How Did Others Fare in the Putnam Exam Yesterday?

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Discussion Overview

This thread discusses participants' experiences and solutions during the Putnam exam, focusing on specific problems from both sessions. The conversation includes reflections on problem-solving approaches, uncertainties in solutions, and hopes for scoring points.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confidence in solving a couple of problems but is uncertain about the scoring, particularly for A1, where they initially believed the curves could only be tangent along the line y=x.
  • Another participant challenges this view, suggesting that tangency can also occur at points where the slope is -1, indicating there are additional values of a.
  • Participants discuss the area of a convex set intersecting hyperbolas in A2, with one participant asserting the area should be 4 but struggled to provide a solid solution.
  • For B1, one participant finds the problem straightforward, while another notes a potential oversight regarding constant functions that could affect the solution.
  • There is a shared concern about the accuracy of solutions for A1, with participants speculating on how many values of a they might have identified correctly.
  • Participants express varying degrees of confidence in their solutions for other problems, with some fearing they may receive low scores due to errors or incomplete answers.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the tangency conditions in A1, with some believing there are only two values of a while others suggest there are four. Additionally, there is uncertainty about the scoring and correctness of various solutions across different problems, indicating a lack of consensus on several points.

Contextual Notes

Participants mention potential errors in their solutions, such as not simplifying answers fully or including extraneous solutions, which may affect their scores. There are also references to specific conditions in problems that may not have been fully addressed.

Who May Find This Useful

Individuals preparing for mathematical competitions, particularly the Putnam exam, may find insights into problem-solving strategies and common pitfalls discussed in this thread.

Doom of Doom
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Did anyone else toil away for six hours yesterday at the Putnam exam?

I'm fairly confident that I got a couple.
________
A1: Find a where y=a x^2 +a x + \frac{1}{24} and x=a y^2 +a y + \frac{1}{24} are tangent.

This was easy, once you recognize that the curves could only intersect (i.e. be tangent) along the y=x line.

_____
A2: Find the least possible area of a convex set in the plane that intersects both branches of the hyperbolas xy=1 and xy=-1. (A set S is called convex if for any two points in S the line segment connecting them is contained in S)

It seems obvious to me that this area would be 4, but I spent quite a bit of time on this problem and wasn't able to turn in a good solution for it.

______
A3: Take the integers 1,2,...,(3k+1) and write them down in a random order. What is the probability that, as you write them down, there is no time when the sum of all of the numbers written up to that point is divisible by 3?

If you take all of the numbers and mod them by 3, you get: k+1 ones, k twos and k zeros. Then, if you ignore the zeros (as long as you don't write one in the first position), the only way you can write the one's and two's is like this (and this is easy to show):

1,1,2,1,2,1,2,...,1,2

So, you can count the number of ways that you can write that out, then count the number of ways that you can insert the zeros into that sequence, and divide out by (3k+1)!.

_______
Those were the problems I got in the first session. The only problem I solved in the second session was B1.

_______
B1: If f(n) is a polynomial with positive integer coefficients, prove that the only case when f(n) divides f(f(n)+1) is when n=1.

This was very easy. All you have to do is mod by f(n), and you get f(n) (mod f(1)) =0 which can only occur when n=1 (because all of the coefficients are positive).

____
B3: Find x_{2007} where x_0=1 and x_{n+1}=3x_n+\left\lfloor x_n \sqrt{5}\right\rfloor

I didn't turn in a solution for this, and I was never able to get this in closed form, but here is what I reduced it to:

x_n=(3+\sqrt{5})^n-(\sqrt{5}-2)\sum_{i=0}^{n-1} x_i

_____

That's what I turned in. So, I hope that I get 30 pts. How did everyone else do?
 
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Doom of Doom said:
____
B3: Find x_{2007} where x_0=1 and x_{n+1}=3x_n+\left\lfloor x_n \sqrt{5}\right\rfloor

I didn't turn in a solution for this, and I was never able to get this in closed form, but here is what I reduced it to:

x_n=(3+\sqrt{5})^n-(\sqrt{5}-2)\sum_{i=0}^{n-1} x_i

_____

That's what I turned in. So, I hope that I get 30 pts. How did everyone else do?

Trying these for fun.

This thing grows fast!
1,5,26,136, 712, 3728, 19520, 102208, 535168, 2802176, 14672384...

x_100 = 7.53313*10^71
 
Doom of Doom said:
A1: Find a where y=a x^2 +a x + \frac{1}{24} and x=a y^2 +a y + \frac{1}{24} are tangent.

This was easy, once you recognize that the curves could only intersect (i.e. be tangent) along the y=x line.
Unfortunately this isn't true (I said the same thing in my solution). It's also possible for them to be tangent at a point where the slope is -1. This gives two additional values of a. :(
 
Xevarion said:
Unfortunately this isn't true (I said the same thing in my solution). It's also possible for them to be tangent at a point where the slope is -1. This gives two additional values of a. :(

Ah. I see that now. I wonder if my answer of a= 2/3, 3/2 will be sufficient to get me any points at all on that problem. For that matter, I wonder how many people tried solving this one and got only two values for a (instead of all four).
 
However, my solutions for B1 and A3 were pretty much spot on. Now I'm hoping for 20 points. :)

What other problems did you solve?
 
Well as I said I did the same thing you did on A1. I think we'll be pretty lucky even to get 1 point for that.

I did A2 and A3 for sure. I also had a proof for A4, but (1) I didn't simplify my answer all the way and (2) I included one extraneous solution. So I'm not going to hope for much.

On B, I had B1 and B2, but for B1 I didn't point out the error in the problem statement (it doesn't work for constant functions). Also I might have made a sign error in my work in B2 (which didn't matter in the end because we take abs value). I don't remember. So it's possible that I won't get much at all on the second half.

I'll be rather disappointed if I get only 20. But I wouldn't be too surprised if I get no more than 2 on all of the ones that aren't perfect.
 

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