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Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.7 AIME, harder Prove it United States

Problem:
Let $a_{1}, a_{2}, \ldots, a_{2005}$ be real numbers such that
$\begin{array}{ccccccccccc} a_{1} \cdot 1 & + & a_{2} \cdot 2 & + & a_{3} \cdot 3 & + & \cdots & + & a_{2005} \cdot 2005 & = & 0 \\ a_{1} \cdot 1^{2} & + & a_{2} \cdot 2^{2} & + & a_{3} \cdot 3^{2} & + & \cdots & + & a_{2005} \cdot 2005^{2} & = & 0 \\ a_{1} \cdot 1^{3} & + & a_{2} \cdot 2^{3} & + & a_{3} \cdot 3^{3} & + & \cdots & + & a_{2005} \cdot 2005^{3} & = & 0 \\ \vdots & & \vdots & & \vdots & & & & \vdots & & \vdots \\ a_{1} \cdot 1^{2004} & + & a_{2} \cdot 2^{2004} & + & a_{3} \cdot 3^{2004} & + & \cdots & + & a_{2005} \cdot 2005^{2004} & = & 0 \end{array}$
and
$a_{1} \cdot 1^{2005}+a_{2} \cdot 2^{2005}+a_{3} \cdot 3^{2005}+\cdots+a_{2005} \cdot 2005^{2005}=1 .$

Solution

Solution:
$1 / 2004!$

The polynomial $p(x) = x(x-2)(x-3) \cdots (x-2005) / 2004!$ has zero constant term, has the numbers $2, 3, \ldots, 2005$ as roots, and satisfies $p(1) = 1$ . Multiplying the $n$ th equation by the coefficient of $x^{n}$ in the polynomial $p(x)$ and summing over all $n$ gives
$a_{1} p(1) + a_{2} p(2) + a_{3} p(3) + \cdots + a_{2005} p(2005) = 1 / 2004!$
(since the leading coefficient is $1 / 2004!$ ). The left side just reduces to $a_{1}$ , so $1 / 2004!$ is the answer.

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