## Task $8 / 82$

Determine the smallest number $n$ that in the decimal system has the form

$$
n=\sum_{i=1}^{k} 10^{2 i-1} x+\sum_{i=0}^{k} 10^{2 i} y
$$

where $(x ; y \in \mathbb{N}, 00)$ and is divisible by 264 without a remainder.
Due to $264=2^{3} \cdot 3 \cdot 11$, for $n$ the known divisibility rules must hold:

$$
\begin{array}{r}
3 \mid k(x+y) \\
8 \mid 101 y+10 x \\
11 \mid k(x-y)
\end{array}
$$

From (3), it immediately follows that due to $x \leq 9, x \neq y, y \leq 9$, $11 \mid k$. Since the smallest number $n$ is sought, we initially try with $k=11$.

Then, according to (1), $3 \mid (x+y)$. The smallest $x$ is $x=1$; then $y=2$ or $y=5$ or $y=8$, but none of these combinations satisfy condition (2). On the other hand, with $x=2, y=4$ ( $y=1$ is also ruled out), a usable pair is obtained: $8 \mid 101 \cdot 4 + 10 \cdot 2 = 424$. Thus, the solution is found:

$$
n=\sum_{i=0}^{k} 10^{2 i+1} \cdot 2 + \sum_{i=0}^{k} 10^{2 i} \cdot 4 = 24242424 \ldots 24
$$