The numbers: {1, 2, 4, 12, 14, 21, 24, 41, 42, 124, 142, 214, 241, 412, 421}.
If repeats are allowed, the above plus: {11, 22, 44, 111, 222, 444} plus infinitely many more by preceding the numbers given so far by various combinations of {1, 2, 4} in 1, 2, 3, ... digit groups making 4, 5, 6, ... digit numbers.
45
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
9915 1159 5555 9999 1111
25 digits
6
45
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
Using only arrangements of the given digits: without repeating digits, 15. with repetition, 39. The answer does not include numbers such as 2^3 = 8 or 3^(2+1) = 27
9915 1159 5555 9999 1111
25 digits
6
1,956 different numbers can be made from 6 digits. You can calculate this by using the permutation function in a summation function, like this: Σ6k=1 6Pk = 6P1+6P2+...+6P5+6P6 What this does is calculate how many 1 digit numbers you can make from 6 digits, then how many 2 digit numbers can be made from 6 digits and adds the amounts together, then calculates how many 3 digit numbers can be made and adds that on as well etc.
18 can.
162
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77
binary.
9