# Multiplying two decimals using (camouflaged) binary representation

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 $\color{red}{6} \;\; \color{red}{54}$.

• This seems to be Russian peasant multiplication. – Joel Reyes Noche Jun 12 '17 at 1:29
• @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 1:36
• 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 4:14
• 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 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 2:24

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.