Forward and Reverse Engineering AWT & RMP data

Nine data points will be forward and reverse engineered as ancient
scribes would have recognized. Five data points are found in the
Akhmim Wooden Tablet, 1/3, 1/7, 1/10, 1/11 and 1/13, and four
are found in the RMP, 1/6, 1/20, 1/40 and 100/70, following the
modern remainder arithmetic form:

(64/64)/n = Q/64 + R/(n*64)

with Q = quotient and R = remainder.

For example, AWT 1 divides 1/3 of a hekat as:

(64/64)/3 = 21/64 + 1/(3*64),

then the scribe converted the remainder data (often mentally) into
a fixed common divisor 1/320, a number that scribes named ro.
To better explain the scibal method of compulation, which may
have looked like this to the scribe:

(64/64)/3 = (4' 16' 64')hekat + ( 1 3")ro, consider the following,

A. Forward computation: follow a remainder arithmetic structure,
computing a quotient (Q) and a remainder (R), a two-part statement:

1. AWT 1:

A Foward Engineeriing

(64/64)/3 = 21/64 (Q) + 1/(3*64)(R)
= (16 + 4 + 1)/64) + (5/5)*1/192)
= (4' 16' 64') hekat + 5/960
= (4' 16' 64') hekat + (5/3) ro
= (4' 16' 64') hekat + (1 3") ro


Note, the 2nd remainder step multiplied 5/5 * 1/(3*64) = 5/(3*320)
thereby creating one common divisor multiple, 1/320, for all five
AWT solutions.

B. Reverse engineering computation: begin with (4' 16' 64') hekat
+ (1 2")ro and find the original problem. First, it must be noted that
2' = 32/64. 4' = 16/64, 8' = 8/64, 16' = 4/64, 32' = 2/64 and
64' = 1/64, or:

(1) Quotient: (4' 16' 64') = (16 + 4 + 1)/64 = 21/64


(2) Remaidner, Egyptian fraction: ( 1 2/3) = 5/3

(3) Remainder, including ro: 5/3 ro = 5/(3*320)
= 5/960
= 1/192

C: Double check the result consists of taking a Q value like 21/64
and an R values, such as 1/192 and adding them:

21/64 + 1/192 = (63 + 1)/192 = 64/192 = 1/3.

2. RMP 83 - 1

A. Forward Engineering

(64/64)/6 = 10/64 + 4/(6*64)
= (8 + 2)/64 + (20/6 * 1/320)
= (8' 32') hekat + ( 3 3') ro

B. Reverse Engineering

(1) Quotient part (8' 32') = (8 + 2)/64 = 10/64

(2) Remainder part (3 3') ro = 10/3 * 1/320 = 10/960 = 1/96

C. Proof ( Q + R):

10/64 + 1/96 = (30 + 2)/192 = 32/192 = 1/6

which means that B (2) actually should read:
(3 3') ro = 20/6 * 1/320 = 20/1920 = 1/96

3. AWT 2

A. Forward Engineering

(64/64)/7 = 9/64 + 1/(7*64)
= (8 + 1)/64 + (5/7)* 1/320
= (8' 64') hekat + (2' 7' 14') ro

B. Reverse Engineereing

(1) quotient part: (8' 64') = (8 + 1)/64 = 9/64

(2) remainder part: convert Egyptian fraction

2' 14' 28' = (7 + 2 + 1)/14 = 10/14 = 5/7, or

(3) remainder part including ro : 5/7 ro = 5/(2240) = 1/448

C.Proof: (Q + R):

9/64 + 1/448 = (63 + 1)/448 = 64/448 = 1/7

4. AWT 3

A. Forward Engineering

(64/64)/10 = 6/64 + 4/(10*64)
= (4 + 2)/64 + 20/(10)* 1/320
= (16' 32') hekat + 2 ro

