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Egyptian Multiplication: One possible result of a night of Egyptian sex.

But on a more mathematical note, the style of multiplication used in ancient Egypt is a rather interesting process. What we know of this system came from the translation of the Rhind Papyrus. The actual procedure they used for multiplying utilized no single operation more difficult than doubling. Also, as a big fan of powers of 2, seeing this method really blew my mind.

First, a general algorithm to multiply A*B

Step 1: Create a vertical column starting at 1 (20); list all powers of 2 down the paper (unless you happen to have a sheet of authentic papyrus lying around) stopping at the largest value that does not exceed B.

Step 2: Make a second vertical column to the right of the first, starting with the value A. In each row below A write the double of the row above it. Continue in the manner until this column has as many rows as the column in step 1.

Step 3: (A little knowledge of binary makes this step a breeze) mark off the rows such that the figures in the left column that are marked total to exactly B.

Step 4: Sum up the marked numbers from the right column and you're done! Easy as pi.

Now, a specific example of this algorithm, using actual numbers. Let us multiply together 18 and 43
Step 1: the powers of 2

	1
	2
	4
	8
	16
	32

Step 2:  the matching doubles of A

	1	18
	2	36
	4	72
	8	144
	16	288
	32	576

Step 3: flag the desired rows (1+2+8+32 == 43)

	1	18  -
	2	36  -
	4	72
	8	144 -
	16	288
	32	576 -

Step 4: add up the flagged figures from the second column:

18+36+144+576 = 774

See also:
Egyptian Division
Egyptian mathematics

Valtteri Suomalainen's book "Sauna Syyriassa" describes a similar method being used by the Syrians, using e.g. rocks. Say we want to multiply 10 by 6:

  1. Make two piles of rocks (or what ever material is being used) side by side, 10 rocks in one and 6 in the other. Let's name the piles A0 and B0 (the order doesn't matter). In this case, let's call the bigger pile A0.
       A   B
    0 10   6
    
  2. Make another pile (A1) under pile A0, containing half as many rocks as there is in A0, rounding down; in this case, 5 rocks (floor(10/2) rocks). Then make a new pile (B1) under pile B0, with twice the number of rocks in pile B0; in this case, 12 rocks.
  3.    A   B
    0 10   6
    1  5  12
    
  4. Repeat the previous step until you get a pile An with only 1 rock, and a corresponding pile Bn. In this case, Bn will contain 48 rocks.
  5.    A   B
    0 10   6
    1  5  12
    2  2  24
    3  1  48
    
  6. Remove all the lines k where Ak is even. In our example, these are lines A0 (=10) and A2 (=2).
  7.    A   B
    0 10   6
    1  5  12
    2  2  24
    3  1  48
    
  8. Add together all the remaining B's:
    B1 + B3 == 12 + 48 == 60 == 10 * 6
    QED.

Any information on the origins of this particular method would be most welcome.

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