10. (40 points) Let $P(n)$ denote the product of the digits of a natural number $n$. For what largest natural number $k$ does there exist a natural number $n>10$ such that

$$
P(n)<P(2 n)<\ldots<P(k n) ?
$$
# Answer: 9.

Solution: First, let's show that $k<10$. Indeed, for $k \geqslant 10$, the set of numbers includes $P(10 n)$. It equals zero because the number $10 n$ ends in 0. Then $P(n)<0$, which is impossible.

Now let's provide an implementation for $k=9$. We can take $n$ as a number of the form $\underbrace{\overline{11 \ldots 1}}_{j}$ for any $j \geqslant 2$. In this case, $P(s n)=s^{j}, s=1, \ldots, k$, and the numbers $s^{j}$ clearly increase.