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Apparently multiplication is the cool topic of the week for math bloggers.

Dave over at Math Notations is holding a poll about how multiplication should be taught.

And there is a cool new blog called Relearning Math where Math Learner is trying to improve his mental math skills, and has posted several posts on tricks and methods for paper and mental multiplication.

Russian Peasant Multiplication

In keeping with the multiplication theme, I am blogging about Russian Peasant Multiplication today. Russian Peasant Multiplication (sometimes just called Peasant Multiplication, and probably many other names as well) is a cool multiplication method I learned about only recently. The only skills required to carry it out are doubling, halving and addition.

Suppose I want to multiply 27 x 58. I would put 27 and 58 at the tops of two columns. Then I would repeatedly halve 27, dropping any fractional part, and double 58. To get the product, I add all of the numbers in 58’s column that are next to an odd number in 27’s column, like this:

 A B add 27 58 58 13 116 116 6 232 3 464 464 1 928 928 Answer: 1566

Why does it work? Wikipedia explains it in terms of powers of two. But I think there’s another explanation that is more intuitive and easier to understand (although far less concise).

If we weren’t dropping the fractional parts, then whenever we halve what is in column A and double what is in column B, the product of column A x column B does not change. (We’re just moving a factor of two around, but multiplication is commutative and associative, so that doesn’t change anything.) But since we are dropping fractional parts, we lose part of the product whenever we halve an odd number. And the amount we lose from that product is precisely the amount that was in column B next to the odd number before we halved it.

To see this algebraically, imagine that we have some odd number $2n+1$ in column A, and some other number $m$ next to it in column B. Then the product of the two columns at that point is $(2n+1)(m)$ or $2mn + m$. When we move to the next row, column A will contain $n$, and column B will contain $2m$, and their product will be $2mn$, which is $m$ less than the previous product.

At the bottom of the chart, we have 1 in column A and some number in column B. The product of the two original numbers is equal to 1 times <whatever is at the bottom of column B>, plus all the $m$‘s we dropped as we went through. By adding all up all of the numbers next to odd numbers, including the one next to the number 1 at the bottom, we add all the dropped pieces back into the answer, and get the correct product.

Edited to add: I have just found a nice Dr. Math page that illustrates an explanation of why it works, similar to my explanation above, but with more pretty pictures.

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16 Comments leave one →
1. Pseudonym permalink
January 27, 2008 11:11 pm

One that I haven’t seen mentioned is Karatsuba multiplication. This is, incidentally, a really serious algorithm; it’s one of the method used by computers to multiply insanely large numbers.

The idea is from ring theory, but for our purposes, all you need to know is the following decomposition:

(qa+b) (qc+d)
= q^2 ac + q((a+b)(c+d) – ac – bd) + bd

(For two digit numbers, for example, you’d use q=10.)

Because you can share the computation of ac and bd, you only need to perform three multiplications instead of the standard four.

Taking 27*58 as our example, we need:

ac = 2*5 = 10
bd = 7*8 = 56
(a+b)(c+d) = 9*13 = 117
Then 27*56 = 100*10 + 10*(117 – 10 – 56) + 56

The cool thing about Karatsuba multiplication is that you can apply it recursively. You can multiply two four-digit numbers by using q=100. This gives you three basic multiplications of two digit numbers, each of which can be performed using q=10. In total, you perform nine basic multiplications, compared with the 16 that you’d do in standard long multiplication or the lattice method.

2. January 27, 2008 11:17 pm

Very cool, Pseudonym, especially the recursiveness. Thanks for contributing it! It does look like a nicer algorithm for computers than humans, though, as there seems to be a lot to keep track of. However, I bet someone could invent a tabular or lattice type representation of it to make it human-friendly.

3. Pseudonym permalink
January 29, 2008 12:20 am

It would be nice if you could do that, yeah. If you’re only doing one step, it’s straightforward to do on paper, but it’s the recursion, particularly on the middle term, that makes it unwieldy.

Incidentally, if you do construct a lattice, and you colour in each base multiplication that you have to perform, you get a fractal:

http://blog.wolfram.com/2007/09/arithmetic_is_hardto_get_right.html

4. January 29, 2008 12:49 am

I had actually seen that fractal before — it is pretty cool. Thanks again for commenting and contributing a cool algorithm.

5. Monique Trundy permalink
September 21, 2008 1:29 pm

Hello,

I am an Education Major at the University of Maine, Farmington. I was wondering if I could use your image of the man with the mathematical symbols in a presentation I’m giving about Google Earth. I would greatly appreciate it.

Thanks for your time.
~Monique Trundy

6. September 21, 2008 1:46 pm

Monique,

The image is not mine, it is free clip art from pppst.com (as indicated in the image). The exact page that image came from was http://www.phillipmartin.info/clipart/math.htm and it indicates that the clip-art is free for non-profit use.

Hope that Helps!

7. October 29, 2008 5:06 am

Dear Mrs.Ryan,
I like the way you put up the math probloms I need so work on it I appeartitte your work.

Love,
Madeleine.

P.S I love as a teacher and friend.
P.S.S Loves and kisses, Madeleine

8. June 20, 2009 2:57 pm

Hey, nice tips. I’ll buy a bottle of beer to that person from that forum who told me to visit your blog :)
p.s. Year One is already on the Internet and you can watch it for free.

9. January 1, 2011 6:38 pm

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12. August 2, 2013 12:53 am

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14. April 24, 2014 4:50 am

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