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This post is largely copied from a response to the 17 Camels Trick MO posting. Here I ask:

Q. Does anyone (in any country) teach this binary method of decimal multiplication?

It strikes me as easy as (if not easier than) the traditional method of multiplication.

It is a method for multiplying two decimals, say $13 \times 27$. Form two columns headed by $13$ and $27$. Halve $13$ and discard any remainder, and double $27$. Continue halving the first column and doubling the second until the first column reaches $1$:

\begin{array} \mbox{13} & 27 \\ {\color{red}{6}} & {\color{red}{54}} \\ 3 & 108 \\ 1 & 216 \end{array}

Now discard every row for which the first column is even ($\color{red}{6}$ above). Sum the remaining elements of the second column: $$27 + 108 + 216 = 351 = 13 \times 27 \;.$$ This is of course using the binary representation $13 = 2^0 + 2^2 + 2^3$, and excluding the $\color{red}{2^1}$ row $\color{red}{6} \;\; \color{red}{54}$.

Joseph O'Rourke
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  • Seems highly unlikely to be taught this way. It's fighting against the base-10 representation itself. – Daniel R. Collins Jun 12 '17 at 01:18
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    This seems to be Russian peasant multiplication. – JRN Jun 12 '17 at 01:29
  • Wikipedia calls it Ancient Egyptian multiplication. – JRN Jun 12 '17 at 01:34
  • @JoelReyesNoche: Spot on---Thanks for identifying that! – Joseph O'Rourke Jun 12 '17 at 01:34
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    @DanielR.Collins: Good point, it is fighting against the decimal system. It is an interesting mix of decimal and binary. – Joseph O'Rourke Jun 12 '17 at 01:36
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    I know it arose once at my school this year (since a coworker of mine tweeted about it at me!) and it has arisen before on MESE in an answer of Robert Talbert's here. I can see this being explored in teacher education courses, but -- at least in the US -- I share Daniel R. Collins' skepticism. – Benjamin Dickman Jun 12 '17 at 04:14
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    This approach would only make sense it doubling a number is easier than multiplying it by anything else. This is, however, not the case: Multiplying by ten is a lot easier, multiplying by five might also be not that hard. A similar approach to your algorithm is square and multiply, to compute not $13 \cdot 27$ but rather $27^{13}$. Here, taking a square can indeed be considered easier than taking other exponents. But for classical multiplication, I would always go with $13\cdot 27 = 10\cdot 27 + 3 \cdot 27$ and thus use decimal representation. – Dirk Jun 12 '17 at 09:18
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    @Bemte: Actually, I do find doubling a number easier than multiplying it by any other single digit number. But perhaps that's just me. Perhaps there is research data on this... – Joseph O'Rourke Jun 12 '17 at 13:52
  • The point is not whether you find it easier, but whether a student who's learning to multiply finds it easier. I'd say that doubling arbitrary numbers is barely easier than multiplying arbitrary numbers. Not to mention, you also need to know how to divide by two. – Javier Jun 12 '17 at 20:59
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    @Javier: Doubling an arbitrary number can be done by adding it to itself. –  Jun 12 '17 at 22:17
  • @Javier: I think it is an interesting question whether doubling/halving is more easily learned. The answer is not obvious to me, and certainly cannot be decided by individual introspection. Perhaps I'll post a separate question on that issue sometime... – Joseph O'Rourke Jun 12 '17 at 22:45
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    Found a claim by Sebastian Hayes that Russian Peasant multiplication "is one of the tools employed by traditional ‘lightning calculators’ and mathematical idiots savants." – Joseph O'Rourke Jun 12 '17 at 22:49
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    I'm not sure if it counts, but I know I almost covered it in a Math for Primary Educators course. I used notes from someone who I believe managed to at least work through an example in the past. I'll try to remember to ask them if they know of it actually coming up in a primary/secondary classroom. – pjs36 Jun 12 '17 at 23:12
  • I think this method is very well-suited for mental multiplication, which is probably why it is associated (in name at least) with "peasants", who presumably perform their mathematics in the marketplace without recourse to writing materials. – mweiss Jun 13 '17 at 02:21
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    I also don't think it's necessarily helpful to think of this as being about binary representations. A more basic way of thinking about it is that each time you move down the table (halving the 1st number and doubling the second) the product remains constant, unless you have an odd number in the left column and have to round down; at that point you lose an amount equal to the number in the right column. So iterate until you get down to $1 \times n$, and then add those lost amounts back in to get the product. – mweiss Jun 13 '17 at 02:24
  • It's nice to have lots of ways to multiply. I like $13\times 27=(20-7)(20+7)=20^2 - 7^2$, which gets us quickly to $351$ – G Tony Jacobs Jun 15 '17 at 02:58
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    I once tutored a (college) student in precalc. When we had to multiply and divide polynomials, he was immediately lost. I eventually was able to work out of him that his schooling had been at a traditional Vedic school and they had ONLY learned the above method for multiplying (and it's related method for dividing). So he was lost as soon as something (e.g. polynomials) didn't have a binary representation. – Aeryk Aug 21 '17 at 18:43

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I am someone (in a country), and I teach it in certain (college) classes.

In a discrete math class I teach it and then use the fact that if you multiply $n\times 1$ with this method (halving $n$ and doubling $1$), you get $n$ as a sum of powers of $2$. This then justifies a common algorithm for converting a base-10 representation of an integer to binary.

In a number theory class, I take these ideas further and use them to develop the the successive squaring and reducing algorithm for computing $a^b \pmod{m}$ efficiently.

I also teach it in a math for elementary teachers class as an alternative multiplication algorithm. (Some of the students in this class seem amazed and delighted, and are willing to work through why it works.)

paw88789
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