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RMP40: Ahmes' attested method
Posted:
Mar 14, 2009 3:09 PM


Ahmes' attestable solution to RMP 40 requires the creation of a proportional 5term arithmetic proportional, with a difference of 5 1/2, series that summed five members to 60.
Was such a 5term, S = 100 series, with an unknown d, easy to find?
Luca Miatello published in HM a method that may correct the traditional 'false position' aspect of 100 years of reporting scholar method to read RMP problems discussed trial and error (false position) and algorithms.
This math forum post offers a direct method that show how and why Ahmes selected the d = 5 1/2, S = 60, 5term series, without using trial and error.
Ahmes has been shown to have used a well defined method contained in the Kahun Papyrus (KP), published years earlier, as applied in RMP 64. This is the method that Ahmes used to calculate the needed 5term series, with d = 5 1/2 and S = 60.
Discussion: The KP and RMP 64 method finds the largest term (xn) in an arithmetic progression, by citing the number of terms (n), the share difference (d) and the sum (S), written as:
(d/2)(n1) + S/n = xn (formula 1.0)
Three of the four equation parameters are needed to solve for the 4th unknown parameter. Formula 1.0 was a flexible method. For example, knowing x1, d and n, S can be found (and so forth).
Formula 1.0 and its Kahun P. and RMP 64 useages, was confirmed by John Legon, and others, several years ago.
Returning to the question: could have Ahmes applied formula 1.0 to create a series in RMP 40 based on d = 5 1/2, n= 5, and S = 60?
Plugging in the parameters,
xn = x5 = (5 1/2)x (5  1) + 60/5 = 11 + 12 = 23
Ahmes did write out
x5 = 23 x4 = 15 1/2 x3 = 12 x2 = 6 1/2 x1 = 1
At this point, it may be clear to the reader that Ahmes also knew that x3 = S/5 = 12.
Secondarily, the third fact needed to apply formula 1.0 was written in terms of the sum of the first two terms (x1 + x2) was 1/7th the sum of the last three terms (x3 + x4 + x5).
But was this required fact built in as 5 1/2 = d was selected? The answer is yes.
Proof: by plugging in another value for d, say 5, and S = 60 5term series reports impossible data:
x5 = 5*(5  1)/2 + 60/5 = 12 1/2 + 12 = 24 1/2. x4 = 19 1/2 x3 = 14 1/2 x2 = 9 1/2 x1 = 4 1/2
since,
x1 + x2 = 4 1/2 + 9 1/2 = 14 x3 + x4 + x5 = 14 1/2 + 19 1/2 + 24 1/2 = 54 1/2
14 x 7 = 98, was too large, and did not contain a fraction. Therefore, Ahmes knew to select a fraction when selecting d.
Try 5 1/3
x5 = (5 1/3)* (5  1) + 60/5 = 21 1/3 x4 = 16 x3 = 10 2/3 x2 = 5 1/3 x1 = 0
which was also rejected, this time for two reasons:
1. x1 could not be zero 2. x3 had to equal 12
Again, another impossible answer. That is, 5 1/2 is the only value for d that creates a 5term series that summed to 60 by using formula 1.0.
Thirdly, why did Ahmes choose S = 60? This issue should be discussed.
Ahmes chose S = 60 because he knew:
x2 + x3 + x3 = 60
an
x1 + x5 = 40 x2 + x4 = 40 x3 = 20
in any 5term series, with S = 100.
Finally, the third clue stated in the problem, that the first 2terms were 1/7th the last twoterms, is mute. It was not needed to find the 5term S = 60, d = 5 1/2 series. It is only needed to confirm an interesting fact related to both proportional 5term series.
Q.E.D.



