Using the digits in 1996 and any operations (but not mathematical constants), try to write equations that have the numbers from 0 to 100 as the answer.
For example with 1995:
0 = 1*(9-9)*5
2 = (19-9)/5
Provide as many as you can. Digits 1,9,9 and 6 do not have to appear in order. (But each digit has to be used - 1 and 6 once, 9 twice.)
This is more of a game than a puzzle
(In reply to re(3): Missing numbers
Well, levik has given us a solution for 41 that doesn't need the idea of floor or ceiling.
And I just thought of one for 91 on the way home:
91 = 91+6-((sqrt(9))!)
So, if we can just find something for 29 we'll have done all 101 without using any exotic operators. Let's go for it - there's only one left after all...