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.

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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.

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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, $23=9+8+6$.)