The sum of the digits is 45, which is divisible by 9. So any such number will be divisible by 9. In order to be divisible by 11 (and therefore by 99), the sum of the even-numbered digits must differ from the sum of the odd-numbered digits by a multiple of 11. The only ways to write 45 in this way are 17+28 and 6+39. You can't have four or five distinct digits with sum 6, so we are looking for *either* four digits with sum 28 (so that the other five have sum 17), *or* five digits with sum 28 (so that the other four have sum 17). A bit of thought shows that the best you can do in the first case is 28=4+7+8+9 and 17=1+2+3+5+6, giving the number 142738596. In the second case you can have 28=1+3+7+8+9 and 17=2+4+5+6, giving the number 123475869.