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Berkeley Math Circle: Monthly Contest 7

Algebra Difficulty 5.6 AIME, harder Prove it United States

Problem:
Let a1,a2,…,a1000a_{1}, a_{2}, \ldots, a_{1000} be real numbers such that
a1⋅1+a2⋅2+a3⋅3+⋯+a1000⋅1000=0a1⋅12+a2⋅22+a3⋅32+⋯+a1000⋅10002=0a1⋅13+a2⋅23+a3⋅33+⋯+a1000⋅10003=0⋮a1⋅1999+a2⋅2999+a3⋅3999+⋯+a1000⋅1000999=0a1⋅11000+a2⋅21000+a3⋅31000+⋯+a1000⋅10001000=1. \begin{array}{r} a_{1} \cdot 1+a_{2} \cdot 2+a_{3} \cdot 3+\cdots+a_{1000} \cdot 1000=0 \\ a_{1} \cdot 1^{2}+a_{2} \cdot 2^{2}+a_{3} \cdot 3^{2}+\cdots+a_{1000} \cdot 1000^{2}=0 \\ a_{1} \cdot 1^{3}+a_{2} \cdot 2^{3}+a_{3} \cdot 3^{3}+\cdots+a_{1000} \cdot 1000^{3}=0 \\ \vdots \\ a_{1} \cdot 1^{999}+a_{2} \cdot 2^{999}+a_{3} \cdot 3^{999}+\cdots+a_{1000} \cdot 1000^{999}=0 \\ a_{1} \cdot 1^{1000}+a_{2} \cdot 2^{1000}+a_{3} \cdot 3^{1000}+\cdots+a_{1000} \cdot 1000^{1000}=1 . \end{array}
What is the value of a1a_{1} ?

Solution

Solution:
The key idea is to multiply on weights which are polynomial coefficients. Specifically, let P(n)=n(n−2)(n−3)…(n−1000)=∑k=01000cknkP(n)=n(n-2)(n-3) \ldots(n-1000)=\sum_{k=0}^{1000} c_{k} n^{k}, say. Note that c1=0c_{1}=0 and c1000=1c_{1000}=1.
Now take the kkth equation and multiply it by ckc_{k}, then add all equations together. From this we obtain
a1P(1)+a2P(2)+⋯+a1000P(1000)=c1000=1. a_{1} P(1)+a_{2} P(2)+\cdots+a_{1000} P(1000)=c_{1000}=1 .
But P(1)=(−1)999⋅999!P(1)=(-1)^{999} \cdot 999! and P(2)=⋯=P(1000)=0P(2)=\cdots=P(1000)=0. Therefore, a1=−1999!a_{1}=\frac{-1}{999!}.

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