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*(99)*5
2 = (199)/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(4): Missing numbers by friedlinguini)
From the site where I originally saw this posted, 41 looked like one of the tougher ones, since they had a few versions for most of the numbers, but for 41 only the following:
41 = (√9)! + ((√9)! * 6)  1

Posted by levik
on 20020815 09:46:03 