The MMP is held in the Pushkin State Museum of Fine Arts in Moscow, Russia. Its contents cover 25 problems. The MMP has been called the Golenischev Mathematical Papyrus. Its first modern owner was Egyptologist Vladimir Golenidenov. Based on the palaeography of the hieratic text, it probably dates to the Eleventh dynasty of Egypt. Approximately 18 feet long and varying between 1 1/2 and 3 inches wide, its format was divided into 25 problems with solutions by the Soviet Orientalist Vasily Vasilievich Struve in 1930. It is one of the three well-known Mathematical Papyri along with the Rhind Mathematical Papyrus (RMP) and the Egyptian Mathematical Leather Roll (EMLR). The Moscow Mathematical Papyrus is about the same age as the EMLR, and 150 years older than the RMP. Gillings writes up the algebra correctly. Claggett reproduces Peet's 1920's and Gillings 1972 views, but adds little new beyond either.

The 25 MMP problems are written in arithmetic, algebraic, and geometric metaphors with several of the ancient definitions, and references included in Marhall Clagett's 1999 translation, and Gillings 1972 review of the text. Clagett discussed Struve's 1930 German analysis, and Peet's 1923 review, but little of the latest discoveries on the text.

MMP 6 reports the solution to a 2nd degree equation

1. (3/4)x^2 = 12

Scribal algebra solved the problem by

2. (4/3) (3/4)x^2 = (4/3)(12)

3. x^2 = 16

4. x = 4

The MMP scribe's second step actual wrote invert (3/4) and multiply by 12. That is, scribal division inverted and multiplied, a modern arithmetic technique.

More importantly, MMP 10 considered the area (A) of a circle by solving for diameter D, setting pi = 256/81 such that a formula that Gillings and earlier scholars incompletely read by stressing (8/9)(8/9) may have been involved in finding the area and circumference of a circle,

The actual MMP 10 calculations mentioned an area of an egg, a semi- circle, The circle's diameter 9 was input into an undefined formula reporting

1. A^1/2 = (8/9 x 9) = (72/9) x (72/9)/2

writing 9 - 9/9 = 8 x 8 = 64, which meant

2. A = (72/9)(72/9) = 64/2 = 32 (cubit ^2)

meant that the MMP scribe began with

3. A = 256/81 x (D/2)(D/2) (area of a circle, replacing pi with 256/81 and r by D/2)

4. A = (64/81)( D)(D) was reduced to

5. A = (8/9)(8/9)(D)( D) (algebraic geometry formula 1.0)

RMP 41 solved, D =9, height (H) 10 using the same MMP 10 formula

6. A = (8/9)(9)(8/9)(9) = (8)(8) = 64 cubits^2

writing

7. (9 - 9/9) = (8 x 8) = 64 cubits^2

8. Volume (V) = (64) (10) = 640 cubits^3

In RMP 42 3/2 of a cubit^3 became a khar unit citing D = 10, H =10

9. A = (8/9)(10)(8/9)(10) writing

10. A = (10 - 10/9) = (80/9)(80/9) = 6400/81 cubit^2

11. V = (6400/81)(10) = 64000/81 = 790 10/81

12 . V = (3/2)(H)(8/9)(D)(8/9)(D) = khar unit (algebraic geometry formula 2.0)

a third volume formula scaled algebraic geometry formula 2.0 by 3/2 in RMP 43 and the Kahun Papyrus created a Khar unit by

14. (3/2)V =(3/2)(3/2)(H)(8/9)(8/9)(D)D) = (4/3)(D)(4/3)(D) such that

15. V = (2/3)(H)(4/3)(D)(4/3)(D) = Khar unit (algebraic geometry formula 2.1)

used in RMP 43. Line 5. RMP 43's raw data

4096/9 = 455 1/9 in line 3,

1/20 of 455 1/9= 22 1/2 + 1/4 + 1/180 (not 1/45) in line 4

5. 4090/180 = (22 + 1/2 + 1/4 + 1/180) hekat

and line 5 by writing the remainder 1/180 of 100 hekat unit to a hekat unit to:

100/180 = (5/9) and

(5/9)(64/64) = 320/576 =

(288 + 18 + 9)/576 + [5/288(5/5) = (25/9)ro)

(1/2 + 1/32 + 1/64)hekat + (2 + 7/9) ro, with

Ahmes recorded 7/9 ro as (1/2 + 1/4 + 1/36)ro

information that is pertinent to MMP 10 related to area and volume formulas that

created cubit, khar, 1/20 of khar = 100 hekat and hekat units related to the

Akhmim Wooden Tablet and over 40 RMP uses.

