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Berkeley Math Circle Take-Home Contest #6

Algebra Difficulty 4.8 AIME Prove it United States

Problem:

Let $g_{1}(x), g_{2}(x), \ldots, g_{5}(x)$ be polynomials with integer coefficients. Suppose that their product $f(x) = g_{1}(x) g_{2}(x) \cdots g_{5}(x)$ satisfies $f(1999) = 2000$ . Prove that for some $i \in \{1,2,3,4,5\}$ , the sum of the coefficients of $g_{i}(x)$ is odd.

Solution

Solution:

For some $i$ , $g_{i}(1999)$ is odd. Indeed, if this were false, each $g_{i}(1999)$ would be divisible by $2$ , so their product, $f(1999) = 2000$ , would be divisible by $2^{5}$ , which is not the case.

Now write $g_{i}(x) = c_{n} x^{n} + c_{n-1} x^{n-1} + \cdots + c_{0}$ . Then
$g_{i}(1999) - (c_{n} + \cdots + c_{0}) = c_{n}(1999^{n} - 1) + c_{n-1}(1999^{n-1} - 1) + \cdots + c_{0}(1 - 1),$
which is even since $1999^{k} - 1$ is even for each $k$ . Since $g_{i}(1999)$ is odd, $c_{n} + \cdots + c_{0}$ is also odd, as desired.

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