Why can I not figure out this parlour-trick? 
Which ball are you using?
djm
This has been discussed here before, but I can’t find it… Note that 0
and all the multiples of 9 are the ones with the symbol that you get
from the ball. Start with 11 and try counting up, doing the equation
each time. They always come out to 9. That’s because as you add 1
to get to the next number, you are only adding to the one’s place, so
when you add the digits together, you’re increasing the result of
that addition by 1 each time you move to the next number. When
you subtract, the increment of the number itself and the sum of its
digits negate each other.
At 20, it moves to the next multiple of 9, which is 18.
Feeling stupid…as expected! 
Still, pretty clever!
I forgot: they throw in a few extra of the asterisk
symbol at non-multiples of 9 as a red-herring…
In the other version of this, the symbols randomly changed to help keep it “mysterious,” even though the numbers which would work always had the same symbol as each other.
Am I wrong or does 22 seem to be thrown in there as a red herring?
Fearfaoin…what a great parlour trick! You read my mind!
The Swiss one.
(in case anyone caught it–yes, I put Swedish here first. Meatballs on the brain.)
(10t + d) - (t + d) = (10t - t) + (d - d) = 9t
So whatever number you pick, the result is always a multiple of 9. Specifically, 9 times the tens digit.
You’re speaking my language MTGuru. 
I know how you feel, it took me way too long to figure out 3-cup.
Is this mathematically impossible?
Yes, it is. It’s a dirty, dirty little trick.
Yep, the starting position of the demo is the inverse of the starting position of the puzzle. And the demo can be solved in one move: turn 2 cups. I feel dirty.
No wonder he looks down on us! 
Well then, explain this… (a mathematical illusion)
let a<b
then a=b+c
then mutiply both sides with (a-b) and subtract ac - →
a(a-b)-ac=(b+c)(a-b)-ac →
a2-ab-ac=ab-b2-bc
break out a and b → a(a-b-c)=b(a-b-c)
and finaly divide both sides with (a-b-c) → a=b (!!!)
ah…cheep trick employing bad math?
yep
(b+c)(a-b)-ac is not the same as (b+c) ((a-b)-ac)
And for the true mathamagician there is this one.
a = b
multiple both sides by a — a^2 = ab
subtract b^2 from both sides — a^ 2 - b^2 = ab - b^2
factor — (a -b)(a + b) = b(a-b)
divide both sides by a-b — ((a -b)(a + b))/(a -b) = (b(a-b))/(a-b)
reduce (or whatever it is called) — a + b = b
and since a = b, substitute b for a — b + b = b
add the bs together — 2b = b
divide both sides by b — 2b/b = b/b
2 = 1
The same principle is in operation with this one as with falkbeer’s. Quote this post and see the entry below –
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Both problems involve dividing by zero, which is not allowed in polite or even Christian society.