# Fast arithmetic, without a calculator?

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?

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Related... –  Guess who it is. Sep 8 '11 at 19:18
"Is it really crucial to be able to crunch numbers on the fly?" - well it's sorta kinda embarrassing for me to take long in figuring out how much change I ought to be getting, or reckoning how much to tip the waitress. Otherwise... –  Guess who it is. Sep 8 '11 at 19:21
Well, that's not really a problem... But more intensive operations, indeed, drag me down. –  Curiosity Sep 8 '11 at 19:24
Crucial for what? It can be useful in some circumstances, but it really depends on your career goals. A mathematician doesn't really need it, although occasionally mental arithmetic can come in handy in a calculus lecture. –  Robert Israel Sep 8 '11 at 23:53

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.
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.