[Iberoamerican $1998]$ Let $\lambda$ be the positive root of the equation $t^{2}-1998 t-1=0 .$ Define the sequence $x_{0}, x_{1}, \ldots$ by setting $ x_{0}=1, \quad x_{n+1}=\left\lfloor\lambda x_{n}\right\rfloor \quad(n \geq 0) $ Find the remainder when $x_{1998}$ is divided by 1998
$1998<\lambda=\frac{1998+\sqrt{1998^{2}+4}}{2}=999+\sqrt{999^{2}+1}<1999$
\begin{aligned} x_{1}=1998, x_{2}=& 1998^{2} . \text { since } \lambda^{2}-1998 \lambda-1=0 \\ \lambda=1998+\frac{1}{\lambda} & \text { and } x \lambda=1998 x+\frac{x}{\lambda} \end{aligned}
for all real numbers $x$.
since $x_{n}=\left\lfloor x_{n-1} \lambda\right\rfloor$ we have
$ x_{n}<x_{n-1} \lambda<x_{n}+1, \quad \text { or } \quad \frac{x_{n}}{\lambda}<x_{n-1}<\frac{x_{n}+1}{\lambda} $
since $\lambda>1998,\left\lfloor\frac{x_{n}}{\lambda}\right\rfloor=x_{n-1}-1 .$
how they found $\left\lfloor\frac{x_{n}}{\lambda}\right\rfloor=x_{n-1}-1$ ???
From $$\frac{x_{n}}{\lambda}<x_{n-1}<\frac{x_{n}+1}{\lambda}$$ we have$$ \left\lfloor\frac{x_{n}}{\lambda}\right\rfloor<x_{n-1}\leq \left\lfloor\frac{x_{n}}{\lambda}+\frac1\lambda\right\rfloor\leq\left\lfloor\frac{x_{n}}{\lambda}\right\rfloor+1$$ The first inequality is strict because $x_{n-1}$ is an integer.