Relevant AE mathematical and metrological and other terms (per Bruce Friedman):

===================================================

"1/2 + 1/4 pesu kind of beer" is a weak strength beer made with a cheaper less potent grain.

Beer A has an exchange rate that defies the mathematical looking name.

Beer B is exchanged at half the value of beer A made with the same pesu of the stronger grain.

See problem MMP#5 [121509_see also #8 and #9] or an example of exchange of beer A and B.

The Horus-eye symbols for hekat fractions ["<" for 1/2 and "X" for 1/4] are strung as "X <" for the beer name. Consider that beer B is made with typical recipes but with poor quality ingredients per Clagett's note 7 on pages 229-230. "Weak-grained beer", "*lesser beer, or beer B" Malt-Date beer is specified a beer B most clearly in the hieratic of MMP 8. See below.

"Bank" or "idb" is the same as our rise over run calculations in modern construction. *"Bank" is not the same as "seked" see below. **See usage and an example of the "bank" in problem MMP#7 below. "Beer" is an AE staple food. See also "bread." *Typically measured in jugs or "des-jugs."

**Typically with a "pesu" of less than 8. "Bread" is another primary AE staple food. *Typically measured in loaves. **Some surviving AE bread molds suggest loaves may have been a standard issue [size or quality]. ***121509_See new entry below; Grain "Des jug" is a standard quantity of beer (and other fluids?). Size unclear. 121509_"Grain" Used to define the value and to produce the AE staple foods of beer and bread. a. Upper Egyptian grain is harvested from the southern nile delta (nearer to Aswan than to Cairo). This is the top quality grain which sets the exchange standard for all other grains. b. Other grains noted in AE papyri: TBD "Hekat" or "hkt" or "heqat" is a quantity of volume that is typically applied to grain administration and for discussion purposes is about a bushel. "Kathete" means "meryt" or [height] or vertical length (see MMP#7) "Khet" refers to a unit of length equal to [a coil of rope] 100 cubits long. *In hieratic script "khet" is one symbol which looks like the hieroglyphic coil meaning a quantity of 100. **The "breadth" or run or base length is shown in MMP#7 as the khet symbol followed by the water determinative (meaning perhaps: length in the direction of flowing water, i.e. = the base in khet). See "teper." ***Gillings says this is about 57 yards. "Meryt" or "meryet" or "meret" or "mryt" refers to the vertical altitude AKA height. See "kathete." "Pefsu" is sometimes "pesu" or "pfesu" or "cooking ratio" or "baking ratio." *If "pesu" is shown as equal to 2, this means two loaves of bread can be made of one common unit quantity of grain. **If "pesu" is shown as equal to 4, this means four loaves of bread can be made of one common unit quantity of grain [a hekat]. ***As "pesu" increases the quality (and / or size) of the product decreases. "Seked" sometimes called the slope is the cotangent of our modern slope angle as the AE defined by: the run in palms [sevenths of a royal cubit] over one [royal] cubit of rise. "Setat" or "setjat" is an area 100 cubits x 100 cubits = 1 khet x 1 khet = 1 square khet = 10000 square cubits. *The square root of the number of square cubits in one setat is equal to the number of cubits in one khet, i.e. 100. **Gillings says this is about 2/3 of an acre. "Teper" or "tpr" or "tepro" or "mouth" is the base length. See also my note about breadth lister under "khet" above. 122009: "Thousand of land" is 10 setat ANALYSIS: #1: (Clag. V3; p.213 and fig. IV.6a at column I at right and bottom right of page 386- partial fragment and another tiny fragment) Called an Aha! problem and "false position" problem by Clagett. Clagett is often misleading by referencing the work of T. E. Peet on Aha problems and "pick a number" problems. Peet, in the RMP/1923 text, is not appreciative of the specific numbers that are "picked" and misses the point of the AE method of solving for unknowns. *See 1895 Hultsch-Bruins methods resolved before Peet muddled his analysis. This MMP problem is severely fragmented and what follows is only a possible reconstruction. The meaning of "....." is the same as "missing ." ....X-Y.... .....N-Y.... ....