topics: |  math | number-theory

An ethiopian multiplication process

A Colonel in Ethiopia needed to buy 7 cows at 22 Maria Theresa dollars each,
but no one could compute it so a priest was called.  The priest's boy came
and dug two rows of holes.  Into the top (which was called the doubling row)
he put 7 pebbles; and into the other he put 22 (divide-by-two row).  From the
22 in row 2, 11 pebbles were put on the next hole, then 5 (11/2 rounded up)
in the next, then 2  in the next and finally 1.

Correspondingly in the doubling row, 14, 28, 56, and 112 pebbles were put in
the next four holes.  Now, the houses with even numbers (in the dividing row)
are evil (both, "2" and "22"); and all the pebbles are removed from such
evil houses in both rows.  So we have:
	   h1  h2  h3  h4  h5
	   7   14  28  56  112	 doubling row
	   22  11   5   2   1	 dividing row
Since 22 and 2 are evil the "7" and "28" are removed.  The sum of the
remaining pebbles, 14+28+112 is 154, the desired answer.

This works because 10110 is the binary equivalent of 22, which means
7 x 22 = 7 x [2⁴ * 1, 2³ * 0, 2² * 1, 2¹ * 1, 2⁰ * 0] which is
the result.

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