number-systems
I was studying the conversion technique between Hexadecimal and Binary where
For
example
$4C2_{16} = 010011000010_{2}$
Can be done by substituting 4 bits for each Hexadecimal digit.
Why is it so easy to achieve this conversion to binary by combining a series of 4 bits to represent each Hexadecimal digit?
This obviously does not work for decimal to binary conversions, but I am not too sure why either.
For example
$15_{10} ≠00010101_{2}$
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.
HINT:
Let's consider the case of the Gaussian integers of the form $n+im$ and let the base be $-1+i$.
In general, if one has a base and a number to expand in that base then the one will divide the number repeatedly by the base and will take notes of the remainders. I claim that in the case of the base $-1+i$ the remainder is $0$ or $1$ if the number to be expanded is a Gaussian integer.
So, take a Gaussian integer $n+im$ and divide it by the base chosen:
$$\frac{n+im}{-1+i}=\frac{(n+im)(-1-i)}{2}=\frac{m-n}{2}-i\frac{n+m}{2},$$ where the nominator and the denominator have been multiplied by $-1-i$ (the complex conjugate of the base).
If both $n$ and $m$ are even or both of them are odd then $m-n$ and $m+n$ both are even and the result of the division is a Gaussian integer, $$\frac{m-n}{2}-i\frac{n+m}{2}$$
and the remainder is $0$.
If only one of $n$ or $m$ is even (the other one is odd) then one can consider the following version of the quotient:
$$\frac{n+im}{-1+i}=\frac{m-n+1}{2}-i\frac{n+m-1}{2}-\frac{1}{2}(1+i).$$ In this case the result of the division is the following Gaussian integer:
$$\frac{m-n+1}{2}-i\frac{n+m-1}{2}$$ and the remainder is $1$. Indeed $$\big(-1+i\big)\big( \frac{m-n+1}{2}-i\frac{n+m-1}{2} \big)=n+im-1.$$
One can repeat this prescription with the resulting Gaussian integers until, as a result of the repeated divisions by $2$, the original number disappears and the sequence of the remainders stays...
Let's illustrate the algorithm by the example given in the OP: $$24-i11.$$
If $$(m-n)-i(n+m)=(24-(-11))-i(24+(-11)=-35-i13$$
(See the second pair of columns in the table below) was divisible by $2$ then the third pair of columns would contain the same without any modification and the remainder would be zero (See the column called Remainder). Since $-35-i13$ is not divisible by two, there is a modification according to the formula $$(m-n+1)-i(n+m-1)=-34-i12$$ and the remainder is $1$. The first red arrow points at the result of the first division. (Division by $2$ is performed now.) Then the procedural step detailed above will be repeated until the corresponding value in the third pair of columns has become $0+i0$.
$ \color{white}{bbi}$
If one multiplies the remainders by the different powers of the base $i-1$ and sums up the results then the result will be the complex number which was to be expanded:
$ \color{white}{bbbbbbbbb}$
Finally, we have $$(24-i11)_{10} =(110010110011)_{-1+i}.$$
I hope that this hint above will help the OP solve the other problems. This is not impossible since the skeleton of the method is the same as that of the one that we use in case of the simplest base conversions in "the every day life."
Best Answer
Conversion from $2$ to $16$ is simple since $16=2^4$ so every $4$ digits count for one digit.
Meanwhile, bases $10=2\cdot5,16=2^4$ don't have a pattern for exponents of $10$ and $16.$
There is a way to convert bases, though.