=5 Gillings offers nothing more. Case closed. #2 (Clag. V3; p.213 and fig. IV.6a at column II of page 386- badly fragmented) Called a "calculation of a ship's rudder" problem by Clagett. Clagett correctly shows that all the numerical content of this problem is lost to time. Gillings offers nothing more. Case closed. #3 (Clag. V3; p.213 and fig. IV.6a at column III of page 386- badly fragmented) Called a "calculation of a ship's mast" problem by Clagett. Clagett correctly shows a fair translation of Struve and I below demonstrate this algebraically: Given a cedar log of 30 cubits Find X = 30* (1/3 [+] 1/5) [Find 30* 8/15] X=16 Note that IMHO the reconstructed algebra above is not certain due to the damage to the papyrus. *Interpretation of the multiplier (8/15) seems fairly secure but the "answer" [16] is reconstructed by Struve and is NOT on the surviving fragment. Note that the Scribe of the MMP did not show his calculations [duplations] unless these too were lost to time. Gillings offers nothing more. Case closed. #4 (Clag. V3; p.213-4 and fig. IV.6b at column IV-V of page 387- badly fragmented with figure of a triangle) The MMP scribe's figure appears to be scalene or less possibly, isosceles. Called a "calculation of area of a triangle" problem by Clagett. Triangle is "redrawn" by Struve in such a way as to leave the scalene question unanswered. Given "mryt" [height measured perpendicularly from base] = 10 "khet" Given "teper" [base] = 4 "khet" 4 x 1/2 = 2 2 * 10 = 20 [square khet] My algebra above is a reasonable reconstruction but again, the damage is severe and the content uncertain. **CORRECTED 121509_*20 square "khet" is equal to 20 "setat" AND 2 "thousands of land." **An equivalent method to the modern formula for the area of [any type of] a triangle: A= 1/2 base x height, appears to be in use. Gillings improperly reproduces the sketch from Struve. Gillings states that a khet is about 57 yards. Gillings specifies this is a right triangle [he denies the AE a true definition of "meryet"] yet offers nothing more. See very similar problem RMP#51. Case closed. #5 (Clag. V3; p.214 and fig. IV.6b at column VI-VII of page 387- badly fragmented) Calculation of exchange of loaves of bread of high quality for jugs of lesser quality beer, as per Clagett. Given 100 loaves of bread of "pesu" 20 (see note directly below) *This means that 20 such loaves could be made from flour produced from one hekat of common grain. Exchange these 100 loaves for jugs of "*lesser beer*" of "pesu" 4 (see both notes directly below) *A "pesu" of 4 usually means that 4 jugs of beer could be made of one hekat of grain but NOT in this case. **This *lesser beer* (as defined above) could be exchanged for half as much [of any product] as better beer would "buy." ***Because of the cheaper ingredients, the exchange rate is halved. The MMP scribe finds the quantity of grain needed to produce his 100 loaves of pesu 20; answer is 5 hekats. To find the exchanged quantity of the (half valued) lesser beer he reduces the requisite quantity of grain by one half. 1/2 of 5 hekats = 2 [+] 1/2 hekats. 2 1/2 x [pesu of *lesser beer*] 4 = 10 [des-jugs] *Note that Struve's reconstructed portions of the fragment are implied by the structure of MMP#8 (below). **Gillings calls this a pesu problem and offers nothing more. Case closed. #6 (Clag. V3; p.214-5 and fig. IV.6c at column VIII of page 388- fragmented with a figure of a rectangle.) Called a "example of calculating a rectangle" problem by Clagett. Given an area of 12 "setat". Given a ratio of breadth equal to 3/4 the length. In modern format the question is: find sides given area [12] = length x 3/4 length = 3/4 x (length squared) Find the reciprocal of 3/4 [shown as 1/2 [+] 1/4] is 4/3 [shown as 1 [+] 1/3] Multiply area by ratio reciprocal: 12 x 4/3 to find 16. Find square root of 16 to be 4. 4 is length. 3 [described as 3/4 x length] is breadth. Scribe shows "proof" that 3 x 4 equals 12. checked 1 [is to] 4 checked 2 [is to] 8. Total [12] is not shown. *An equivalent method to the modern formula for the area of a rectangle: A= base x height, appears to be in use. **Additionally, an equivalent method to the modern formula for finding the sides of a rectangle given the area and sides' ratios: Side one [length] = square root of (A x the inverse ratio of sides) or: L=sq rt (12 x 4/3); B= 3/4L *Gillings states correctly that this problem and MMP#7 are examples of solving simultaneous equations. **See Clagett's note 11 on p. 230 where he discusses the "finding of two unknowns.". ***121509_See subtleties of concepts of ratio and proportion better defined in KMP problems with arithmetic progressions/proportions. Case closed. #7 (Clag. V3; p.215 and fig. IV.6c at column IX of page 388- fragmented) Called a "example of calculating a triangle" problem by Clagett. Given area = 20 "setat" with given rise/run ratio ["bank"] of 2 [+] 1/2 Find sides by doubling the area: 2 * 20 = 40 Multiply doubled area by "idb" ["idb" = "bank"]: 40 x 2 1/2 = 100 Find square root of result above = 10 "khet" = the height = vertical length = "kathete." *Note that the hieratic symbol for square root in both MMP problems #6 and #7 is reconstructed by Struve. **I believe Struve was correct to insert the square root function as no other insertion resolves the data. Find reciprocal of 2 [+] 1/2 = reciprocal of 5/2 = 2/5 [shown as 1/3 [+] 1/5 which equals 6/15 and equals 2/5] Multiply reciprocal of bank by square root of (doubled area x bank) = multiply (reciprocal of bank) x length: 2/5 x 10 = 4 "khet" = the height = the altitude = "meryt." *Clagett correctly refers to these workings a evidence of AE awareness of the method of solving the area of any type triangle. **See Clagett's note 14 on p. 231. ***Gillings states correctly that this problem is an example of solving two simultaneous equations. ****Sadly the MMP scribe did not show more of his workings. Case closed. 