Treebeard's Stumper Answer

Crunching 2000
For the last three years (1997, 1998, and 1999), we've had the same challenge: use just the four digits of the year 2, 0, 0, 0 in that order with parentheses and any standard math operations, to make the numbers from 0 to 100. For starters,
0 = 2*(0+0+0) (+,, x, /, and parentheses) 1 = 20!+0+0 (factorial, 0! = 1) 2 = 2^0+0!+0 (^ is the power of, 2^0 = 1, but 0^0 is undefined(?)) I'm sure this is not possible for year 2000. There's just not enough to work with! But what's the first number we can't get for year 2000? It helps that 0! ("zero factorial") equals 1 by definition.
I can't make 11 from the digits of the year 2000, so I think that's the first. It's not far until the next! We've had fun with this stumper for the last few years, but it might be hundreds of years before we can do it again! I also considered a few "illegal operations" including floor (_2.1_ = 2) and ceiling (2.1 = 3) and concatenate (2$(32)=21). Here's the best I can do:
0 = 2 * 0 + 0 + 0 1 = 2 * 0 + 0 + 0! 2 = 2 + 0 + 0 + 0 3 = 2 + 0! + 0 + 0 4 = 2 + 0! + 0! + 0 5 = 2 + 0! + 0! + 0! 6 = 2 * (0! + 0! + 0!) 7 = (2 + 0!)! + 0! + 0 8 = 2 ^ (0! + 0! + 0!) 9 = (2 + 0!) ^ (0! + 0!) 10 = 20 / (0! + 0!) 11 = (0! + 0!) / .2 + 0! 12 = 2 * (0! + 0! + 0!)! 13 = _ sqrt (200) _  0! 14 = _ sqrt (200) _ + 0 15 = (0! + 0! + 0!) / .2 16 =  sqrt (200)  + 0! 17 = 18 = 20  0!  0! 19 = 20  0! + 0 20 = 20 + 0 + 0 21 = 20 + 0! + 0 22 = 20 + 0! + 0! 23 = (2 + 0! + 0!)!  0! 24 = (2 + 0! + 0!)! + 0 25 = (2 + 0! + 0!)! + 0! 26 = .2 ^ (0! + 0!) + 0!
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