Thursday, May 7, 2009

RMP 50 TO 61

RMP 51-60

The units of measures used in the RMP geometry problems will generally be listed first, as a guide to the reader. Several RMP problems, like RMP 53, have garbled one or more units of measures. RMP 53 is one of several problems that are projected to be resolved writing this blog - over the coming weeks.

050309_RMP Geometry problems from prior handwritten notes: notes from 040909-041109

(FROM Bruce Friedman, 5/7/09)

RMP #51: (units mentioned: cubit, khet, setat)

UNSPECIFIED TRIANGLE TYPE WITH BASE OF 4 KHET UNITS AND OTHER DIMENSION OF 10 KHET UNITS

RMP SAYS: YOU ARE TO TAKE HALF OF 4, NAMELY 2 TO FIND ITS RECTANGLE [!].
YOU ARE TO MULTIPLY 10 BY 2.

THIS IS ITS AREA.
i.e. 2 "thousands of land" which = 200,000 cubits squared and = 20 square khet = 20 setat

I WILL REFER TO THIS AS 2 "TOL"
RMP WORKINGS:
1
400 [CUBITS]
1
1000 [CUBITS]
/2
200
[=2 khet]
2
2000

regarding the type of triangle:
If isoceles the result is exact. Same exactness if this is a right triangle.
If scalene, the answer is exact only if 10 khet refers to the height, otherwise only a good approximate. Likely AE FORMULA IN ACTION: 1/2 x base x height = area
IMHO: AHMES INTENDED THIS TO BE AN ISOCELES!

RMP #52: (units mentioned: khet)

RECKONING OF A TRUNCATED TRIANGLE OF LAND - A TRAPEZOID.
20 KHET IN HEIGHT, 6 KHET AT BASE AND 4 KHET AT CUT SIDE [TOP]
RMP SAYS: COMBINE TOP AND BASE [ADD 4+6] RESULT 10; HALVE THE 10 [NAMELY 5]

TO FIND ITS RECTANGLE.

YOU ARE TO MULTIPLY 20 FIVE TIMES*

* HERE AHMES FINDS AN ERRANT RESULT OF 20 TOL.

THE CORRECT ANSWER IS 10 TOL.

likely AE FORMULA IN ACTION: 1/2 x (base PLUS TOP) x height = area OF TRUE TRAPEZOID WITH PARALLEL BASE AND TOP SIDES.


RMP #53: (units mentioned: setat, mh and COL)

Three areas problem deal with two of triangles, and a third undefined shape.

RMP WORKINGS:

CHECK 1

4 /2 [SETAT]

THESE CHECKED MULTIPLICANDS 1, 2, AND 1/4

CHECK 2

9

YIELDS TOTAL: 4 1/2 1/8 (45/8) SETAT (TRIANGLE #1)

A TRIANGLE WITH 4 1/2 khet (base), and 9/4 (altitude) khet (SOLVED)

AHMES ADDED ONLY 1 AND 1/4 and 4 1/2 SETAT

AND FOUND 5+5/8 SETAT = 45/8 SETAT


FURTHER DISCONNECTED? RMP WORKINGS:

1/10 OF IT IS (1+ 1/4 + 1/8 cubit strips (MH^2)) AND 10 "CUBITS OF LAND" [HENCEFORTH COL]

AHMES FURTHER SHOWS:

1/10 OF IT SUBTRACTED, THEN THIS IS THE AREA.

AREA SOLUTION INVOLVES 9/10 OF A QUANTITY…
THIS MAY REFERS TO THE CONVERSION FROM SETAT TO TOLAL.
SINCE 10 SETAT = 1 TOTAL;

AHMES TOOK 1/10 OF 11/8 mh AND ADDED 10 CUBITS OF LAND (10 mh, or 10 COL)


TO FIND THE TOTAL TWO ALTERNATIVES ARE CONSIDERED:

MORE RMP WORKINGS:

1
7 [SETAT]

CHECK 2

1 TOL AND 4 SETAT

/2
3
/2

CHECK

/4

1 /2 /4

TOTAL [dmd]

1 TOL + 5 /2 /4 SETAT = 15 3/4, SETAT = 126/8 SETAT

AHMES HALVES THE TOTAL TO FIND: 7 /2 /4 /8 SETAT = 63/8 SETAT.

