Multiplication

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 in column A, and some other number next to it in column B. Then the product of the two columns at that point is or . When we move to the next row, column A will contain , and column B will contain , and their product will be , which is 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 ‘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.