Pretty short one, really. Given your two hands, what's the largest number you can represent?

Some example representations:

1 finger on each hand: that's 2. 2 fingers on one hand, and 3 on the other: that's 5.

Only rule is that, if you represent x, you should be able to represent all numbers smaller than x using the same method. No negative numbers, only zero and above.

Edit(4 Jan): Pretty impressive answers by everybody. I'd like to clarify this question, however: you'll need to represent each number in a manner that's clearly and easily distinguishable. For example, bending at an angle can be ambiguous - how will you tell apart a representation of a finger bent at 50* and at 55*? At most you could use a straight finger for one number and bent finger for another.

  • $\begingroup$ Message to downvoter: the answer is not 10. $\endgroup$ – cst1992 Jan 1 '16 at 7:21
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    $\begingroup$ I suppose the downvote is because of the ambiguity of the question... $\endgroup$ – Dr Xorile Jan 1 '16 at 7:31
  • $\begingroup$ I have to have some ambiguity. Otherwise I'd be giving the answer away. As I said, the only rule is that 0 to that number should be representable; no gaps. $\endgroup$ – cst1992 Jan 1 '16 at 7:33
  • $\begingroup$ So is my answer not the one you're looking for? $\endgroup$ – Dr Xorile Jan 1 '16 at 7:34
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    $\begingroup$ Print out a sheet of paper with an arbitrarily large list of numbers. Indicate a number by pointing to it. If the sheet isn't allowed, use an imaginary one. $\endgroup$ – f'' Jan 1 '16 at 8:14

Say you have 4 fingers on each hand; all your fingers have three joints which can all be extended or collapsed(representing a 0 and a 1) independent of each other.

You now have 3 bits of binary information in one finger, but you gotta remember, you have 8 in total (8*3=24), so you have 24 binary bits of information in your 8 fingers.

So what's the greatest number you can represent with this system? 111111111111111111111111(base 2)=16777215(base 10)

So using my system you could display over 16 million different numbers, in just 8 fingers, but hang on...

You have 5 fingers on each hand, but your thumbs act differently than your other fingers. They only have 2 joints, so that's another 4 bits of information to our already huge 24 bit number, we now have 28 bits to work with.

So what's the new highest number?

1111111111111111111111111111(base 2)=268435455(base 10).

Wow- a silly 4 bits added to the end allow you to display over 260 million different numbers.

That's a lot of information from just 10 simple fingers, don't ya think?

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    $\begingroup$ You must have some remarkable fingers, if you can bend the first and third joint (but not the second) of your middle finger. Good answer, and welcome to puzzling! $\endgroup$ – frodoskywalker Jan 2 '16 at 10:13

Something around 2,500,000,000,000,000.

If we imagine a plane an arbitrarily far but non-infinite distance in front of me, I can point each of my fingers such that the ray extended from it will intersect that plane at a single point. Let any parallel rays be considered - for this purpose - to be hitting the same point. Based on this, I can point my ten fingers to generate a set of dots in an arbitrary order in both the X and Y axes, with points possibly being equivalent along one or both axes. I may also choose not to point with some or all of my fingers, generating smaller arbitrarily ordered sets of points.

This describes the following set of options:

If no points are the same, there are 10! ways of ordering along a single dimension = 3,628,800.

If a pair of points are the same, then there are 10 choose 2 = 45*9! orders. If a triplet are the same, there are 10 choose 3 = 120 * 8! orders. We continue this sequence on down for a total of (per wolfram Alpha):

sum_2^10binomial(10, X) (11-X)!+10! = 26,065,011

But then we have the cases where one set is already paired off and another set gets paired. This is the same problem, but with (10 choose 2) * 8 choose from 2 up to 8 with a factorial on the end, noting that we need to divide the first case in half to avoid duplication. This gives...

binomial(10, 2) sum_2^8binomial(8, X) (9-X)!-binomial(10, 2) binomial(8, 2)×(7!)/2 = 5,436,405

We do this again for 10 choose 3 * 7 choose from 2 up to 7 etc etc:

binomial(10, 3) sum_2^7binomial(7, X) (8-X)!-binomial(10, 3) binomial(7, 3)×(5!)/2 = 2,184,120

And so on and so forth, keeping in mind that this gets trickier as we have more than two clusters, since we get phrases like Bi(10,2)Bi(8,2)Bi(6,3). We also get a bunch more from the cases where I only point with nine of my fingers (10*9!, plus a few more for each possible set of clusters), the ones where I only point with eight fingers, etc. I can't do the math at 4:00 am, so I'll approximate that we are at about 50,000,000 total cases. But this is just one axis - we can do the same on the other axis, for a total of ~ 50,000,000^2 = 2,500,000,000,000,000 distinct situations. Index them and order to taste.

Old Answer


If I have my fingers extended, each could reasonably touch any finger or (with some effort) combination of fingers on the other hand. Thus we have five fingers, each with (1+5+10+10+5+1) possible finger combinations they could be touching for a product of 32^5 values. Further, each hand could be either side up, and above or below the other - that's 8 more per option - 8*(32^5), which is 2^28 = 268,435,456 options. Some of these may be tricky to physically achieve, but they should all be possible. Pick your favorite ordering.

  • $\begingroup$ This is a remarkable answer, but unfortunately representation in this form is pretty difficult to interpret for somebody who's standing in front of you, for example. $\endgroup$ – cst1992 Jan 4 '16 at 16:26
  • $\begingroup$ @cst Since when was readability for someone else part of the requirements? $\endgroup$ – Deusovi Jan 4 '16 at 16:27
  • $\begingroup$ @Deusovi You're right, there was a communication gap. I have had to edit the question; it was attracting answers of the type that I wasn't looking for. $\endgroup$ – cst1992 Jan 4 '16 at 16:30

Any number!

And you need only 1 finger.

Consider your straight index finger as the number $0$ and your completely bend finger as $x$ where $x$ is any number you like. Now measure the angle that your finger bend at one of the joints of that finger in comparison with the stretched finger and call it $a$. Now every other number is constructed by bending your index finger slightly less and measuring the angle that you bent it and call it $b$. and then your number is $\lceil {b \over a} \cdot x \rceil$.

Of course measuring this angle can only be up to some accuracy but that doesn't matter. The question is not how accurately you can know what the number is but if you can represent every number, and this is true

  • $\begingroup$ You did put me in a soup there... $\endgroup$ – cst1992 Jan 4 '16 at 16:04

Assuming your fingers can only be up or down, there are only 1024 combinations. Using simple binary you can do all the numbers from 0 to 1023. Which would make 1023 the biggest number.

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    $\begingroup$ Now add in up/left/right/down as an aditional 2 bits per hand, and you have 16k, do the dame with angling it forward/backwards and you have 256k+ $\endgroup$ – DrunkWolf Jan 1 '16 at 9:15

$g_5$, as in the notation of Graham's number.

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    $\begingroup$ I'm intrigued to hear how you plan to represent all the numbers less than this as well. $\endgroup$ – Zerris Jan 2 '16 at 2:18
  • $\begingroup$ @Zerris, d'oh, misread the challenge. I will leave this up in case someone much smarter than me thinks of a way. $\endgroup$ – user1717828 Jan 2 '16 at 2:23

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