([]){{}({}[((((()()()){}){}){}){}]<>)<>([])}{}<>([][(()()()()){}]){({}[()]){([]){{}{}([])}}}{}([{}])({}({}){})({}({})({}){})({}(({}){}){})({}(({})({})){}{})({}(({})({}){}){})({}((({}))({}){}){}{})({}((({}){}){}){})(({}(((({})){}){}){}{}{}<(((()()())()){}{})>)){{}({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{}[{}]<(())>){((<{}{}>))}{}(<()>)}{}
Truthy is 1, Falsy has a 0 on the top of the stack.
Try it Online!
I'm pretty sure there is a shorter way to do the multiplication in a loop, but I haven't found it yet.
#reverse and subtract 48 from all numbers (ASCII -> decimal)
([]){{}({}[((((()()()){}){}){}){}]<>)<>([])}{}<>
([][(()()()()){}]) #height - 8
{({}[()]){ #if not 0 subtract 1
([]){{}{}([])} #if still not 0 pop everything
}}{} #this loop pops everything unless there are 8 or 9 digits
([{}]) # -I
({}({}){}) # H*2
({}({})({}){}) # G*3
({}(({}){}){}) # F*4
({}(({})({})){}{}) # E*5
({}(({})({}){}){}) # D*6
({}((({}))({}){}){}{}) # C*7
({}((({}){}){}){}) # B*8
(({}(((({})){}){}){}{}{} # A*9 pushed twice with:
<(((()()())()){}{})>)) # 11 under it
{{} #if not 0
({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{} # mod 11
[{}]<(())>){((<{}{}>))}{} # logical not
(<()>) # push 0 to exit loop
}{}
# implicit print
