Galperin G.A.

A natural number $N$ is 999... 99 ( $k$ nines) times greater than the sum of its digits. Specify all possible values of $k$ and for each of them, provide an example of such a number.
Answer: such a number exists for any $k: N_{k}=9 k \cdot\left(10^{k}-1\right)$. Let $9 k=\overline{s_{1}^{1}} \overline{-\ldots} S_{t}^{-} \overline{0}-.0$ ( $s_{t}$ is not equal to 0, there may be no zeros at the end). We will check that the sum of the digits of the number $N_{k}$ is equal to $9 k$. Write the difference between the numbers $9 k \cdot 10^{k}$ and $9 k$, taking into account that $9 k<10^{k}$ for any $k$ :

![](https://cdn.mathpix.com/cropped/2024_05_06_75db48f86176e8f64efag-33.jpg?height=188&width=792&top_left_y=312&top_left_x=10)

In the bottom row, the number $N_{k}$ is written. It is easy to see that the sum of its digits is

$s_{1}+\ldots+s_{t}-1+9+\ldots+9+(9+1)-s_{1}-\ldots-s_{t}=9 k$.

(in the left part of the equation, there are $k$ nines)