(second triangle, altitude of 7 khet and base of 9/4 khet)

USING THE FORMULA OF THE AREA OF A TRIANGLE (1/2 BASE TIMES ALTITUDE)


PEET MENTIONS THAT QUANTITES OF 7 AND 10 APPEAR IN RMP #54 AND THAT

2. 1/10 OF IT….10 COL

3. 1/10 OF IT ….AREA

4. 1/2 X 2 1/4 X 7 SETAT

which is read in 2009 as:

RMP #54: (setat and COL)

TO DIVIDE 7 SETAT OF LAND INTO 10 FIELDS

RMP WORKINGS:

1
10

THIS SEEKS/FINDS (1/2 + 1/5) SETAT TIMES TEN = 7 SETAT

/2

5

/5

2

RMP proof WORKINGS:

1

/2 /8 SETAT + 7 /2 COL

THIS FINDS (1) TIMES (7/10) SETAT

CHECK 2

1 /4 /8 SETAT + 2 /2 COL

10 parts: (7/10) *(4/4) = 28/40 = 25/40 + 3/40 = 5/8 setat + 15/2 COL


THIS FINDS (2) TIMES (7/10) SETAT 4 2 /2 /4 setat + 5 COL

5 parts: (14/10)*(4/4) = 56/40 = 55/40 + 1/40 = 11/8 setat + 5/2 COL


THIS FINDS (4) TIMES (7/10) SETAT CHECK 8 5 /2 SETAT + 10 COL

5/2 parts: (28/10)*(2/2) = 56/20 = 55/20 + 1/20 = 11/4 setat + 5 COL

THIS FINDS (8) TIMES (7/10) SETAT UNSHOWN TOTAL 7 SETAT

5/4 parts:(56/10*) = 56/10 = 55/10 + 1/10 = 11/2 setat + 10 COL

THIS FINDS (2+8) TIMES (/2 SETAT + /8 SETAT + 7.5 COL) = 7 SETAT

1 part: 7 setat (1/2 + 1/8 = 5/8 setat + 15/2 COL )?


RMP #55: ( SIMILAR TO RMP #54, with 100 mh = 100 COL = 1 SETAT)

RMP DIVIDEs 3 SETAT TO 5 FIELDS

RMP WORKINGS:

1
5

THIS SEEKS/FINDS (1/2 + 1/10) [TIMES 5 = 5/2 + 5/10 = 3 SETAT]

/2 2 /2

/10 /2

RMP proof WORKINGS:

CHECK 1

/2 SETAT + 10 COL

THIS FINDS

(1) TIMES (3/5) SETAT 17/8 SETAT + 7 /2 COL

(11/5) *(8/8) = 88/40 = 85/40 + 3/40 = 17/8 setat + 7 1/2 COL

THIS FINDS (2) TIMES (3/5) SETAT

CHECK 2 /2 /4 /8 SETAT + 2 /2 COL = 3 setat

THIS FINDS (4) TIMES (3/5) SETAT not SHOWN TOTAL 3 SETAT

THIS FINDS (1+4) TIMES (/2 SETAT + 10 COL) = 3 SETAT


RMP #56:

NOTE: RATIO OF HEIGHT TO BASE IS 25:36

RMP SHOWS: PYRAMID WITH SQUARE BASE SIDE = 360 [CUBITS] AND GIVEN CENTRAL VERTICAL HEIGHT OF 250 [CUBITS]

TAKE HALF OF 360 [BASE] FIND 180

RECKON 250 TO FIND 180 [I.E. SOLVE 250X=180; X = 180/250]

AHMES SHOWS [/2 /5 /50] TIMES 250 = 180

"A CUBIT BEING 7 PALMS, YOU ARE TO MULTIPLY BY 7"

CUBITS

PALMS
1
7

THIS SEEKS/FINDS (1) TIMES 1 CUBIT = 7 PALMS

/2
3 /2

THIS SEEKS/FINDS (1/2) TIMES 1 CUBIT = 3 1/2 PALMS

/5
1 /3 /15

THIS SEEKS/FINDS (1/5) TIMES 1 CUBIT = 1 + 1/3 + 1/15 PALMS

/50
/10 /25

THIS SEEKS/FINDS (1/50) TIMES 1 CUBIT = 1/10 + 1/25 PALMS UNSHOWN TOTAL

THIS FINDS (1/2 + 1/5 + 1/50) TIMES 7 PALMS = SKD OF 5 + 1/25 [PALMS]

