Solution:
By condition (a), the number $n$ cannot have more than 10 digits. Write $m$ for the number formed by reversing the digits of $n$.

The first digit of $n$ is the last digit of $m$, and as such must be even, and thus at most 8. Assume that the first digit is 8.

Then the second digit of $n$ is the tens digit of $m$, and must be chosen to make the last two digits of $m$ divisible by 4. As 98 and 78 are not divisible by 4, but 68 is, we conclude that the second digit is at most 6. Assume that the second digit is 6.

Then the third digit must be chosen so that the last three digits of $m$ are divisible by 8, but since 968 is divisible by 8 we derive no information from this.

Appending the remaining digits, selecting the greatest possibility at each turn, we conclude that $n$ is at most 8697543210. As this number indeed satisfies the conditions, it is the answer.