Putnam for Amusement: 1998 B1 and 1991 A2

  • Published on Oct 2, 2018
  • Let's have some fun with slick algebraic manipulations.
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Comments • 26

  • How it feels like
    How it feels like 5 months ago

    When you get 3(x+1/x) in the first problem you can make a quadratic function by letting the expression equal to a constant...then to get real values you need the delta x (b^2-4ac) >= o so then you get that the expression will be greater than or equal 6 ๐Ÿ™‚

  • puzzle collector
    puzzle collector 6 months ago

    wow this is fun!

  • Mohd Tauheed Khan
    Mohd Tauheed Khan 7 months ago

    Put x=1

  • Mohd Tauheed Khan
    Mohd Tauheed Khan 7 months ago

    I used minimum value of (x+1/x) =2

  • Amber Deshbhratar
    Amber Deshbhratar 11 months ago +3

    this is a very good USclip channrl

  • Arvind Kumar
    Arvind Kumar 11 months ago

    The first one has also come in Regional Mathematics Olympiad of India also

  • Paco Libre
    Paco Libre 11 months ago

    If you know paschalโ€™s triangle, expanding the 1998 problem isnt so bad. [x^6+6x^4+15x^2+20+15x^-2+6x^-4+x^-6] and [x^3+3x+3x^-1+x^-3] I have the fourth row memorized, and getting to the sixth row is just some addition.

  • etotheipi
    etotheipi 11 months ago

    I'm actually surprised that these are Putnam problems. I've always viewed Putnam problems as USAMO/IMO level, but I would have thought as the first one as an AMC 10 difficulty and the second one as easier than a JMO 1. I guess the Putnam has gotten a lot harder over the years.
    Thanks a lot for making these vids! They're well explained as always.

  • Felipe Lorenzzon
    Felipe Lorenzzon 11 months ago

    I tried to add the functions inside each parenthesis on the first question, but it didn't help me out

  • Quahntasy - Animating Universe

    Who does Putnam problem for amusement lol.

  • ืœื™ื—ื™ ื’'ื•ืœื™ืืŸ

    I did the first one in a different i think easier way.
    I said let x0 be the value for which this expression is minimal.
    Then i said lets substitute t=1/x0, which we can because x0!=0.
    It's easy to see that the expression remains the same for t=1/x0 and has therefore the same value x0 for to substitute instead of t to make it minimal so for t=x0 the expression is minimal but t=1/x0 so x0=1 for the minimal value of the expression, substituting in the expression gives you 6.

    • Luka Mernik
      Luka Mernik 10 months ago +4

      I don't think this is correct. You just got lucky. In general a function does not have a unique minimum, so x_0 and 1/x_0 can be two different minimums and are not necessarily the same point. I.e. if the function has a minimum at 2 and 1/2, then the value f(2)=f(1/2), but x_0 and 1/x_0 are not equal.

  • Randall Kim
    Randall Kim 11 months ago

    I have a request for a problem. Please take a look at the 2006 AIME II (Alternate AIME) #13. Keep up the great videos!

  • adandap
    adandap 11 months ago +2

    I like the first problem. I wasn't quite as slick as you, but the binomial approach isn't a dead end. You can use is recursively to express x^6 + x^(-6) in terms of 4th and 2nd powers, and then keep going until everything is in terms of powers of (x+1/x). You end up in the same place. (I hope that was clear.) Now onto the other one...

  • Ryan C.
    Ryan C. 11 months ago +1

    This is weird, the first problem was on the 1998 Putnam (mks.mff.cuni.cz/kalva/putnam/putn98.html), but the exact same problem was also on the 1979 AHSME (artofproblemsolving.com/wiki/index.php?title=1979_AHSME_Problems/Problem_29). I find it really strange that the same exact problem would be on tests of incredible different difficulty and also tests nearly 20 years apart.

  • el tapa
    el tapa 11 months ago +1

    Oh mate, amazing

  • 77Chester77
    77Chester77 11 months ago +30

    As always: At the beginning I have no clue at all but at the end you made it look so easy. Bravo.

  • Nikolajus Elmutis
    Nikolajus Elmutis 11 months ago

    How comes they are so easy u.u

  • Sarthak Sharma
    Sarthak Sharma 11 months ago

    Sir 1998 question
    Can be simply solved by using AM>=GM inequality

  • Avishek Tiwari
    Avishek Tiwari 11 months ago

    Upload more question

  • Jeff Ahn
    Jeff Ahn 11 months ago

    Woow. So elegant.

  • Siddhartha Shree kaushik
    Siddhartha Shree kaushik 11 months ago +2

    I got 3 notifications within 2 minutes
    Are you serious?

  • Siddhartha Shree kaushik
    Siddhartha Shree kaushik 11 months ago

    Please stop sending multiple notifications
    It's fucking annoying

    • LetsSolveMathProblems
      LetsSolveMathProblems  11 months ago +1

      I did not send multiple notifications. USclip is programmed to issue notifications to subscribers who wish to receive them, and I am not in control of it (as far as I am aware). I apologize for the possible inconvenience.