Bob,

From the responses I guess you can conclude that your implementation is
optimal. However, there may be one way to improve it - I just don't have
time to follow it through. If you write the 16bit numbers:

 N1 = A*2^8 + B
 N2 = C*2^8 + D

Where A,B,C, and D are 8-bit numbers, then:

 N1 * N2 = AC*2^16 + (BC+AD)*2^8 + BD

The middle term can be re-written:

BC+AD = (B-A)(C-D) + (AC + BD)

Or

 N1*N2 = AC*2^16 + ((B-A)(C-D) + (AC + BD))*2^8 + BD

Notice that the products in the (AC+BD) term now appear twice in the
expression. So for the cost of more additions, you can replace one of
the multiplications.

At first glance, this approach doesn't seem like it'll buy anything - but
it's worth a shot.

Scot

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