Find the largest number $n$ having the following properties:

(a) No two digits of $n$ are equal.

(b) The number formed by reversing the digits of $n$ is divisible by 8 .

Remark. $n$ cannot start with 0 , but it can end with 0 .
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.

Remark. Note the logical structure of this solution. Everything but the last sentence amounts to a proof that no number $n>8697543210$ satisfies the conditions, by considering the ways that a number $n$ can exceed 8697543210: it could have more than 10 digits, the first digit could be 9 , the first two digits could be 89,88 , or 87 , and so on. When all these possibilities have been eliminated, it only remains to verify that the established upper bound, 8697543210, is indeed a number satisfying the conditions (a) and (b).