Tuesday, May 30, 2006

Polynomial On The Floor? Topic: Algebra/Polynomials. Level: AIME.

Problem: (2005 Putnam - B1) Find a nonzero polynomial $ P(x,y) $ such that $ P(\lfloor a \rfloor, \lfloor 2a \rfloor) = 0 $ for all reals numbers $ a $.

Solution: What's the relation between $ \lfloor a \rfloor $ and $ \lfloor 2a \rfloor $? We can see that either

$ \lfloor 2a \rfloor = 2 \lfloor a \rfloor $

or

$ \lfloor 2a \rfloor = 2 \lfloor a \rfloor + 1 $

for all reals $ a $ (yes, even negative ones). So it suffices to have a polynomial with zeros at $ y-2x $ and $ y-2x-1 $, such as

$ P(x,y) = (y-2x)(y-2x-1) $.

QED.

--------------------

Comment: This is about the easiest question you'll see on a Putnam test. Each question is worth 10 points (120 total) and about half of the people who take it get 0 points.

--------------------

Practice Problem: (2005 Putnam - A1) Show that every positive integer is a sum of one or more numbers of the form $2^r3^s$, where $r$ and $s$ are nonnegative integers and no summand divides another. (For example, $23=9+8+6$.)

3 comments:

  1. Woo, induction.

    ReplyDelete
  2. If y=2x or y=2x+1, shouldn't the function's zero be written
    f(x,y)=(y-2x)(y-[2x+1])=(y-2x)(y-2x-1)?

    ReplyDelete
  3. hmm, boo... you're right lol.

    ReplyDelete