121509: #8 (Clag. V3; p.215-216 and fig. IV.6d at column X-XI of page 389- largely complete* and footnote 15 on page 231) The workings and arrangements are in harmony with less complete MMP #5. The totals are identical and only the more complete description of the "lesser beer" make this problem worthy of any scrutiny. Perhaps Malt/date beer simply was less delicious than the standard more valuable beer. See preliminary definitions. #9 (Clag. V3; p.216-217 and fig. IV.6e-f at columns XII-XVII of page 390-391 -largely complete) Called a "calculation of grain for bread and beer" problem by Clagett. *Unlike the RMP where Ahmes tends to do the difficult problems first and show some easier problems to clarify his advanced procedures, this complex MMP problem follows easier problems (MMP 5 and MMP 8). A given quantity of 16 hekats of grain is to be consumed in the production of 100 loaves of bread with pfesu 20 AND the rest of the grain is to be used to create three equal quantities of lesser beer with pfesus of 2 and 4 and 6. **Keep in mind that (as defined in problems 5 and 8) one jug of lesser beer pfesu 2 is exchanged for half as much value as one jug of standard beer pfesu 2. This working gets tricky! First the scribe deducts the 5 hekats of grain needed for the 100 loaves of bread pfesu 20. Scribe does not show his work! With the 11 remaining hekats of grain the scribe performs this calculation: Asking himself how much grain is needed for one jug of STANDARD beer of pfesu 2 - he finds = 1/2 hekat Asking himself how much grain is needed for one jug of STANDARD beer of pfesu 4 - he finds = 1/4 hekat Asking himself how much grain is needed for one jug of STANDARD beer of pfesu 6 - he finds = 1/6 hekat He adds the 1/2 and 1/4 and 1/6 and finds 2/3 + 1/4 hekats. This = 11/12. He doubles the required grain because the value of lesser beer is one half of standard beer. and find 1 + 2/3 + 1/6. This = 11/6 Scribe divides 11 remaining hekats of grain by 11/6 to find 6. *Division working is not shown. Scribe states the quantity of output from 16 hekats of grain is: 100 loaves pfesu 20 and 6 jugs lesser beer pfesu 2 and 6 jugs lesser beer pfesu 4 and 6 jugs lesser beer pfesu 6 ***I again would describe this as a working on simultaneous equations and as a result of the geometric progression of beer pfesus I would say this demonstrates arithmetic and geometric progressions too. Case closed. #10 Most curious is that an item is described in two dimensions. I have offered all conceivable reconstructions for comparison and logical elimination. 122009: MMP#10 [This has to be a basket but does NOT have to be a hemispherical basket! It is not a cylinder or semi-cylinder.] (Clag. V3; p.218-219 and fig. IV.6g at column XVIII-XX of page 392- largely complete but poor vague script; See footnotes 18-20 on pages 231-4; also see fig IV.7 on page 407) Called a "calculation of basket" problem by Clagett. Peet's figure on V3 page 407 shows a reasonable recreation of how a hemisphere could be described with two dimensions [the diameter is the mouth and the radius is the depth]. Except that I have never seen a hemispherical AE basket in any museum, this seems correct. All of Peet's alternates to a basket are not viable. *Struve's work modified only slightly by Clagett and freely translated by myself: Scribe gives a basket with a mouth of 4 [+] 1/2 (unspecified linear units but I would not say cubits - perhaps spans or palms) Find the area. This is not a volume calculation. Find 1/9 of 9: Scribe is clearly performing an operation involving a circle - certainly. Scribe operates on 9 instead of 4.5 because he has doubled SOMETHING. Struve and Clagett include this reconstructed line: The basket is half an "inr." INR is assumed to be egg or sphere but this is not certain as the papyrus is damaged here. The result is 1 [one ninth of nine - work not shown]. Calculate the remainder as 8 [nine minus one is eight - work not shown]. Find 1/9 of 8 equals 2/3 [+] 1/6 [+] 1/18 [no workings]. Again, we know how this must involve a second circular consideration but we do not know much else. *Struve shows the 1/18 incorrectly in his glyph tr. as 1/10 [+] 1/8 - the scribe did