Performing mental arithmetic without a base I'm sure many of you will be aware of the amazing ability for some people to 'see' mathematical calculations as shapes, and to perform mental arithmetic with very little conscious effort, simply by manipulating the imaginary shapes of each number together. See this YouTube for a fascinating documentary about somebody who does just this.
Since these calculations happen on an unconscious level (the answers just 'come to them'), it seems unlikely that their methodology (albeit subconscious) would be rooted in the decimal system (i.e. base 10 math), since this system is simply a convenient way for us to conceptualise large numbers by breaking into groups that we can represent using our fingers.
We often hear of the advantages of performing mental arithmetic using alternative bases, such as 8 (octal) or 16 (hexadecimal); but how about systems that do away with bases completely? Does anyone know of any methods for mental arithmetic which are unconstrained by having to work in a base? Obviously the answers would need to be converted back into decimal at the end in order for us to conceptualise them, but surely some work must have been done into ways of manipulating numbers like this?
 A: The video shows someone doing large numbers of digits.  This is a base-like calculation.  
Computers (ie people who can do rapid calculations), can do this with some ease, but it takes practice.  I can do several places of criss-cross calculations in various bases, like 120, and 10, but the mind does not hold the digits easily.  
I'm pretty sure that some of the super feats done by the Trachenberg system gives the same sorts of results.  In any case, the person might have a greater difficulty with a different base, because, like the chinese stone-cage, the process uses features of base 10.
My grandfather could add money columns (£,s,d) as fast as one can run a finger down the column.  I suppose, it's not so much the mathematics, or the form of the number, but the kid in the video can just hold hundreds of digits and do lots of simple arithmetic slides.
On Bases
The only 'real' numbers are the digits.  Anything else is represented as a path of remainders.  So for example, '7' is a real thing, but '73' is a pathway to a number.  The same numbers in dozenal are '7' and '61'.  
Something like '7!' is a different path to a number.  One can for example, learn 7! in ten or dozenal or 120, to get eg 5040, 2E00, and 4200.  But these are not 'conversions' or even calculations.  They're the outcome of rote-learnt tables.  Other numbers do not lend themselves to such fast conversion.  A similar-sized 5120, i do not know its dozenal form, but its twelfty form is 4280.  I suppose you can write it as 5040 + 80, and this allows one to directly write 2E68 for the dozenal.
In all of the examples above, the numbers represent a series of remainders, but i did not go through the remainders to find the number.  I saw 5040 as 7!, and wrote 7! in the various bases.  
Some numbers i know only in base 10, some in base 12, some in base 120.  The order of a gosset's E8, is 3.43.24.00.00.  If you know further that this is 1.72*10!, you could work it out in dozenal, by noting that 10! = 1270000, and that 127 is dec 175, and this divided by 9 gives 19 4/9.  From this one gets 1754 00000.  The decimal, you could get by multiplying 6912 * 1008 * 100, which i think is 696729600.  
Many years ago, i did a project to find the index of primes 2-19 for the primes: this is for a primitive root g, then eg $g^i=2 \pmod p$.  This of course, suited the factor-model, because you would see something like, eg 507, and you say, $3.13^2$, and the like.  I found it hard going when $p$ got to 56.00 (6720).  I could not look at a number like say 3135 and say immediately, that it is 3*5*11*19.  (this one i could, but there are others i couldn't that had fairly small divisors.)  The factor method is not really something one does.
Some features are learnt through the fingers.  A typist might tell that a word is typed wrong because the feeling from the keys is that.  I know more that eg 696729600 feels right, rather than looking at the digits.  Likewise, i can type in the various short-chords of the polygons, (the chord which makes the third side of a triangle of two edges), usually without thought, eg 1.801937736 or 1.93185165259 for {7} and {12}.  I have typed these in quite often.  The heptagon was done on a ten-digit calculator, the dodecagon is older, was done on a 12-digit calculator.
A: Yes, it is possible to do a mental calculation on a problem of numbers described a totally different way than any positional number base. In some cases, answering the problem is much easier than determining what those numbers are in any base. Let's define two functions from $\mathbb{N}$ to $\mathbb{N}$ $f$ and $g$ recursively as follows.


*

*$f(0) = 1$

*$g(0) = 0$

*$\forall n \in \mathbb{N}f(n + 1) = f(n) + 2g(n)$ and $g(n + 1) =  f(n) + g(n)$
Suppose you're then asked "What is $(f(8))^2 - 2(g(8))^2$?" It's very easy. There's no need to first calculate $(f(8))^2$ and $(g(8))^2$ in any base. All you need to do is show by induction that for any even natural number $n$, $(f(n))^2 - 2(g(n))^2 = 1$ and for any odd natural number $n$, $(f(n))^2 - 2(g(n))^2 = -1$. This can be done as follows. For any natural number $n$, $(f(n + 1))^2 - 2(g(n + 1))^2 = (f(n) + 2g(n))^2 - 2(f(n) + g(n))^2 = (f(n))^2 + 4f(n)g(n) + 4(g(n))^2 - 2(f(n))^2 - 4f(n)g(n) - 2(g(n))^2 = -((f(n))^2 - 2(g(n))^2)$
