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This has been on my mind for quite a while now... Is it really crucial to be able to crunch numbers on the fly?
I have considerate difficulty making out the quotient of $1 / 0.732 $ for example. I can approximate it well, but still (~1.3...). When I go for precision, I am too slow.
Are there techniques to accelerate arithmetic operations?
Doerfler's book Dead Reckoning: Calculating Without Instruments contains algorithms specifically designed for computing nontrivial things mentally. But it's still not easy and requires quite a lot of practice, I guess. To begin with, one must be really good at multiplying two-digit numbers (so I'm lost there already...).
The first step should be to simplify as much as possible. Start by getting rid of the decimals: $$ \frac{1}{0.732}=\frac{1000}{732} $$ Knowing divisibility tests, it's easy to see that the numerator and denominator are divisible by 4, so let's reduce the fraction: $$ \frac{1000}{732}=\frac{1000\div4}{732\div4}=\frac{250}{(700\div4)+(32\div4)}=\\ \frac{250}{175+8}=\frac{250}{183} $$ The numerator is larger than the denominator, so we can strip out whole numbers: $$ \frac{250}{183}=\frac{183}{183}+\frac{67}{183}=1\frac{67}{183} $$ At that point, you technically have your answer, $1\frac{67}{183}$, but you wanted the decimal places.
For decimals, there's a simple method for denominators which end in 9 that I refer to as Leapfrog Division. There's also a slightly more advanced version for denominators ending in 1.
I can see that the $183$ will easily scale up to end in a 9 by multiplying by 3, so we'll multiply the numerator and denominator by 3: $$ 1\frac{67}{183}=1\frac{67\times3}{183\times3}=1\frac{201}{549} $$ Now, we're ready to use the leapfrog division trick to get the decimals. Start by writing down the $1$ followed by decimal point, just to get it out of the way. Next, divide $201$ by $55$ ($55$ is used because $549$ is next to $550$) so as to get a quotient and a remainder: $$ 201\div55=3, \ remainder \ 36 $$ Write down the quotient, $3$, to the right of the decimal point: $$ 1.3 $$ Next, create a new problem by taking the remainder, $36$, and "leapfrog" it in front of the quotient, $3$, to make a new number, $363$. Just as before, you'll divide this number by 55: $$ 363\div55=6, \ remainder \ 33 $$ Write this new quotient to the right of your previous results: $$ 1.36 $$ Again, leapfrog the remainder, $33$, in front of the quotient, $6$, to make a new number, $336$, which you'll divide by $55$. Keep repeating this approach for as many decimals as you deem appropriate, or until you notice they're repeating: $$ 336\div55=6, \ remainder \ 6 \\ Result: \ 1.366 \\ 66\div55=1, \ remainder \ 11 \\ Result: \ 1.3661 \\ 111\div55=2, \ remainder \ 1 \\ Result: \ 1.36612 \\ 12\div55=0, \ remainder \ 12 \\ Result: \ 1.366120 \\ 120\div55=2, \ remainder \ 10 \\ Result: \ 1.3661202 \\ 102\div55=1, \ remainder \ 47 \\ Result: \ 1.36612021 \\ 471\div55=8, \ remainder \ 31 \\ Result: \ 1.366120218 \\ $$ ...and so on. I stopped at $1.366120218$, but you shouldn't have any problem understanding how to continue. According to Wolfram|Alpha, this calculation is right as far as it goes.
There are methods which apply a few short-cuts for speedy calculations. I would suggest you to go through Trachtenberg's Speed System or Vedic mathematics.