B. Reverse Engineering

(1) quotient: (16' 32') = (4 + 2)/64 = 6/64

(2) remainder: 2 ro = 2/320 = 1/160

C. Proof (Q + R):

6/64 + 1/160 = (30 + 2)/320 = 32/320 = 1/10


5. AWT 4

A. Forward Engineering

(64/64)/11 = 5/64 + 9/(11*64)

= (4 + 1)/64 + (45/11)* 1/320

= (16' 64') hekat + (4 11') ro

B. Reverse Engineering

(1) quotient: (16' 64') = (4 + 1)/64 = 5/64

(2) remainder, Egyptian fraction: (4 11') = ( 44 + 1)/11 = 45/11

(3) remainder including ro: 45/11 ro = 45/11 * 1/320 =
45/(11*320) = 9/(11*64) = 9/704

C. Proof: (Q + R):

5/64 + 9/704 = (55 + 9)/704= 64/704 = 1/11

6. AWT 5

A. Forward Engineering

(64/64)/13 = 4/64 + 12/(13*64)
= 16' + (60/13)* 1/320
= 16' hekat + (4 8/13 ) ro
= 16' hekat + (4 2' 13' 26') ro

B. Reverse Engineering

(1) quotient: 16' = 4/64

(2) remainder, Egyptian fraction:

4 2' 13' 26' = (104 + 13 + 2 + 1)/26
= 120/26 = 60/13
(3) remainder including ro

(60/13)*ro = 60/(13*320) = 12/(13*64)

C. Proof: (Q + R):

4/64 + 12/(13*64)= (52 + 12)/(13*64)
= 64/(13*64) = 1/13

7. AWT 83 - 2

A. Forward Engineering

(64/64)/20 = 3/64 + 4/(20*64)
= (2 + 1)/64 + (20/20)*1/320
= (32' 64') hekat + 1 ro

B. Reverse Engineering

(1) quotient: (32' 64') = (2 + 1)/64 = 3/64

(2) remainder, Egyptian fraction 1, insufficient data

(3) remainder including ro : 1 ro = 1/320

C. Proof: (Q + R):

3/64 + 1/320= (15 + 1)/320
= 16/320 = 1/20

which means the Egyptian fraction step B (2) should be 20/20,
or 4/(20*64)

8. RMP 83 - 3

A. Forward Engineering

(64/64)/40 = 1/64 + 24/(40*64)
= 64' + (120/40)*1/320
= 64' hekat + 3 ro

B. Reverse Engineering

(1) quotient: 64' = 1/64

(2) remainder, Egyptian fraction 3, insufficient data

(3) remainder including ro: 3 ro = 3/320

C. Proof (Q + R):

1/64 + 3/320 = (5 + 3)/320= 8/320 = 1/40

which means that the Egyptian fraction was 120/40, or
24/(40*64)

9 . RMP 47

A. Forward Engineering

(6400/64)/70 = 91/64 + 30/(70*64)
= (64 + 16 + 8 + 2 + 1)/64
+ (150/70)*1/320

= (1 4' 8' 32' 64')hekat
+ (2 7') ro

B. Reverse Engineering:

(1) Quotient: (1 4' 8' 32' 64') = (64 + 16 + 8 + 2 + 1)/64
= 91/64

(2) Remainder, Egyptian fraction: (2 7') = 15/7

(3) Remainder including ro: (15/7) ro = 15/(7*320)
= 3/(7*64)

C. Proof: (Q + R)

91/64 + 3/(7*64) = (637 + 3)/(7*64)
= 640/(7*64)
= 10/7


The 100/70 fact seems hidden. However, the point that is found
in the scribal narrative, meaning the initial problem was 100/70.

In summary the forward and reverse processes work equally well
for the 33 additional data points found in RMP 69 and 81. Oddly
scholars reporting RMP 81 only cited hin data. A hin was 1/10th
of a hekat, and implementing forward and reverse engineering takes
the hin fact into account, while exactly scaling to 1/3200th in the
remainder term as scribes first calculated the data.