HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

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 $a_{1} \cdot 1^{2005}+a_{2} \cdot 2^{2005}+a_{3} \cdot 3^{2005}+\cdots+a_{2005} \cdot 2005^{2005}=1$ What is the value of $a_{1}$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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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