Presented by Bruce C. Friedman
Page images created: January 2002
Last modified: July 1, 2003
Minor Correction: April 16, 2005
The data, as it appears in,
columns 11 and 12 of the Kahun IV, 3 "fragment".
I believe the analysis by R. Gillings correctly identifies the numbers on the
page but is otherwise in error.
Column 11:
| Line # | SUM/Series | Multiplier | Checked: Yes or No | ||||
| I | 1/12, 1/3 | 1 | Yes | ||||
| II | 1/6, 2/3 | 2 | No | ||||
| III | 2/3, 1 | 4 | No | ||||
| IIII | 1/3, 3 | 8 | Yes | ||||
| V | 1/12, 2/3, 3 | 9 | Yes | ||||
Read the above as follows:
[See reference Line # one (I)] Column eleven is a demonstration of Egyptian
multiplication by doubling. In this case 1/12+1/3 is doubled three times to
produce:
[ref II] 1/6 + 2/3
[ref III] 2/3 + 1
[ref IV] 1/3 + 3
Note the diagonal slash [checked Y] denoting that REF I and IV have been combined
to yield:
[ref V] 1/12 + 2/3 + 3 = 9 X (1/12 + 1/3)= 3.75 = The Mean Variant
We do not have clear evidence of how the [ref I] value was chosen or why the
goal of column 11 is to find 9 times this value. But we do see this is part
of the scribes effort to distribute the quantity [100] into the ten portions
shown in column 12.
The [ref V] value: 1/12 + 2/3 + 3 = 3.75
The first portion listed in column 12 is 1/12 + 2/3 + 13 = 13.75 [This is the
greatest distribution]
The ref I value multiplied by nine [the number of gaps between 10 portions]
give us the mean variant!
MV = 3.75
The first numbers in Column 12 (twelve) are 10, 100.
Although this is often read as one number, 110, my analysis is predicated on
this as 10 (ten portions) and 100 (a quantity of one hundred).
The list of ten portions from column 12:
(Hi)=13.75 shown as 13, 2/3, 1/12
The other portions shown as the Hi portion less 5/6 in successive steps which
ends with the Lo portion.
(Lo)=6.25 shown as 6, 1/6, 1/12
This 5/6 (or 10/12) is the typical gap of 2/3, 1/6.
TG = 5/6
Line
# |
Decimal
Sum |
Unit
Fraction Series |
Checked:
Yes to all |
I |
13.7500 |
1/12
+ 2/3 + 13 = 165/12 |
Yes |
II |
12.9166 |
1/12
+ 1/6 + 2/3 + 12 = 155/12 |
Yes |
III |
12.0833 |
1/12
+ 12 = 145/12 |
Yes |
IIII |
11.2500 |
1/12
+ 1/6 + 11 = 135/12 |
Yes |
V |
10.4166 |
1/12
+ 1/3 + 10 = 125/12 |
Yes |
VI |
9.5833 |
1/12
+ 1/6 +1/3 + 9 = 115/12 |
Yes |
VII |
8.7500 |
1/12
+ 2/3 + 8 = 105/12 |
Yes |
VIII |
7.9166 |
1/12
+ 1/6 + 2/3 + 7 = 95/12 |
Yes |
IX |
7.0833 |
1/12
+ 7 = 85/12 |
Yes |
X |
6.2500 |
1/12
+ 1/6 + 6 = 75/12 |
Yes |
TOTAL |
100.0000 |
95
+ 60/12=1200/12 = 100 |
NA |
The total of the above ten
entries is 100.
The integers total 95 and all the other fractions total 60/12.
Given that we only have one way [actually there's another way] to distribute
this quantity using a 3,4,5 triangle, can you explain how this was done or if
this was even an actual consideration made by the scribe?
Again look at the actual fragment!
Column 11 Column 12
Note that the special/unusual double symbol, which here appears to represent the quantity 1/6, is also seen on the ACHMIM wood tablets.
GRID NAME |
TOPIC/LINK |
OLD NAME |
RMP/EMLR |
HOME |
|
EMLR |
CLEAN |
|
EMLR |
UGLY |
|
LINKS |
LINKS &
SPHINX |
|
EMLR |
QUEST |
|
HIERATIC |
H. QUESTIONS |
|
HIERATIC |
H. ANSWERS |
|
KAHUN |
KAHUN INTRO |
|
KAHUN |
KAHUN GEOMETRY |
|
KAHUN |
MORE KAHUN |
|
GEOMETRIC ANALYSIS OF THIS FRAGMENT