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

Not quite full solution by Nick Reed)

51 = ((9+1)*6)-9

52 = (6*9)+1-sqrt(9)

53 = (6*9)-(1^9)

54 = ((6-1)*9)+9

55 = ((1/9)+6)*9

56 = (6*9)-1+sqrt(9)

57 = ((9-1)*6)+9

58 = (6*9)+1+sqrt(9)

59 = 69-9-1

60 = 61-(9/9)

61 = 61+9-9

62 = (6*9)+9-1

63 = (9*1)+(6*9)

64 = (6*9)+1+9

65 = (9*9)-16

66 = ((9-1)*9)-6

67 = 69+1-sqrt(9)

68 = 69-(1^9)

69 = 69*(1^9)

70 = 69+(1^9)

71 = 69-1+sqrt(9)

72 = ((6+1)*9)+9

73 = (6*9)+19

74 = (9*9)-1-6

75 = (9*9*1)-6

76 = (9*9)+1-6

77 = 96-19

78 = ((9-1)*9)+6

79 = 69+9+1

80 = (9*9)-(1^6)

81 = (9*9)*(1^6)

82 = (9*9)+(1^6)

83 = 99-16

84 = ((9+1)*9)-6

85 = (9*9)+sqrt(16)

86 = (9*9)+6-1

87 = (9*9)+(6*1)

88 = (9*9)+6+1

89 = ((6!)/9)+(9*1)

90 = 9*(9+(1^6))

91 = ???

92 = 99-1-6

93 = 99-(6*1)

94 = 99+1-6

95 = 96-(1^9)

96 = 96*(1^9)

97 = 96+(1^9)

98 = 99-(1^6)

99 = 99*(1^6)

100 = 99+(1^6)

(Note: "^" is "to the power of" and "!" is factorial...)