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...)
re: Not quite full solution
