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

etc.
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 by TomM)

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...