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*

I'm missing 3 values - 29, 41 and 91... (plus I've probably got a typo/mistake somewhere else...)

0 = (9-9)*6*1

1 = ((9-9)*6)+1

2 = ((9+9)/6)-1

3 = ((9+9)/6)*1

4 = ((9+9)/6)+1

5 = ((9*6)/9)-1

6 = (1*6)+9-9

7 = 1+9+6-9

8 = (9/9)+6+1

9 = 9*(1^(6+9))

10 = 9+(1^(6+9))

11 = 9+9-1-6

12 = (9+9-6)*1

13 = 9+9+1-6

14 = 9+6-(1^9)

15 = (1+9)*9/6

16 = 19-9+6

17 = 9+9-(1^6)

18 = (9-6-1)*9

19 = 9+9+(1^6)

20 = (9*9)-61

21 = (6+(1^9))*sqrt(9)

22 = 19+9-6

23 = 9+9+6-1

24 = (1*9)+6+9

25 = 1+6+9+9

26 = ((9-6)*9)-1

27 = (9-6)*9*1

28 = ((9-6)*9)+1

29 = ???

30 = (9*6)-((sqrt(9)+1)!)

31 = (96/sqrt(9))-1

32 = (96/sqrt(9))*1

33 = (96/sqrt(9))+1

34 = 19+9+6

35 = (6*9)-19

36 = (9+1-6)*9

37 = 91-(6*9)

38 = 99-61

39 = ((9-1)*6)-9

40 = ((9-sqrt(16))!)/sqrt(9)

41 = ???

42 = ((6-1)*9)-sqrt(9)

43 = 61-9-9

44 = (9*6)-9-1

45 = (9*6)-(9*1)

46 = (9*6)-9+1

47 = (9*6)-((sqrt(9))!)-1

48 = 6*(9-(1^9))

49 = 61-9-sqrt(9)

50 = 69-19

(values 51-100 follow)