Thanks Sparky what is the logic you followed?Unfortunately I don't know what the right answer is.
here we go,
The last digits can be 6,7,8, or 9
so if it is 9, for example, there are 2 more digits before it that need to be chosen from 6,7, and 8 (order matters here) so 3!/1! permutations = 3!=6
so there will be 6 numbers with 9 in the end, 6 numbers with 8, etc
total 6x4 = 24 numbers
so sum of last digits will be (6+7+8+9)*3!
do the same the middle and the first digits
and you get the number
Dan did the same thing basically but he accounted for repeats, so in his case 999 is possible but it isn't possible in my solution