Determine the largest multiple of 36 that has all even and different digits.
For a number to be divisible by $36=4 \times 9$, it must be divisible by 4 and by 9. Therefore, the sum of the digits of the number $n$ we are looking for must be divisible by 9.

On the other hand, since all the digits are even, the sum of the digits is also even. Thus, the sum of the digits is at least 18. Since $0+2+4+6+8=20$, the number $n$ must be formed by the digits $0, 4, 6$, and 8.

The largest number we can form with these digits, without repeating, is 8640, which is also divisible by 4, ensuring that this is the number we are looking for.