25. The sequence $\left(x_{n}\right)$ is defined by the conditions: $x_{1}=1$, and for each natural number $n$, the number $x_{n+1}$ is the largest number that can be obtained by rearranging the digits of the number $x_{n}+1$. Find the smallest $n$ for which the decimal representation of the number $x_{n}$ has exactly 2017 digits.
25. Answer: the smallest such number $n$ is $18298225=9(1+2+3+\ldots+2016)+1$.

To understand how this sequence is constructed, let's write out the first few terms. Initially, the sequence contains single-digit numbers:

$$
1, \quad 2, \quad 3, \quad \ldots, \quad 9 .
$$

Next come the two-digit numbers:

![](https://cdn.mathpix.com/cropped/2024_05_06_8ebb97158f1b93c124d5g-2.jpg?height=62&width=974&top_left_y=426&top_left_x=541)

Then the three-digit numbers:

$$
100,110,111, \quad 211,221,222, \quad 322,332,333, \ldots, 988,998,999 .
$$

What is the pattern here? Let's think about how many terms in the sequence can have exactly $k$ digits.

The first $k$-digit number can appear in this sequence only as a result of the operation $99 \ldots 9 + 1 = 100 \ldots 0$ (no digit rearrangement is required). Next, we claim that the sum of the digits of each subsequent $k$-digit number is 1 greater than the sum of the digits of the previous number. Indeed, digit rearrangement does not change the sum of the digits, and adding 1 always increases the sum of the digits by 1, except in the case where the last digit is 9. However, in the representation of all numbers in our sequence, the digits are arranged in descending order, and if the last digit is 9, then the number consists entirely of 9s, and the next term in the sequence is a $(k+1)$-digit number.

Thus, the sums of the digits of consecutive $k$-digit terms in the sequence are $1, 2, 3, \ldots, 9k$, i.e., the sequence contains exactly $9k$ $k$-digit numbers.

It remains to calculate the answer. We have found 9 single-digit terms in the sequence, $9 \cdot 2$ two-digit terms, $9 \cdot 3$ three-digit terms, and so on, $9 \cdot 2016$ terms with 2016 digits. Therefore, the number of the first 2017-digit term in the sequence is

$$
9(1+2+3+\cdot+2016)+1=\frac{9 \cdot 2016 \cdot 2017}{2}+1=18298225 .
$$