"IT'S BATTER [SKD] IS 5 /25 PALMS"

I.E. FOR EVERY 1 CUBIT OF VERTICAL RISE, THE MASON IS TO JOIN TO A POINT
5 + 1/25TH PALMS HORIZONTALLY DISTANT. Likely AE FORMULA IN ACTION:

SKD = 7 TIMES BASE OVER TWO TIMES HEIGHT

RMP #57:

NOTE: RATIO OF HEIGHT TO BASE IS 3:4

RMP SHOWS: PYRAMID WITH SQUARE BASE SIDE = 140 [CUBITS]

FIND HEIGHT!

AND GIVEN A SKD OF 5 PALMS PLUS 1 FINGER = 21 FINGERS = 5.25 PALMS

"YOU ARE TO DIVIDE ONE CUBIT BY SKD DOUBLED"

I.E. 5 /4 PALMS DOUBLED IS 10 /2 [PALMS]

"RECKON WITH 10 /2 TO FIND 7 [PALMS]"

[A CUBIT BEING 7 PALMS]
CUBITS
PALMS
//3
7

THIS SEEKS/FINDS (2/3) TIMES 10.5 PALMS = 7 PALMS

//3
93 /3

THIS SEEKS/FINDS (2/3) TIMES 140 [BASE] = 93 /3 CUBITS

[IT'S HEIGHT IS 93 + 1/3 CUBITS]

I.E. FOR EVERY 140 CUBITS OF PYRAMID SIDE LENGTH WITH GIVEN SKD OF 5 /4, THE MASON IS TO JOIN THE SIDES TO A CENTRAL POINT WITH HEIGHT 93 + 1/3 CUBITS VERTICALLY DISTANT.

2
5.25
140
7
2 X 5.25 = 10.5
140 X 7 = 980
980/10.5
93.33333333

likely AE FORMULA IN ACTION: CENTRAL VERTICAL HEIGHT = 7 TIMES ENTIRE BASE OVER TWO TIMES SKD

*PEET CALLS STYLE OF CALCULATIONS IN RMP #57 AND SIMILAR #45 "ILLOGICAL!"
PEET IS MISTAKEN!

RMP #58:
NOTE: RATIO OF HEIGHT TO BASE IS 2:3

RMP SHOWS: PYRAMID WITH HEIGHT = 93 + 1/3 [CUBITS] AND GIVEN BASE SIDE = 140 CUBITS YOU ARE TO FIND SEKED HALF 140 = 70, RECKON WITH 93 /3 TO FIND 70

I.E. (93+1/3)X[SKD] = 70 [CUBITS]; SKD = 5 PALMS AND ONE FINGER
[A CUBIT BEING 7 PALMS] HEIGHT IN CUBITS

1
93 /3

THIS SEEKS/FINDS (1) TIMES 93 + 1/3 CUBITS = 93 + 1/3 CUBITS

CHECK /2
46 //3

THIS SEEKS/FINDS (1/2) TIMES 93 + 1/3 CUBITS = 46 + 2/3 CUBITS

CHECK /4

23 /3

THIS SEEKS/FINDS (1/4) TIMES 93 + 1/3 CUBITS = 23 + 1/3 CUBITS

UNSHOWN TOTAL

70

THIS SEEKS/FINDS (1/2 + 1/4) TIMES 93 + 1/3 CUBITS = 70 CUBITS

SCRIBE SHOWS (1/2 + 1/4) X 93 + 1/3 (CUBITS) = 70 CUBITS AND RESOLVES 3/4

[THE SEKED IN THE UNACCEPTED FORMAT]

AS FOLLOWS:

