Egyptians & Russian Peasants To The Rescue!
I was never the world’s most natural mathematician. I wish I’d been taught the tricks used to solve multiplication problems by the Egyptians and Russian peasants.
Consider the sum 58 x 93. That would take me a while to work out normally. I don’t even know where to start, truth be told.
However, if I were an Egyptian this problem would resolve itself fairly easily.
You have to create a column starting at 1 and doubling in turn. From these numbers, keep those that add up to the multiplier; those that are unnecessary will be jettisoned. For the multiplier 58, we get the sequence 1, 2, 4, 8, 16, 32. (We go no higher, since the next row would be 64, which is bigger than the multiplier.) We’ll note that 1 and 4 won’t be needed, since the other digits added together make 58.
We take the same approach in the second column, only this time starting with the multiplicand and doubling in steps. In our case this gives us 93, (2 x 93 =)`186, (4 x 93 =) 372, (8 x 93 =) 744, (16 x 93 =) 1488, (32 x 93 =) 2976.
Remember that we don’t need to use the rows which contain the values 1 and 4 in column 1. If we add the remaining rows in column 1 we get 58, the multiplier, as we already know. If we add the corresponding rows in column 2, we get (186 + 744 + 1488 + 2976 =) 5394 … which your handheld calculator will confirm is the product of 58 and 93 
Not to be outdone, Russian peasants also had a pretty swell method of solving unwieldy multiplication problems.
The procedure is fairly straightforward. You list the multiplier in one column and the multiplicand in another. As you divide one column repeatedly by two you double the other.
You’re only concerned with what lies to the left of the decimal point, so 3½ and 3.5 will be listed as 3. You repeat until your division leads you to 1.
We are only interested in those rows which contain odd numbers in column one. Sum their equivalents in column two (the sequential doubling of the multiplicand) and you get … 5394. Magic
An additional bonus is that the technique holds whichever you put in the divisor column; here’s a table where I’ve swapped the order around, halving the multiplicand and doubling the mulitplier. The result, happily, is the same
December 12th, 2008 at 10:58 am
Erm… At the risk of sounding dumb, I don’t understand either of those… Surely our version of long multiplication is a lot quicker and more straight-forward?
December 12th, 2008 at 2:27 pm
I worded everything really badly. I’ve rewritten a chunk of it and included a couple of tables to demonstrate what’s going on. See what you think now
December 12th, 2008 at 3:01 pm
Aha. It turns out I’m not thick at all, twas all your fault Babel for explaining it in a muddled way
Both methods make much more sense now you’ve added the tables, and both look pretty cool! Cheers!
March 15th, 2009 at 9:32 pm
[...] Russian peasants and the Ancient Egyptians were kind enough to impart the knowledge of how to handle these numbers and come up with an answer with very little effort [...]