Does there exist a six-digit number $A$ such that the last six digits of any number in $A, 2A, 3A, \cdots, 500000A$ are not all the same?
[Solution] Let $A$ be a six-digit number.
If $A$ is even, then $500000A$ ends with six 0s.
If $A$ is a multiple of 5, then $200000A$ ends with six 0s.
Therefore, we might as well assume that the last digit of $A$ is $1, 3, 7, 9$. At this point, for any odd digit $d$, there exists a single-digit number $b$ such that the last digit of $Ab$ is $d$.
First, choose $b_{0}$ such that the last digit of $Ab_{0}$ is 1. Let the last two digits of $Ab_{0}$ be $c_{1}1$.

Choose a single-digit number $d_{1}$ such that $d_{1} + c_{1}$ equals 1 or 11. Choose $b_{1}$ such that the last digit of $Ab_{1}$ is $d_{1}$. Therefore, the last two digits of $A(10b_{1} + b_{0})$ are 11. Let the last three digits of $A(10b_{1} + b_{0})$ be $c_{2}11$, choose a single-digit number $d_{2}$ such that $d_{2} + c_{2}$ equals 1 or 11, and there exists $b_{2}$ such that the last digit of $Ab_{2}$ is $d_{2}$. Therefore, the last three digits of $A \times (100b_{2} + 10b_{1} + b_{0})$ are 111. Proceeding similarly, there exists
$$
B_{1} = 10^{5}b_{5} + 10^{4}b_{4} + 10^{3}b_{3} + 10^{2}b_{2} + 10b_{1} + b_{0}
$$

such that the last six digits of $AB_{1}$ are all 1.
For $2 \leqslant k \leqslant 9$, let $B_{k}$ be the last six digits of $kB_{1}$, then the last six digits of $AB_{k}$ are all $k$.
If $A = 999999$, then $B_{1} = 888889$.
Conversely, if $A = 888889$, then $B_{1} = 999999$. $B_{k}$ is the last six digits of $kB_{1}$, note that
$$
\begin{aligned}
kB_{1} = k \cdot (10^{6} - 1) & = k \cdot 10^{6} - k, 1 \leqslant k \leqslant 9, \text{ thus } \\
B_{k} & = 10^{6} - k > 500000.
\end{aligned}
$$

This shows that when $A = 888889$,
$A, 2A, 3A, \cdots, 500000A$ do not have the same last six digits for any of these numbers.