268*. Consider the sequence of numbers $x_{n}=(1+\sqrt{2}+\sqrt{3})^{n}$. Each of them can be expressed in the form

$$
x_{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}
$$

where $q_{n}, r_{n}, s_{n}, t_{n}$ are integers. Find the limits

$$
\lim _{n \rightarrow \infty} \frac{r_{n}}{q_{n}}, \quad \lim _{n \rightarrow \infty} \frac{s_{n}}{q_{n}}, \quad \lim _{n \rightarrow \infty} \frac{t_{n}}{q_{n}}
$$

13th All-Union Olympiad, 1979 (Tbilisi)

| Class | І- |  |  |  | day |  | 2nd day |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 8 | 269 | 270 | 271 | 274 | 275 | 276 | 277 |  |  |  |
| 9 | 269 | 272 | 271 | 278 | 279 | 280 | 281 |  |  |  |
| 10 | 273 | 272 | 271 | 276 | 275 | 282 | 283 |  |  |  |
268. Let's consider together with the number $\lambda_{1}=1+\sqrt{2}+\sqrt{3}$ its "conjugates," which differ from it by the signs of the radicals: $\lambda_{2}=1-$ $-\sqrt{2}+\sqrt{3}, \lambda_{3}=1+\sqrt{2}-\sqrt{3}$ and $\lambda_{4}=1-\sqrt{2}-\sqrt{3}$. If

$$
\lambda_{1}^{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6},
$$

then, we can sequentially check for $n=1,2,3,1, \ldots 3$

$$
\begin{aligned}
& \lambda_{2}^{n}=q_{n}-r_{n} \sqrt{2}+s_{n}-t_{n} \sqrt{6} \\
& \lambda_{3}^{n}=q_{n}+r_{n} \sqrt{2}-s_{n} \sqrt{3}-t_{n} \sqrt{6} \\
& \lambda_{4}^{n}=q_{n}-r_{n} \sqrt{2}-s_{n} \sqrt{3}+t_{n} \sqrt{6}
\end{aligned}
$$

By adding these four equations, dividing the sum by $4 \lambda_{1}^{n}$, and taking the limit, we get $\lim \left(q_{n} / \lambda_{1}^{n}\right)=1 / 4$, since $\left|\lambda_{j}\right|<\lambda_{1}$ for $j=2,3,4$, and therefore $\lim \left(\left|\lambda_{j}\right|^{n} / \lambda_{1}^{n}\right)=0$. By adding the first equation to one of the other three and subtracting the two remaining, we obtain: $\quad \lim \left(r_{n} \sqrt{2} / \lambda_{1}^{n}\right)=\lim \left(s_{n} \sqrt{3} / \lambda_{1}^{n}\right)=\lim \left(t_{n} \sqrt{6} / \lambda_{1}^{n}\right)=$ $=1 / 4$ (thus, in the sum $q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}$ for large $n$, all terms are approximately equal to each other!) From this, we find

$\lim _{n \rightarrow \infty}\left(r_{n} / q_{n}\right)=1 / \sqrt{2}, \quad \lim _{n \rightarrow \infty}\left(s_{n} / q_{n}\right)=1 / \sqrt{3}, \lim _{n \rightarrow \infty}\left(t_{n} / q_{n}\right)=1 / \sqrt{6}$.

$\nabla$ Note that the answer in this problem would be the same if instead of $1+\sqrt{2}+\sqrt{3}$, any number $a+b \sqrt{2}+c \sqrt{3}+d \sqrt{6}$ with natural $a, b, c, d$ were used.