5. 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?
5. Let $A$ be a six-digit number. If $A$ is even, then $500000A$ ends with six zeros; if $A$ is divisible by 5, then $200000A$ ends with six zeros. Therefore, we can assume that the last digit of $A$ is $1, 3, 5, 7,$ or $9$. For any odd digit $d$, there exists a one-digit number $b$ such that the last digit of $Ab$ is $d$.

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 one-digit number $d_{1}$ such that $d_{1} + c_{1} = 1$ or 11. There exists $b_{1}$ such that the last digit of $Ab_{1}$ is $d_{1}$, so 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 one-digit number $d_{2}$ such that $d_{2} + c_{2} = 1$ or 11. There exists $b_{2}$ such that the last digit of $Ab_{2}$ is $d_{2}$, so the last three digits of $A \times (100b_{2} + 10b_{1} + b_{0})$ are 111. By continuing this process, 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 $2 \leqslant k \leqslant 9$, if $B_{k}$ is the last six digits of $kB_{1}$, then the last six digits of $AB_{k}$ are all $k$. If $A = 999999$, then $B_{1} = 888889$. Let $A = 888889$, then for $1 \leqslant k \leqslant 9, B_{k} = 1000000 - k > 500000$, so 888889 satisfies the problem's requirements.