The Russian Peasant Method and the Egyptian Method are identical, except that the Russian Method uses a math trick to determine which factors of 2 must be used.
As a recap, to solve 9x13 using the Egyptian Method we find 1x9=9, 2x9=18, 4x9=36, 8x9=72. We then determine which combination of these equations gives us 13 9's. Finally, we sum up the right hand side of all of these equations. The end result is 13x9 = 1x9 + 4x9 + 8x9 = 9+36+72 = 117.
The issue in the method is that we must determine which combinations of these equations gives us 13 9's.
When we first look at the Russian Method's algorithm, there are two steps that seem chaotic: Only paying attention to the odds on the left, and ignoring all the decimal remainders. BUT, these two steps actually compensate each other perfectly. The remainders are actually accumulated in the odd rows while we simplify the expression.
Let's take advantage of the fact that x*y = 1*x*y = 2/2 * x*y=x/2 * 2*y.
In equation 1, we have expanded 13 x 9 = 6 x 18 + 1 x 9. Note that we have converted this multiplication problem to a multiplication and addition problem, where we added a remainder. The most important note is that the remainder is EQUAL TO THE RIGHT NUMBER IN THE ROW ABOVE.
Equation 2 does not introduce a remainder, only halves and doubles the two numbers in the multiplication.
Equation 3 converts the multiplication aspect into two additions, one being a remainder and the final being the largest doubling we reach.
In the Russian Method, we add the Right numbers in the rows where the Left numbers are odd. But what we are actually doing is adding half of the Right numbers in rows that are not integers. It just happens that these two statements are equivalent.
Even rows will not create a remainder. Even rows with a remainder are just represented using the row above, and this row above must be odd because it created a remainder.
Therefore, the Russian Method is the same as the Egyptian Method but uses this neat remainder trick to determine which combinations of equations are needed!
Hi Evan! OK, I've been through your explanation a couple of times, and I almost (but not quite) see the generality of it. I like your expansion of the original question to x*y = 1*x*y = 2/2 * x*y=x/2 * 2*y -- that is a neat way to approach the doubling and halving! I can see that odd numbers in the left hand column produce remainders, and even numbers don't -- but I am still having a slight problem seeing how the remainders are accounted for in the odd-numbered lines. It would be helpful if you could show me in class today! I'm missing seeing the generality of this one important step. We'll hold off on doing a 'reveal' with the rest of the class as others are still working on it till next Wednesday, and it will be interesting to see a variety of approaches!
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