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When I was in middle school, I remember once discussing with my friends different methods of counting with one's fingers, that is, assigning a numerical value to each finger so one could count the natural numbers up to a maximum number. For example, the configuration$\dots$ $$ \begin{array}{c|l l l l l} \text{Side\Finger} & \text{Pinky} & \text{Ring} & \text{Middle} & \text{Index} & \text{Thumb}\\ \hline \text{Left} & 1 & 1 & 1 & 1 & 5 \\ \text{Right} & 10 & 10 & 10 & 10 & 10 \end{array} $$ $\ldots$has a maximum number of $59$, and it is clear that one can represent every number between $1$ and $59$ using only one's fingers using this configuration.

My question is this:

What is the maximum number $M$ one can count to using only one's fingers, such that there exists a hand representation for each $n \in \mathbb{N}, n \leq M$? Also, why does this given configuration of numbers and fingers have a larger maximum than other configurations?

Unlike digital representations in base $n$, each number need not have a unique hand representation, although I suspect the correct answer will have a unique representation for each number.

I do not know the answer to this question and so don't truly expect a conclusive answer to be given any time soon, if at all.

Feel free to provide the best hand configuration that you could come up with, as well as any insights which may lead us all to the correct answer.

Thanks for reading and good luck!

simonv
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    I can count to 1023 using only my fingers. Each finger represent a power of 2 from 0 to 9. – dan post Nov 06 '18 at 06:50
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    What restrictions are you placing on the use of fingers? For example, the ancient Mesopotamians purportedly had a base-60 system based on pointing to joints of one hand with different fingers of the other. If you only want to count whether each finger is “included” or not, then you only get $2^n$ as an information-theoretic upper bound. – platty Nov 06 '18 at 06:53
  • Nice question! I think some Asian countries use a method of dividing fingers into different parts even, so one finger can have three or four different states. That gives us already $3^{10}$ or $4^{10}$ options. I guess what it comes down to, is to how many parts you can divide a finger ... – Matti P. Nov 06 '18 at 06:54
  • Less challenging (both gymnastically ans regarding the conversion of the finger positions) than methods for $2^{10}$ or more would be to adjust your scheme to make the right thumb worth 50, allowing you to count to 99. – Hagen von Eitzen Nov 06 '18 at 07:19
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    I think anyway we have to see each of the ten fingers as a digit (notice the pun!). The question then boils down to, what is the largest base we can use, or how many states a finger can have. If we can have $n$ states per finger, then we can represent $n^{10}$ numbers using both hands. – Matti P. Nov 06 '18 at 07:48

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