> No the Bellboy has $2 > The Till has $25 > The Clients paid $9 each > Regardless there is still $1 missing >
I am going to make the assumption that you are not just trying to pull our legs.
After all, if we accepted your version of the problem, there would be 2 + 25 + (3 X 9) = 54 dollars in existence. Not missing just 1 dollar, but 24. :-)
Reminds me of an Abbot and Costello routine, where the clever Bud Abbott cheats his friend Lou out of money, by borrowing and repaying in quick succession. The trick is to start off owing money, borrow to repay it, pay back and then borrow more, so that it appears the original amount is "missing" and no longer owed. I expect that some similar scam long ago led to the invention of double entry bookkeeping.
The innkeeper riddle is simple. I.e., 30-5+3+2=30. One gets confused by the shell game operation of dividing out among the customers and bellboy -- but looking at the arithmetic in the equation above, one can clearly see that no money has disappeared; 3 + 2 is still 5 and -5 + 5 is still 0. The original $30 is all accounted for. 30-5+3+2-30=0
A deeper point can be made of this simplicity. When the arithmetic is written in a clear numerical equivalence, the properties of identity, reflexivity and transitivity that characterize an arithmetic -- or algebraic -- equation, are manifestly obvious. One shouldn't think of these terms as mere mathematical jargon; they underpin the logic of problem solving in both literary and mathematical language. They map abstract symbols to facts. Learning how to do that in ever more meaningful ways may prevent one from making irrational conclusions about missing money that isn't really missing. Or, more importantly, about "Jewish mathematics."