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Now, I am sorry if this seems like a simple question in the title, but I really have not found any answer to this.

Say we have the calculations

64563456435 / 4545432 = x

6456734 / 454543432 = x

645634.56435 / 4545.432 = x

6456.3456435 / 45454.32 = x

or any other two large numbers.

Now what I learned back in school was only manual dividing of numbers through numbers up to 999. But now I would like to know the FASTEST way of manually calculating things like this. Is there any specific algorithm out there to deal with problems like this?

uncanny
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  • There are complex computer algorithms that can shorten the number of steps for really large multiplication and division problems but I doubt that there are any methods that will allow you to do this manually for numbers like the ones you wrote down that are faster than the method you know for three digit numbers. Why do you need to do this rather than using a computer? – Ethan Bolker Oct 10 '17 at 23:46
  • The division algorithm for $x/y$ in decimal arithmetic says the following. Find the largest integer $k$ between $1$ and $9$ and the largest integer $n$ such that $x-(k \cdot 10^n)y \geq 0$. Then write $k$ down in the appropriate position as a digit, and repeat on the problem $(x-(k \cdot 10^n)y)/y$. in decimal arithmetic we will usually choose to take $n$ to be the largest $k,n$ such that $x \geq (k \cdot 10^n)y$ instead, but the principle is the same. – Ian Oct 10 '17 at 23:53
  • It's pretty hard to beat the division algorithm short of having useful lookup tables (which indeed is what people did a century or so ago, using logarithms). – Ian Oct 10 '17 at 23:54
  • @Ian could you explain this in detail? I know of lookup tables, but only for sinus/cosinus etc. – uncanny Oct 11 '17 at 00:08
  • @EthanBolker Well, how would you use the method for 3 digit numbers on such a large divisor? Just guessing and calculating it again? Also, well, computers are not available everywhere and I enjoy doing things in simple, old-fashioned ways. Old-fashioned as in "The fastest way to do things without modern technological assistance". You would not believe how much more inspired and creative you become once you restrain yourself. This is exactly how many of the old formulas were invented. This was very subjective, it is just a reason I like it. – uncanny Oct 11 '17 at 00:09
  • @uncanny $x/y=\exp(\ln(x)-\ln(y))$. So tables of $\ln$ and $\exp$ can be used to perform division by hand. – Ian Oct 11 '17 at 00:21
  • @Ian Thanks, I will try that out! You say the general algorithm is probably second fastest to just looking it up the way you described. Could you please point me to an explanation of it? – uncanny Oct 11 '17 at 00:28
  • Of the division algorithm? It is what you learned in school, just formalized. Of lookups? It is what I just said about $\exp$ and $\ln$ once you have tables of those two in front of you. A fancy way that people used to implement these tables is by slide rule. – Ian Oct 11 '17 at 01:24
  • @Ian Oh, okay, I get it now. I was just asking about the algorithm, but... slide rule? What do you mean by that? Sorry, all this is very interesting to me. – uncanny Oct 11 '17 at 01:28
  • A slide rule is a device with markings on it that can be used to quickly perform the kinds of lookups that you use to do calculations through logarithms. It was the gold standard for such work before computers became widespread. – Ian Oct 11 '17 at 01:29
  • @Ian Thanks again! Yeah I forgot about that device totally. Anyway, your log algorithm works perfectly on, well, computer. I do own lookup tables from 1911 for log but not ln, and I have never heard of any exp table. Do you know how ln and exp tables are structured and how to use them so I can make one myself? Or do you have a link for something where this was already done? – uncanny Oct 11 '17 at 01:58
  • Some tables are based on $\log_{10}$ and $10^x$ instead of $\ln$ and $\exp$. They work the same way, e.g. $x/y=10^{\log_{10}(x)-\log_{10}(y)}$. – Ian Oct 11 '17 at 02:11
  • @Ian Base 10 instead of Base e sounds even a lot more reasonable to me for normal calculating without lookup-tables, am I correct? Thanks, I could have guessed myself that the base was interchangable – uncanny Oct 11 '17 at 02:17
  • No, base e is far easier to use if you're working from scratch. In a sense when you are working from scratch you really convert everything to base e, do your calculation, and then convert back. – Ian Oct 11 '17 at 02:19
  • @Ian Now this amazes me. Why? A complex number like this sounds way harder to work with than simple 10. I am still talking about manual calculation without computers by the way. – uncanny Oct 11 '17 at 02:27
  • It is because it plays nicer with calculus. For example the linear approximation of $e^x$ near $x=0$ is $1+x$ while the linear approximation of $10^x$ near $x=0$ is $1+x \cdot \ln(10)$. Doing these kinds of calculations from scratch is exactly based on this sort of manipulation. – Ian Oct 11 '17 at 02:28

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