not do this incorrectly. Deduct 2/3 [+] 1/6 [+] 1/18 from 8 to find: 7 [+] 1/9 [no workings]. Multiply the (mouth or other?) dimension (4.5) by 7 [+] 1/9 to find 32 *The papyrus is most unclear due to penmanship here at the close of the problem but although 32 is NOT clear, the operation of (4.5) by 7 [+] 1/9 IS CLEAR. So, ignoring the text and modernizing, we see Area (of half of something) = 4.5 x (8/9 x 8/9 x 4.5 x 2) = 4.5 x (64/81 x 9) = 4.5 x (64/9) = 288/9 = 32 Scribe did not show any of his intermediate calculations or duplations so like the other authors we can put various questions to the answer: 32 [unspecified square linear units] is the area of an item with an opening of 4.5 [linear units]. As an example only, let's assume this is a cubic/square 5-sided box! Each of the six sides of a cube would measure 4.5 units squared or 20.25 sq u. Five sides (only) would total 101.25 sq u. *This actual MMP #10 item area measures less than 1/3 of a cube shaped square box. This is important. If we assumed instead the box was half of such a cube: The bottom would be 20.25 and each of the 4 sides would be half this area for a total of 60.75 sq u. Not 32. If this was a short cylinder (NOT as per Peet) and we gave a freely reconstructed formula for the area as AE might have conceived: A (of open topped cylindrical container with diameter of 4.5 and height of 4.5 also) = Sum of bottom of container (2.25^2)256/81 plus unrolled side of container 256/81 x 4.5 (mouth) x 4.5 (height) = (9/4 x 9/4 x 256/81) [+] (9/2 x 9/2 x 256/81) = (256/16) [+] (256/4) = 16 [+] 64 = 80 Again, not 32. If the above short cylinder had a shorter height of half of 4.5 and was dimensioned otherwise as noted: A (of open topped cylindrical container with diameter of 4.5 and height of 2.25) = Sum of bottom of container (2.25^2)256/81 plus unrolled side of container 256/81 x 4.5 (mouth) x 2.25 (height) = (9/4 x 9/4 x 256/81) [+] (9/2 x 9/4 x 256/81) = (256/16) [+] (256/8) = 16 [+] 32 = 48 - See Peet's semi cylinder at same total as the unrolled curved side only. Again, not 32. If we EXCLUDED the base as above to only find the area of the SIDE of an OPEN cylinder with mouth 4.5 and height of half of mouth: A (of cylinder side) = C x h = 256/81 x 4.5 x 2.25 = 32 This unrolled stubby tube agrees in total with the problem only and is not consistent with the text in any way. If this was a semi cylinder (BISECTED through both diameters at ends as per Peet; see Gillings figure 18.1 on page 197) *The image of a BBQ grill, its basin (or cover) made out of a 55 gallon drum, serves me more completely than the term "semi-cylinder." A freely reconstructed formula for the TOTAL area as AE might have conceived: A BBQ with a square grill opening with sides of 4.5 = Peet's semi-cylinder Because the 2 semicircular "ends" combine to equal the circular part of the answer and the rest is "unrolled" to find the rectangular part of the answer, we would expect the AE to have operated on at least 4 numbers: 1. Circle determinant (a radius or diameter but not necessarily both) 2. Pi 3. Side of rectangle with Pi as a factor 4. Other side of rectangle We would expect also: Combined sum of areas of the 2 semicircular "ends" of BBQ [(2.25^2)256/81] plus unrolled side of BBQ "basin": 256/81 x 2.25 (halfmouth) x 4.5 (height) = (9/4 x 9/4 x 256/81) [+] (9/4 x 9/2 x 256/81) = (256/16) [+] (256/8) = 16 [+] 32 = 48 Again, not 32, but *Peet was only interested in the curved surface, which as above is 32 [sq. units]. Peet (who surely would have exhausted the above options as I have done) abused the integrity of MMP #10 and selected the CURVED surface of the semi cylinder over all others. As we know him to be among the Strawmen in favor of manipulating history to suit a culturally motivated agenda, I do not see how Peet has attained his own misguided goals. Even if Peet is correct, the MMP #10 problem suggests the ancient scribe was highly skilled in geometry and associated logic. FYI: If we assume the AE had a formula for area of a (full) sphere: We expect something equivalent to 4 Pi x radius squared where Pi is 256/81. Accordingly for a hemisphere we could halve this and find A = 2 Pi x radius squared where Pi is 256/81. So the operation on the LID of the basket (or the largest enclosed circle in a sphere) would be easily found as A (lid) = Pi x r^2 The 4-2-1 flow of these formulas shares consistency with the AE style of duplation and metrology and OK binary logic and each was PROBABLY known. Not definitely known! Case closed. 