CUBITS

PALMS

1
7

THIS SEEKS/FINDS (1) TIMES 7 PALMS = 7 PALMS

CHECK /2

3 /2

THIS SEEKS/FINDS (1/2) TIMES 7 PALMS = 3 + 1/2 PALMS

CHECK /4

1 /2 /4

THIS SEEKS/FINDS (1/4) TIMES 7 PALMS = 1 + 1/2 + 1/4 PALMS
UNSHOWN TOTAL

5 /4

THIS SEEKS/FINDS (1/2 + 1/4) TIMES 7 PALMS = 5 PALMS ONE FINGER

140
2
93.33333
7
140/2
70
70/93.333333
0.750000027
[SKD]
.75*7
=3/4 OF ONE CUBIT = 3/4 OF 7 PALMS = 3/4 OF 28 FINGERS = 21 FINGERS = 5 PALMS ONE FINGER

likely AE FORMULA IN ACTION: SKD = 7 TIMES BASE OVER TWO TIMES HEIGHT
RESULT IS TO BE CONVERTED INTO PALMS AND FINGERS OR PALMS AND FRACTIONS THEREOF. FRACTIONS OF A CUBIT ARE NOT ACCEPTED.

FOR EXAMPLE A VERY FLAT PYRAMID WITH GREAT SEKED AND VERY SLIGHT PITCH.
SIDE = 700 CUBITS AND HEIGHT = 10 CUBITS
700/2=350 CUBITS = RUN
350/10=35 [SEKED IN CUBITS] = [35*7] 245 PALMS = 245 SEKED.


FOR A MORE COMPLEX EXAMPLE ANOTHER VERY FLAT PYRAMID WITH GREAT SEKED AND VERY SLIGHT PITCH. SIDE = 700 CUBITS AND HEIGHT = 11 CUBITS
700/2=350 CUBITS = RUN 350/11=31 + 9/11 [SEKED IN CUBITS] = [(31+ 9/11)*7] =217 PALMS + 63/11 PALMS = 222 + 8/11 PALMS SEKED = =222P + 8/11P = 222
PALMS + 2 FINGERS + 5/22 PALMS - A NEEDLESS CONSIDERATION

CORRECT FORMAT

= 222 + //3 +/22 + /66 [PALMS] SEKED!

= 222 PALMS + 2 FINGERS + (/6 /22 /66) PALMS = 222P + 2F + /6 /22 /66 P = SEKED

RESULTING FRACTIONS OF A FINGER WERE RESOLVED NEEDLESSLY - FYI

/6 /22 /66 P = //3 /6 /66 /22 /66 = //3 /6 /22 /33

SEKED = 222 PALMS + 2 //3 /6 /22 /33 FINGERS


RMP #59:

SEE RMP #58 WITH SAME RESULT

NOTE: RATIO OF HEIGHT TO BASE IS 2:3-AFTER AHMES CONFUSED THE QUESTION AND REVERSED THE GIVEN DIMENSIONS RMP SHOWS: PYRAMID WITH HEIGHT = 12 [CUBITS] AND GIVEN BASE SIDE = 8 CUBITS YOU ARE TO FIND SEKED

**SCRIBE HERE REVERSES THE UNITS AND FINDS FOR HT = 8 AND BASE = 12.
HALF 12 = 6, RECKON WITH 6 TO FIND 8
I.E. (8)X[SKD] = 6 [CUBITS]; SKD = 5 PALMS AND ONE FINGER
likely AE FORMULA IN ACTION: SKD = 7 TIMES BASE OVER TWO TIMES HEIGHT

RMP #59B:

NOTE: RATIO OF HEIGHT TO BASE IS 2:3

RMP SHOWS: PYRAMID WITH BASE SIDE = 12 [CUBITS]
AND GIVEN SKD = 5 PALMS ONE FINGER YOU ARE TO FIND HEIGHT
RECKON 5 1/4

DOUBLED FIND 10 1/2

2/3 X 10 1/2 = 7

//3 RECKONED WITH SIDE OF 12 CUBITS FINDS 8 WHICH IS THE HEIGHT.
I.E. FOR EVERY 12 CUBITS OF PYRAMID SIDE LENGTH WITH GIVEN SKD OF 5 /4, THE MASON IS TO JOIN THE SIDES
TO A CENTRAL POINT WITH HEIGHT 8 CUBITS VERTICALLY DISTANT.
likely AE FORMULA IN ACTION: CENTRAL VERTICAL HEIGHT = 7 TIMES ENTIRE BASE OVER TWO TIMES SKD

RMP #60:
NOTE: RATIO OF HEIGHT TO BASE IS 1:2

UNSPECIFIED OBJECT TYPE WITH BASE OF 15 CUBITS AND HEIGHT OF 30; FIND SKD
PEET SAYS THIS IS A CONE BUT THE AE BUILT NOTHING REMOTELY CONICAL EXCEPT COLUMNS!

