37. Let $a_{1}, a_{2}, \ldots, a_{2005}$ be real numbers such that
$$
\begin{array}{l}
a_{1} \cdot 1+a_{2} \cdot 2+a_{3} \cdot 3+\cdots+a_{2005} \cdot 2005=0 \\
a_{1} \cdot 1^{2}+a_{2} \cdot 2^{2}+a_{3} \cdot 3^{2}+\cdots+a_{2005} \cdot 2005^{2}=0 \\
a_{1} \cdot 1^{3}+a_{2} \cdot 2^{3}+a_{3} \cdot 3^{3}+\cdots+a_{2005} \cdot 2005^{3}=0 \\
a_{1} \cdot 1^{2004}+a_{2} \cdot 2^{2004}+a_{3} \cdot 3^{2004}+\cdots+a_{2005} \cdot 2005^{2004}=0 \\
\end{array}
$$
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 $a_{1}$ ?
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.