
Re: RMP # 48
Posted:
Oct 7, 2007 6:15 PM


Franz,
Your translation of RMP 33 looks more like a transliteration, a method used by linguists, rather than a math problem, written by mathematicians. To a mathematician Egyptian algebra involves working Ahmes' algebra problems and translating Ahmes' work into modern arithmetic, adding nothing, leaving nothing out.
Ahmes' arithmetic is shown on Wikipedia, per:
http://en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus
with the pertinent section copied per:
"The RMP's 84 problems begin with six divisionby10 problems, the central subject of the Reisner Papyrus. There are 15 problems dealing with addition, and 18 algebra problems. There are 15 algebra problems of the same type. They ask the reader to find x and a fraction of x such that the sum of x and its fraction equals a given integer. Problem #24 is the easiest, and asks the reader to solve this equation, x + 1/7x = 19. Ahmes, the author of the RMP, worked the problem this way:
(8/7)x = 19, or x = 133/8 = 16 + 5/8, "
RMP 33 was and is processed in the same manner, or
(97/42)x = 33
x = 1386/87 = 14 28/97
Ahmes broke 28/97 into two parts, 2/97 and 26/97 and used the RMP 2/n table series for 2/97,
found by the HultschBruins method
2/97  1/56 = (112  97)/(56*97) = (8 + 7)/(56*97), or
2/97 = 1/56 + 1/679 + 1/776
(pretty neat, right?)
with Ahmes converting 26/97 by an extension of the HultschBruins method, as noted by:
26/97  1/4 = (104  97)(4*97) = (4 + 2 + 1)/(4*97), or
26/96 = 1/4 + 1/97 + 1/194 + 1/388
such that:
28/97 = 1/4 + 1/56 + 1/97 + 1/194 + 1/388 + 1/679 + 1/776
Is Ahmes' logic and answer clear? Note that I have not worked any of Ahmes' intermediate algebra steps in unit fractions, as you have done. Modern arithmetic translations do not require unit fraction aspects to be written out, since beginning vulgar fraction conversions are the key arithmetic facts.
Best Regards,
Milo

