’m curious though. I know the solutions are already posted, but what was your solution to B3? For some reason I’m still not sure how to start it.

• Simon says:

Here’s a sketch:
First, if a>2, we can take the constant function f(x)=2a/(a-2). It’s easy to check that that works.
Now suppose a<=2. Let M be the maximum value of f(x) on the interval [0,a]. The area is at most aM. As for the perimeter, we have to get from 0 to M and back down again, and then we have to get across a at the bottom (and something at the top as well). Thus the perimeter is more than a+2M. So we have a+2m<perimeter=area<=aM. But a+2m>aM, so we have a contradiction.