How can I prove that the following statement is either true or false ?
$$\forall \space n, k \in \mathbb{N}^* \space \text{s.t.} \space (n+1)\lfloor \log n\rfloor-\frac{10}{9}(10^{\lfloor \log n\rfloor}-1) = kn, k = \lfloor \log n\rfloor-1$$ which is equivalent to: $$10^{k+1} \leq n < 10^{k+2}$$
(N.B. : $\log n$ is indeed the base-$10$ logarithm of $n$.)
Actually, it seems to be true, regarding the first values of $k$ and $n$ :
| Values of $k$ | Values of $n$ | Values of $\lfloor \log n\rfloor-1$ |
|---|---|---|
| $1$ | $108$ | $1$ |
| $2$ | $1,107$ | $2$ |
| $3$ | $11,106$ | $3$ |
| $4$ | $111,105$ | $4$ |
| $5$ | $1,111,104$ | $5$ |
| $6$ | $11,111,103$ | $6$ |
| $7$ | $111,111,102$ | $7$ |
| $8$ | $1,111,111,101$ | $8$ |
and so on (notice the pattern describing $n$ first values). However, I have absolutely no idea how to show it.
Try rearranging the condition $$ (n+1)\left\lfloor\log_{10} n\right\rfloor - \frac{10}{9}\left(10^{\left\lfloor\log_{10} n\right\rfloor} - 1\right) = kn $$
into this form: $$ \left(\left\lfloor\log_{10} n\right\rfloor - k\right) n = \frac{10}{9}\left(10^{\left\lfloor\log_{10} n\right\rfloor} - 1\right) - \left\lfloor\log_{10} n\right\rfloor. $$
Let $x = \left\lfloor\log_{10} n\right\rfloor.$ You have $x \in \mathbb Z$ and $x \geq 0$ (since $n$ is a positive integer). Now for any such value of $x,$ solve for $k$ and $n$ in
$$ \left(x - k\right) n = \frac{10}{9}\left(10^x - 1\right) - x $$
under the condition that $10^x \leq n < 10^{x+1}.$ Note that the condition $x \geq 0$ is necessary but not sufficient for there to be a solution with positive integers $k$ and $n.$