1=2/2 2=2 3= (2/(2/2/2))!/2!/(2/(2/2/2)) = 4!/2!/4
assume that we can get any positive integer from 1->n
now: if n%2==1 =>exist k<=n: k/(2/2/2)=k*2= n+1=>done
if n%2==0 =>exist k<=n k/(2/2/2)=k*2= n+2 => n+1 =(n+2)!/(n+2)/n!
Only using: ! ; ( ; ) ; / and digits: 4 ; 5 Can you make an equations to get result = 2 p/s: we have 45/5 = 9 4!=24 5/4=1.25
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