Maths Olympiad Prep

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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
a11+a22+a33++a10001000=0a112+a222+a332++a100010002=0a113+a223+a333++a100010003=0a11999+a22999+a33999++a10001000999=0a111000+a221000+a331000++a100010001000=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(n2)(n3)(n1000)=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)999999!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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