SEE OBELISKS OF CLEOPATRA AND HAPSHETSUTH; THESE ARE PYRAMOIDAL NOT CONICAL.

"RECKON WITH 15 ITS HALF IS 7 /2"

MULTIPLY 7 /2 FOUR TIMES TO FIND 30

"FOUR IS SKD"

INCORRECTLY**I.E. (30)X[SKD=4] = 7.5 [CUBITS]; LOOKS LIKE AHMES INVERTED SOMETHING AS 30/4 = 7.5

INCORRECTLY**I.E. SKD = RUN [7.5 CUBITS] DIVIDED AGAIN BY RISE [30]; LOOKS LIKE

AHMES INVERTED SOMETHING AS 7.5/30 = /4

RMP SHOWS:

1
15 [CUBITS]
CHECK /2
7 /2
1
7 /2
2
15
CHECK 4
30

SOMETHING IS CORRUPT:
IF SKD = 4 AND HT = 30 THEN
GIVEN: RMP #56:
SKD = 7 TIMES BASE OVER TWO TIMES HEIGHT
7 PALMS X 1/2 OF BASE SIDE = SKD X HEIGHT
OR 3.5 X BASE IN CUBITS = SKD * HT
3.5 X = 4 * 30; 120/3.5=BASE = 34 + 2/7

THIS IS CORRECT BASE OF SKD 4 WITH HEIGHT 30 - NOT BASE OF 15 WHICH WAS GIVEN

SIDE = 2/7 * (4 X 30) = 34 + 2/7

IF THIS WAS A PYRAMID OR A CONE WE WOULD EXPECT BASE (OR DIAMETER FOR CONE) OF 34 /4 /28 [34 + 2/7]


TO FIND BASE WITH GIVEN SKD AND HEIGHT WE COULD USE FORMULA:
DIAMETER OF CONE = BASE OF PYRAMID = 2/7 TIMES HEIGHT TIMES SKD

4 = UNKNOWN SIDE DIVIDED BY HEIGHT IN PALMS = X/210; 4=X/210; X = 840
840 PALMS DIVIDED BY 7 =120

THIS IS CORRECT BASE OF SKD 4 WITH HEIGHT 30 - NOT BASE OF 15 WHICH WAS GIVEN

IF A CONE OR PYRAMID WE WOULD EXAMINE SEKED OF SEMI SECTIONAL TYPICAL TRIANGLE WITH HEIGHT = LARGEST SIDE = 30 AND BASE = 7.5 AND FOUND SKD = 1 PALM 3 FINGERS CORRECT FOR ITEM WITH BASE 15 AND HEIGHT 30

BUT SEKED IS NOT FOUR!

likely AE FORMULA IN ACTION: CENTRAL VERTICAL HEIGHT = 7 TIMES ENTIRE BASE OVER TWO TIMES SKD

H= 7 (15)/8 = 105/8 = 13 /8 IS CORRECT HEIGHT WITH BASE = 15 AND SEKED = 4

CORRECT HEIGHT IS NOT 30!

CONVERSIONS FOR ANCIENT EGYPTIAN UNITS OF MEASURE:

Work in progress as prepared by Bruce C. Friedman. May 12, 2009.

Refer to T. E. Peet’s English 1923 RMP and also Tanja Pommerening’s German article in Berichte Zur Wissenschaftsgeschichte #26, (2003) pp.1-16: for further metrological discussions.

Cubit: 1 LINEAR royal cubit = 7 palms = 28 fingers =

Note that all units are reduced to linear or square or cubic cubits as follows:

(Length as determined by Isaac Newton) ~20.6 inches = ~523.5 mm

See source text:

“Miscellaneous works of Mr. John Greaves [1602-1652]…To which are added: I. Reflections on the Pyramidographia…; II. A dissertation upon the Sacred Cubit of the Jews…; III. Tracts upon various subjects…; IV. A description of the grand seignor’s seraglio. To the whole is prefix’d an historical and critical account of the life and writings of the author, pub. By Thomas Birch [1705-1766].” London, 1737, two volumes. (800 pages)

Including:

I. Pyramidographia: A description of Egypt’s Pyramids.

II.With discourse on Roman foot and Denarius.