122009: #11 (Clag. V3; p.219 and fig. IV.6h at column XXI-XXII of page 393- not well described yet complete* and footnote 20 on page 234-5 about Peet's analysis of Struve's bad numerical glyph tr. and Clagett's English tr.) *The scribe erred in copying [?] in col XXII line 3. Called a "reckoning the work of a man in logs" problem by Clagett. The work of a man in logs is not defined exactly! We'll assume this is one man day of labor applied to a task involving wood. We will not yet assume this is specifically cutting or transporting or any specific combination of tasks. The work, (one man day, henceforth "W") yields 100 logs of "5 handbreadths section" (henceforth 5P for five palms). This statement above is modified to: W = 100 logs of 5P is perhaps an industry standard for anticipated output. We also may think of this as a daily quota in that the day's work was incomplete until the minimum limit is reached. We cannot be certain. As per Clagett, the Scribe continues: "But, he has brought these in logs of 4 handbreadths section." *If the worker's task was simply cutting we might possibly expect a linear solution [if logs were all the same thickness] but the process involves squaring as below. Square 5P find 25 and square 4P find 16. Find 25/16 is 1 [+] 1/2 [+] 1/16 [or 1 + 9/16] Multiply 25/16 x 100 and find 156 [+] 1/4 The "answer" of how much work in logs of 4P section per man per day is 156.25 The task is not defined but must involve cutting or the section would not be revealed or stated or important to the question! If we think about the area of the cross section of the cut of a square timber 5P section we find it is 25 sq P. If we think about the area of the cross section of the cut of a square timber 4P section we find it is 16 sq P. Thinking about how much lumber would need to be sawed through as an abstraction we find W = 2500 sq palms If we divide this by 100 logs we find each log section area is 25 sq. palms If we divide this same 2500 sq palms daily work quota/output rate by 16 sq palms we find 156.25 logs. Trees and logs aren't square!: Lets assume the logs were more specifically described as cylindrical: If section diameter is 5P, section area is 5/2^2 x Pi = 25/4 x 256/81 = (256 x 25)/324 = a little over 19.75 If section diameter is 4P, section area is 2^2 x Pi = 16/4 x 256/81 = (256 x 16)/324 = a little under 12.67 Because the scribe's manipulations involve 25/16 in either case, we are not certain nor need we be concerned about the sectional shape of the logs! I would humbly suggest this more likely refers to "boards" or "beams" than logs and that these "logs" are square in section and that the work is CUTTING. Case closed. #12 (Clag. V3; p.219-20 and fig. IV.6i at columns XXIII-XXIV of page 394- sloppily formatted/copied by the ancient scribe yet complete* and footnotes 21-2 on page 235 about pfesu and linear equations generally) Called a "calculation of grain" problem by Clagett. Actually this is a pfesu calculation. *The scribe begins the problem in column XXIII and in the first 3 lines of column XXIV he concludes the work. The 4th line of column XXIV is the first line of problem #13. Scribe seems to say/writes: 10 [+] 3 hekats, make 18 des jugs of lesser beer with 2 [+] 1/6 hekats required per des jug. (13/6) pesu means (6/13) was returned to one hekat. (Finding 13/ (2[+] 1/6) = 6 we see that we can convert 13 hekats of grain into 6 des jugs of lesser beer-[assumed to be pfesu = 1].) Scribe then divides 18 jugs of beer by 6 above to find 3. 3 is the pfesu of the eighteen (less valuable/smaller) 1/3 jugs created from the six larger jugs/des jugs of pfesu 1. We may consider all the jugs are the same size but the 18 jugs are 2/3 containing a malt date brew in addition to 1/3 of standard beer. Despite the general lack of clarity on the formula for lesser beer and the lack of agreement with the other problems, in this case the recipe seems clear. Note that we would expect from previous lesser beer problems (MMP #5 and #8) that one des jug of lesser beer with pfesu = 1 would require (or be exchanged for) a value of only 1/2 hekats of upper egyptian grain and not 2 [+] 1/6 hekats. (See MMP #24) #13 (Clag. V3; p.220-21 and fig. IV.6i and IV.6j at columns XXIV-XXVI of pages 394-395- sloppily formatted/copied by the ancient scribe yet largely complete* and footnotes 23 on page 235 about the scribal error and similarity to MMP #9) See column XXVI - line 3 scribal error of 12 for 6. Besides the error, this is the same problem as MMP #9. #14 (Clag. V3; p.221 and fig. IV.6j and IV.6k at columns XXVII-XXIX of pages 395-396- concise and largely complete* and footnote 24 on page 235 about the skill of the scribe determining the volume of a truncated pyramid) *See also the somewhat irrelevant possible derivation methods on pages 408-411. *In column XXVIII the scribe offers 6 lines of text. Struve's glyph tr. and Clagett's English tr. show this somewhat compressed/bastardized as only 5 lines of text. Best defined on my link: http://www.mathorigins.com/image%20grid/MOSCOW_010.htm

