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RMP 36, and the 2/n tabl.e
Posted:
Aug 1, 2009 9:27 AM


RMP 36 solved 3x + (1/3)x + 1/5(x) = 1 hekat,
Ahmes considered LCM 15 by:
(45 + 5 + 3)x/15 = 1, and
(53/15)x = 1, such that
x = 15/53
Ahmes converted 15/53 thinking: (15/53)*(4/4) = 60/212
and writing:
(53 + 4 + 2 + 1)/212= (1/4 + 1/53 + 1/106 + 1/212)hekat.
Ahmes also converted 30/53, written as 2/53 + 28/53, 5/53, and 3/53 to unit fraction series by the central 2/n table red auxiliary method within two proofs.
1. First 2/n table related proof:
a. 15/53*(4/4) = 60/212= (53 + 4 + 2 + 1)/212 = 1/4 + 1/53 + 1/106 + 1/212.
b. (15/53)*2 = 30/53 = 2/53 + 28/53 = (2/53)*(30/30) + (28/53)*(2/2) = 1/53 + 1/318 + 795 + 1/2 + 1/53 + 1/106
c. 5/53 = (5/53)*(12/12) = (53 + 4 + 2 + 1)/(12*53)= 1/12 + 1/159 + 1/318 + 1/636
d. 3/53 = (5/53)*(20/20) = (53 + 4 + 2 + 1)/(20*53)= 1/20 + 265 + 1/530 + 1/1060
e. sum: 15/53 + 30/53 + 5/53 + 3/53 = 53/53 = one hekat (unity)
2. Second proof scaled 15/53, 30/53, 5/53 and 3/53 (hekat) by inserting 2/n table red auxiliary data:
a. 20 + 10 + 5 = 35, scaled (15/53  1/4) = (4 + 2 + 1)/212]*(5/5)= (20 + 10 + 5)/1060
b. (35 + 1/3) + (3 + 1/3) + (1 + 2/3) + 20 + 10 scaled 30/53 =(2/3)*[(28/53*(2/2) + (2/53)*(30/30)]
c. (88 + 1/3) + (6 + 2/3) + (3 + 1/3) + (1+ 2/3) scaled (5/3)*[(5/53)(15/15) = (53 + 4 + 2 + 1)]
d. 53 + 4 + 2 + 1 scaled (3/53)*(20/20) s 60/1060 = (53 + 4 + 2 + 1)/1060
e. this RMP 36 proof showed that
15/53 = (1/4 + 1/53 + 1/106 + 1/212)hekat
was the singular focus of the problem.
Discussion: Ahmes' scaling of (15/63)+ (30/53) + (5/53)+ (3/53) = 1
was one proof of 15/53, a fact mentioned by Peet.
A second (70 + 100 + 80) proof was related to 1/4 + 1/265 = (260 + 4)/1060, and its common denominator 1060, offered protonumber theory information were not cited by Peet or Chace.
In 2009 is it clear that 2/n table red auxiliary numerators were reported as LCM's in this problem, a major fact that is generally reported by http://rmprectotable.blogspot.com/ .
Best Regards,
Milo Gardner