III. Tracts; Letters; Poems

IV. Turkish Emporer’s Court

Volume 2 [of 2] Includes Isaac Newton’s Dissertation!

“A Dissertation upon the Sacred Cubit of the Jews and the Cubits of the several Nations ; in which, from the dimensions of the greatest Egyptian Pyramid, as taken by Mr. John Greaves, the antient [ancient] Cubit of Memphis is determined.”/ “Translated from the Latin of Sir Isaac Newton, not yet published.”

Remen: 5 palms;

20 fingers;

5/7 LINEAR cubits;

(*sometimes also one-half setat which = 5,000 SQUARE CUBITS)

A special sign was *sometimes used for 1/2 setat also called rmn (remen) and referring to the fact that 5/7 of

a royal cubit squared is roughly 1/2 square cubit. These VERY different uses of remen can be gleaned easily from local context.

Khar: .0956 CUBIC METERS;

6400 ro;

20 hekats;

5 quadruple hekats;

50 quadruple henu;

two-thirds CUBIC cubits

Hekat: .0048 CUBIC METERS;

1/20 khar;

320 ro;

10 henu;

one-thirtieth CUBIC cubits

Double Hekat: .00956 CUBIC METERS;

1/10 khar;

640 ro (or sometimes 320 double-ro?);

20 henu;

5 quadruple henu;

one-fifteenth CUBIC cubits

Quadruple Hekat: .0191 CUBIC METERS;

1/5 khar;

1280 ro (or sometimes 320 quadruple-ro?);

40 henu;

10 quadruple henu;

two-fifteenths CUBIC cubits


Henu: .0005 CUBIC METERS;

32 ro;

1/200 khar;

One-tenth hekat;

1/40 quadruple hekat;

1/300 CUBIC cubits

Quadruple Henu: .0019 CUBIC METERS;

128 ro;

1/50 khar;

two-fifths hekat;

1/10 quadruple hekat;

1/75 CUBIC cubits

Hin: a/p Tanya: 1/10 hekat or 1/10 other units??; same as henu??

Djar or Dja: from Tanya Pemmerening: 5 ro; 1/1280 khar; 1/64 hekat

Khet: 52.35 LINEAR METERS;

100 LINEAR CUBITS; (sometimes called a reel or cord)

Square Khet: 2740.5 SQUARE METERS;

1 setat;

100 "cubits of land";

1/10 of a "thousand of land";

10,000 SQUARE cubits


Setat: 2740.5 SQUARE METERS;

1 square khet;

100 "cubits of land";

1/10 of a "thousand of land";

10,000 SQUARE cubits

*Note: example of doubling 7 setat from RMP #53 - result is: 1 "thousand of land" and 4 setat

Cubit of Land: 27.405 SQUARE METERS;

a strip 100 LINEAR cubits by 1 LINEAR cubit;

1/100 setat;

100 SQUARE cubits

* AKA "cubit strip"

Thousand of Land: 27405 SQUARE METERS;

1000 "cubits of land";

ten setat;

ten square khet;

100,000 SQUARE cubits

Deben: [monetary] 1400-1500 grains [British modern grains] to be elaborated later.

Cubic Cubit: .1434664_ CUBIC METERS;

3/2 khar;

30 hekats;

7 +/2 quadruple hekats;

300 henu;

75 quadruple henu;

9600 ro;

343 CUBIC palms

Square Cubit: .274052_SQUARE METERS;

2.946944_SQUARE FEET;

784 SQUARE fingers;

49 SQUARE palms

? square setat?

Square Remen: .1398226 SQUARE METERS;

25/49 SQUARE CUBITS;

400 SQUARE fingers;

25 SQUARE palms

I have seen references to use of a double remen linear measure of 10/7 cubits.

Related things to consider:

In the Talmud in Mishnah Kelim [on purification rituals] The Cubit of Moses and the common Cubit of the Jews and the Persian [Sushan or Shushan –from Capital Susa] Cubits used in Jerusalem [when Cyrus the Great [Persian] allowed and partly funded the temple reconstruction] were shown related to each other and also related to a number of barleycorns in length. Similarly in the Sanskrit Vedas, see Lalitavistara Sutra where Buddha describes known units of length from the microscopic to the tangible and that these Indian measures are also related to barleycorns.




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