# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

• ## Meta

Polynomial On The Floor? Topic: Algebra/Polynomials. Level: AIME. May 30th, 2006

Problem: (2005 Putnam – B1) Find a nonzero polynomial such that for all reals numbers .

Solution: What’s the relation between and ? We can see that either

or

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

.

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 , where and are nonnegative integers and no summand divides another. (For example, .)

### 3 Responses to “Polynomial On The Floor? Topic: Algebra/Polynomials. Level: AIME.”

1. Anonymous Says:

Woo, induction.

2. quantum leap Says:

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)?