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Re: Egyptian fraction math only used quotient and remainder statements
Posted:
Dec 23, 2011 9:24 AM
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Written up in:
http://ahmespapyrus.blogspot.com/
e. Problems 39, 40, 62-68, Arithmetic Progression Problems
RMP 40 solved for the smallest term, x1, in a 5-term series that summed to 100. Formula 1.0, from Wikipedia's Kahun Papyrus analysis, found x5 in a five-term series that summed to 60, with a share difference of 5 1/2, per:
x5 = 11/2(1/2)(5 -1) + 60/5 = 11 + 12 = 23.
Ahmes found: x3 = 60/5 = 12, x1 = 1, x2 = 6 1/2, and x4 = 16 1/2.
The 5-term sum to 100 was solved without finding is share difference 9 1/6. Ahmes computed x2, x3 x4 and x5 using the fact that arithmetic proportions contain paired members. In this case, x3 = 100/5 = 20, and x1 + x5 = x2 + x4 = 40, with x1 = 1 2/3 solved the problem (1 2/3 + x5 =40, x5 = 38 1/3, and so forth).
Ahmes allocated precious metal, loaves of bread between mean, division of fat, grain and cattle, the cattle cited as tribute by the same arithmetic proportion formula:
xn = (d/2)(n -1) + S/n, (formula 1.0)
with xn (largest term), d (difference between shares), n (number of terms in series), and S (sum of all terms in the series). Ahmes was usually given three of the four variables ( xn, d, n and S) and solved for the fourth variable. In RMP 40 Ahmes was given two variables, with a third available as a 1/7 relationship to the sum of the first two-terms to the final three-terms, in a five-term series. Ahmes created a proportional 5-term series that set S = x2 + x3 + x4 of the unknown five-term series, as mentioned above.
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