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)
Not quite full solution
