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Ahhhh Putnam Competition Tomorrow!

  1. Dec 2, 2005 #1
    I can't believe I am going to get up at 830 later today and get ready to take the putnam exam. Test is from 10-1 and 3-6. I hope to get at least 1 question right if I am going to spend 6 hours on a test (I probably could have taken the MCAT for fun instead). I haven't studied for it or anything (which I dont think you should do). So is anyone else a sucker for S&M like myself and taking the exam as well? Does anyone know how how high the highest scores were last year? I heard that if the scores for a year go above 100, the exam makers make the exam impossible the next year so the highest scores are only in the 80s. I hope this year we get an easy test.
  2. jcsd
  3. Dec 3, 2005 #2
    Best of luck to you Graveneworld! Let us know how it went, and if you can bring us some of the questions. My Algebra teacher at Berkeley, Bjorn Poonen, took first place in that competition 4 years in a row!!! SO INSANE!
  4. Dec 3, 2005 #3
    your lucky your test starts at 10 when i wrote it it started at 8am BUT free lunch at the expense of the math dept is great.

    In my last year the test was easiest... I hated it when you could solve a question only hours after sitting through the 6 hours scratching your head, I swear they need to make those competitions in coffee houses.
    Good Luck and remmeber probability and induction.
  5. Dec 3, 2005 #4
    When do the results come out (Putnam fellows, winning teams, etc.)? I'm interested to see how a lot of the former IMO medalists do.
  6. Dec 3, 2005 #5
    I don't think you should be posting any of the problems now; there are sunday test-takers also.
  7. Dec 3, 2005 #6
    Are you sure? The website says everyone has to take it today. If anyone in a different country is taking it on Sunday, it is because it is already suday there. It is 1:52 pm right now (east coast US). So i think I am ok to post problems from the 1st half since everyone already has done them.
  8. Dec 3, 2005 #7
    Heres the problem I spent all my time on- Let p(z) be a polynomial with roots having modulus =1 in the complex plane. Let g(z)=P(z)/z^n/2. Prove g'(z)=0 has solutions that have modulus 1.
  9. Dec 3, 2005 #8
    I looked again, and I think it should definitely be safe to post questions. The competition is only for people in Canada and the US, so everyone has already finished the 1st half.
  10. Dec 3, 2005 #9
    I thought these were supposed to be questions that didn't require mathematics knowledge beyond calculus...
  11. Dec 3, 2005 #10

    There is always a lot of advanced stuff on the test. The only basic calculus one was evalute the intergral from 0 to 1 of ln(x+1)/X^2+1 dx.
  12. Dec 3, 2005 #11
    Two questions:
    1) Do "with roots" and "has solutions" mean some roots/solutions or all roots/solutions have modulus 1?
    2) Is it g(z)=(P(z)/z^n)/(2) or g(z)=P(z)/(z^n/2)?
  13. Dec 3, 2005 #12

    Hmm... The only thing I can think of is to rewrite the integrand as:


    ..but that's just throwing it out there; I'm not sure if it's even the right idea. Did anyone actually get this one?
    Last edited: Dec 3, 2005
  14. Dec 3, 2005 #13
    I got the integral by decomposing the integrand into partial fractions like:
    [log(1-i)/(-2i)]/(x+i) + [log(1+i)/(2i)]/(x-i)
    (this may not be exactly correct, but it's easy to understand)
    from there it is simply knowing the logarithm of a complex number and complex multiplication.
    It turns out that the answer is (PI*log 2)/8
  15. Dec 3, 2005 #14
    Interesting. Do you remember any other problems from the exam?
  16. Dec 3, 2005 #15

    1) If I remember the question correctly, all roots/solution have mod 1
    2.) g(z)=P(z)/(z^n/2)

    Here are some questions from the second half that I remember.

    Let m,n be positive integers. Consider the n-tuple (x1,x2,....,xn) (x1,x2,..,xn are any integers). Let the n-tuples have the property |x1|+.....+|xn|<=m. If f(m,n) is the amount of n-tuples that satisfies this property, show that f(m,n)=f(n,m).

    Find all intergers n,k1,k2,...,kn such that k1+k2+....+kn=5n-4
    and 1/k1 + 1/k2 +......+1/kn=1

    Find a polynomial f(x,y) such that f([a],[2a])=0 for any real number a.
    [ ] is supposed to be the floor function.
  17. Dec 3, 2005 #16
    Consider a nxn matrix having only entries +/-1. Each row in the matrix is orthogonal to the other. Prove that there is a axb submatrix consisting of only 1s, such that ab<=n. (I'm not sure if i remembered exactly what the question was, so you may not want to rack your brain over this one.)
  18. Dec 5, 2005 #17
    Let a=b=1. :tongue2:
  19. Dec 5, 2005 #18
    In the same vein as Rachmaninoff's...



    I'll actually work on them later...must do homework right now.
  20. Dec 5, 2005 #19
    They probably have to be nontrivial solutions. Again, I probably didn't remember all the details correctly.
  21. Dec 6, 2005 #20
    Putnam questions can be found here (compiled by Valentin of Mathlinks/AoPS). Some of answers can be found here.
    Last edited: Dec 6, 2005
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