$\mathrm{k}$ is the positive root of the equation $\mathrm{x}^{2}-1998 \mathrm{x}-1=0$. Define the sequence $\mathrm{x}_{0}, \mathrm{x}_{1}, \mathrm{x}_{2}, \ldots$ by $\mathrm{x}_{0}$ $=1, \mathrm{x}_{\mathrm{n}+1}=\left[\mathrm{k} \mathrm{x}_{\mathrm{n}}\right]$. Find the remainder when $\mathrm{x}_{1998}$ is divided by 1998.

The text is already in English, so no translation is needed.
Put $\mathrm{p}(\mathrm{x})=\mathrm{x}^{2}-1998 \mathrm{x}-1$. Then $\mathrm{p}(1998)=-1, \mathrm{p}(1999)=1998$, so $1998\mathrm{kx}_{\mathrm{n}-1}-1$. Hence $\mathrm{x}_{\mathrm{n}} / \mathrm{k}<\mathrm{x}_{\mathrm{n}-1}<\mathrm{x}_{\mathrm{n}} / \mathrm{k}+1 / \mathrm{k}$, so $\left[\mathrm{x}_{\mathrm{n}} / \mathrm{k}\right]=\mathrm{x}_{\mathrm{n}-1}-1$.

$\mathrm{k}=(1998 \mathrm{k}+1) / \mathrm{k}=1998+1 / \mathrm{k}$. Hence $\mathrm{kx}_{\mathrm{n}}=1998 \mathrm{x}_{\mathrm{n}}+\mathrm{x} / \mathrm{k}$. Hence $\mathrm{x}_{\mathrm{n}+1}=\left[\mathrm{kx}_{\mathrm{n}}\right]=1998 \mathrm{x}_{\mathrm{n}}+$ $\left[\mathrm{x}_{\mathrm{n}} / \mathrm{k}\right]=1998 \mathrm{x}_{\mathrm{n}}+\mathrm{x}_{\mathrm{n}-1}-1$. Hence $\mathrm{x}_{\mathrm{n}+1}=\mathrm{x}_{\mathrm{n}-1}-1 \bmod 1998$. So $\mathrm{x}_{1998}=1-999=1000 \bmod$ 1998.