Olympiad Maths Prep

Track / Stage 6 / 48 of 400 #1048 of 2000

Problem 1048

National olympiad, first round
Algebra Difficulty 6.1 Prove it

Prove:
⟨a1,⋯ ,al−1⟩=⟨al−1,⋯ ,a1⟩, i.e. aj=al−j,1⩽j⩽l/2. \left\langle a_{1}, \cdots, a_{l-1}\right\rangle=\left\langle a_{l-1}, \cdots, a_{1}\right\rangle, \quad \text { i.e. } a_{j}=a_{l-j}, 1 \leqslant j \leqslant l / 2 \text {. }

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

12. Utilize Question 5, take ξ0=[d]+d.−1/ξ0=1/(d−[d])=\xi_{0}=[\sqrt{d}]+\sqrt{d} .-1 / \xi_{0}=1 /(\sqrt{d}-[\sqrt{d}])= ⟨a1‾,⋯ ,al−1,2[d]⟩\left\langle\overline{a_{1}}, \cdots, a_{l-1}, 2[\sqrt{d}]\right\rangle. But from Question 5, we know that −1/ξ0′=⟨al−1,⋯ ,a1,2[d]‾⟩-1 / \xi_{0}^{\prime}=\left\langle\overline{a_{l-1}, \cdots, a_{1}, 2[\sqrt{d}]}\right\rangle.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.