At my first stab at this, I can get to 48 78. I avoided using superscript, assuming that that would violate your math rule.
We choose the digits:
5, 6, 7
Which can net us the following numbers:
756, 576, 765, 567, 675, 657
Flip the 6 around to get
759, 579, 795, 597, 975, 957
Tilt the 7 ever-so-slightly to get
156, 516, 165, 561, 615, 651
Flip the 6 and tilt the 7 to get
159, 519, 195, 591, 915, 951
Write the 5 very strongly on the back of the paper to show through and get
726, 276, 762, 267, 672, 627
Then the nonsense with the other two digits again:
129, 219, 192, 291, 912, 921
729, 279, 792, 297, 972, 927
126, 216, 162, 261, 612, 621
Then let's throw in the 1 and 2 digit numbers to the count (thanks @WeatherVane) for a bonus 30:
1, 2, 5, 6, 7, 9
12, 15, 16, 19
21, 26, 27, 29
51, 56, 57, 59
61, 62, 65, 67
72, 75, 76, 79
91, 92, 95, 97
Let's try to find 26 numbers. The regular permutations of non-zero digits gives 15 distinct numbers, which is 11 too few:
5, 6, 7, 56, 57, 67, 65, 75, 76, 567, 576, 657, 675, 756, 765
But then if we
arbitrarily choose the "flip the 6" from before, we get:
9, 59, 79, 95, 97, 597, 579, 759, 795, 957, 975
which gets us up to 26. Let's say that's where the overlap is, and that the OP found more by using one or more of the above tricks that I listed.