I am starting to doubt that there are no solutions for the seven remaining
7 remaining
67
69
83
84
86
87
93
That's not allowed ,just asked
82=((2+2)!!)/(.(0!))+2
10 remaining
86 to 89
91
I need to ask if that's ok or not
We can use infinite nested square roots
15 Remaining
71
74
77
79
82 to 84
Can u think of some of the 80s?
16 to go
Any other ideas?
Tools
5=Sqrt((.2)^(-2))=(.2)^(-0!)
48=((2+0!)!)!!
8=(2+2)!!
100=(.(0!))^(-2)
6=(2+0!)!
15=5!!
Remaining
67 to 69
Maybe we can use infinite square roots