I've recently watched the movie 'Stand and Deliver' and an interesting property of the $9$ times table showed up, specifically in this scene. Here it is, summarized:
For $0\leq n\leq 9$, if you want to calculate $n\times 9$, all you have to do is to raise your ten fingers, starting counting up to $n$ from one side, and exclude (bend) the finger you end up with. Then the result is simply given by the two sets of fingers split by the bended one.
For example, if you want to calculate $3\times 9$, and we count from the left (from one's perspective) we proceed in the following manner, where $s$ and $b$ stand for straight and bent fingers, respectively, $$ sssss\,\,\,sssss - 0\\ bssss\,\,\,sssss - 1\\ sbsss\,\,\,sssss - 2\\ ssbss\,\,\,sssss - 3 $$ Hence, the result is given by the number of $s$'s to the left of $b$ and the number of $s$'s to the right of $b$, joining both digits in that order. That is, $2$ and $7$, thus $3\times 9=27$.
I guess it is relatively easy to show why this is the case for the $9$ times table. Furthermore, if one had $100$ fingers, the same argument could me made for the $99$ times table, and so on.
But, what is really going on here? Why doesn't it work for the $8$ times table, even if we consider less fingers, initially? My guess is that the base we are considering plays a big role in understanding this type of counting, but are there any more insights regarding this interesting trick? Is there a 'mathematical generalization of these fingers' that could make the trick work for other numbers?
Any ideas or insights are appreciated.
The divisibility rule of $9$ is that if the sum of the digits of the number is a multiple of $9$, the number itself is divisible by $9$. Hence for two-digit numbers, if one digit is $x$, the other digit has to be $9-x$.
This goes along with the fact that we have $10$ fingers and if we cover any one, the number of remaining fingers will be $9$. The hand trick merely takes advantage of this coincidence.