MMP #15 - MMP #21

MMP #22 (Clag. V3; p.225 and fig. IV.6q at columns XL-XLI of page 402- largely complete and see footnote 33 on pages 236-7) called a "calculation of grain" problem by Clagett. Actually this too is a pfesu calculation. 10 hekat of upper egyptian grain are to be calculated like (or exchanged for) 100 loaves of bread of unknown pfesu.

The remainder from the 10 hekat after the 100 loaves worth of grain are subtracted is to be used for 10 des jugs of standard beer with pfesu 2 to be converted [?] to jugs of lesser beer.

Here the recipe for the lesser beer is again intimated: Deduct the 5 hekats needed to make 10 jugs of lesser beer pfesu 2 [this should read 2.5 hekats to be consistent with MMP #5 or #8].

This leaves 5 hekats for the 100 loaves. But the scribe adjusts the 5 hekats of grain to 2.5 hekats because the lesser beer recipe requires half as much grain. This leaves 7.5 hekats of grain for the 100 loaves. At this point the problem appears to be truncate

If the scribe was to complete all apparently requested work he would find the brad pfesu by dividing the 7.5 hekats of grain available by 100 loaves to find each loaf has a pfesu of 13 [+] 1/3.

i.e. 15/2 hekat = 100 loaves pfesu X

X = 100/15/2

In RMP 69-78 two-level, and higher level inventory controls of grain and other products were reported by a pesu unit. The pesu was an arithmetic inverse to the amount of grain in the product. On a macro level about 1/3 of grain was used for bread, and about 2/3 was used for several types of beers, including one with dates as an ingredient.

