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1998 OIM Problems/Problem 6

Revision as of 14:57, 13 December 2023 by Tomasdiaz (talk | contribs)

Problem

Let $\lambda$ be the positive root of the equation $t^2 - 1998t - 1 = 0$. The sequence $x_0, x_1, x_2, \cdots , x_n, \cdots$ is defined by:

\[\begin{cases} x_0=1 \\ x_{n+1}=\left\lfloor \lambda x_n \right\rfloor\ , & \text{for }n=0,1,2,\cdots\end{cases}\]

Find the remainder of the division of $x_{1998}$ by $1998$.

Note: The brackets indicate an integer part, that is, $\left\lfloor x \right\rfloor$ is the only integer $k$ such that $k le x < k+1$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

https://www.oma.org.ar/enunciados/ibe13.htm

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