For example RMP 71 showed the second level use of a pesu. Beer was made in besha unit, 1/2 of a hekat of grain. 1/4 of the besha (1/8 of a hekat) was poured off, and replace by water. The pesu calculation began with a besha composed of 1/2 a hekat such that:

1/2 - 1/8 = 3/8 was returned to a hekat unit by multiplying by 8/3 or

3/8 x 8/3 = 1,

with 8/3 = 2 2/3 recorded as the pesu. This calculation is confirmed in Moscow Mathematical Papyrus (MMP) problems 12 and 16 by two strengths of beer, the first confirming the second level 2 2/3 pesu made from a besha.

The majority of RMP pesu problems began with 15 or 16 hekats, and removed 5 hekats for use with making 200 loaves of bread at 20 pesu. The remaining hekats, 10 and 11, were used for one or more beers, a methodology that was repeated several times in the MMP.

MMP #23 (Clag. V3; p.226 and fig. IV.6r at column XLII of page 403- largely complete and see footnote 34 on page 237) called a "reckoning the work of a shoemaker" problem by Clagett.

This like MMP #11 suggests work output or quota was formalized for all bureaucratically administered tasks. If the shoemaker only cuts the material for shoes he can produce 10 pairs per day. If the shoemaker only decorates (or assembles) the material for shoes he can produce 5 pairs per day.

To find how many completed cut and assembled and decorated pairs he can produce in a day:

Find that (as given) in 1 day he can cut for 10 pairs and in 2 days he can complete decorating, therefore in 3 days he can fully cut/decorate/complete 10 pairs.

Therefore 3 [+] 1/3 pairs [10/3] can be cut and completed per day.

Case closed.

MMP #24 (Clag. V3; p.226-7 and fig. IV.6s at columns XLIII-XLIV of page 404- largely complete and see footnote 35 on page 237)

Called a "calculation of grain" problem by Clagett. Actually this too is a pfesu calculation.

Given 15 hekats upper egyptian grain to be made into 200 loaves and 10 des-jugs of lesser beer with 1/10 the pfesu* of the loaves.

*Consider that the beer is 1/10th AS WEAK a the loaves. I.e. A jug of this beer is of 10 times the exchange value of one of the loaves. Find the pfesu of the bread and beer.

The scribe frames his workings to solve this alternate problem which yields the same result for the bread pfesu but has easier manipulations:

Given 15 hekats upper egyptian grain to be made into 200 loaves and 100 des-jugs of lesser beer with the same pfesu as the loaves.

300/15 = 20.

20 is the pfesu of the bread (therefore 200*loaves/20 pfesu requires 10 hekats)

*Scribe may have mistakenly shown 100 loaves instead of 200 but Struve/Clagett correct this.

and 5 hekats remain to make the 10 des-jugs of lesser beer with 1/10 the pfesu of the loaves [pfesu = 2 for beer] OR 100 des-jugs of lesser beer with the same pfesu as the loaves [irrelevant but the pfesu of 100 jugs of lesser beer made from 5 hekats grain would be found to be 20 ONLY in this case].

I believe something is bizarre about this as lesser beer pfesu 2 is otherwise shown requiring half the grain as standard beer but in MMP #24 there is no reduction!

In MMP #5 we see:

2 1/2 [hekats] x [pesu of *lesser beer*] 4 = 10 [des-jugs]

Here in MMP #24 we see:

5 [hekats] x [pesu of A STRONGER/MORE VALUABLE*lesser beer*] 2 = 10 [des-jugs]

In effect, the MMP 24 jugs should be twice or 4 times as large or twice or 4 times as valuable as those in MMP #5.

This remains unclear. See MMP #12 for related chaos.

Case closed.

MMP #25 (Clag. V3; p.227 and fig. IV.6t at column XLV of page 405- complete and see footnote 36 on page 237)

Called a "calculating a quantity" problem by Clagett.

Find X when 2x + x = 9

Solve 3X=9

Workings suggest 2+1=3; 9/3 = 3

Case closed.

## Wednesday, December 9, 2009

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