Moscow Mathematical Papyrus

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The Moscow mathematical papyrus

The Moscow Mathematical Papyrus is also called the Golenischev Mathematical Papyrus, after its first owner, Egyptologist Vladimir Goleniščev. The Papyrus dates to the Middle Kingdom, ca. 1700 BCE. The Moscow papyrus is smaller than the Rhind papyrus, it contains 25 problems in total.

The problems in the Moscow Papyrus follow no particular order, and the solutions of the problems provide much less detail than those in the Rhind Mathematical Papyrus.[1]

Types of problems

  • Problem 2 and 3 are ship’s part problems. One of the problems calculates the length of a ship’s rudder and the other computes the length a ship’s mast given that it is 1/3 + 1/5 of the length of a cedar log originally 30 cubits long. [1]
  • Aha problems involve finding unknown quantities (referred to as Aha) if the sum of the quantity and part(s) of it are given. Problems 1, 19, and 25 of the Moscow Papyrus are Aha problems. [1]
  • Most of the problems are pefsu problems: 10 of the 25 problems. A pefsu measures the strength of the beer made from a heqat of grain. A higher pefsu number means weaker bread or beer. The pefsu number is mention in many offering lists.[1]
  • Problems 11 and 23 are Baku problems. These problems calculate the output of workers. [1]
  • Seven of the twenty-five problems are geometry problems and range form computing areas of triangles, to finding surface area of a hemisphere (problem 10) and finding the volume of a frustrum (a truncated pyramid). [1]


The Problems

Problem 3 MMP.Calculate the length of a ship’s rudder and the other computes the length a ship’s mast given that it is 1/3 + 1/5 of the length of a cedar log originally 30 cubits long. [1]


Problem 6 MMP. Given a rectangular enclosure of 12 units area and the ratio of the sides as 1 : 3/4, find the lengths of the sides. This problem seems to be identical to one of the Kahun Papyri in London. The problem is also interesting because it is clear that the Egyptians were familiar with square roots. They even had a special hieroglyph for finding a square root. It looks like a corner and appears in the fifth line of the problem. We suspect that they had tables giving the square roots of some often used numbers. No such tables have been found however.

Problem 6 from the Moscow Mathematical Papyrus
[line 1] Example of Calculating a rectangle
[line 2] If someone says to you: A rectangle is 12 setjat [in area] [has a breadth 1/2 1/4 [i.e. 3/4] of its length.[Calculate it]
[line 3] Calculate 1/2 1/4 to get 1. The result is 1 1/3
[line 4] Take this 12 setjat 1 1/3 times. The result is 16.
[line 5] Calculate its square root. The result is 4 for its length [and] 1/2 1/4 of it is 3 for the breadth

At the bottom (sixth line) is an illustration showing a rectangle with area 12 written in the interior, 4 at the top (in Clagett's book it's 3 at the top, bt that clearly should be a 4) and 3 at its side.

The steps of the solution given in both papyri is the same:

1 : 3/4 = 4/3; \ 4/3 * 12 =16; \ \sqrt{16} = 4 \ (=x);\ 4 * 3/4 =3 \ (=y) [2]


Problem 7 MMP. "A triangle of given area is such that it altitude is 2.5 times its base; find both." [2]. More specifically the problem states: “[There is a triangle with area 20 [setjat] and 'bank' (idb – ratio of height to base) of 2 ½; find the length and the breadth of the triangle. [1]

Problem 8 MMP translates as:

(1) Example of calculating 100 loaves of bread of pefsu 20
(2) If someone says to you: “You have 100 loaves of bread of pefsu 20
(3) to be exchanged for beer of pefsu 4
(4) like 1/2 1/4 malt-date beer
(5) First calculate the grain required for the 100 loaves of the bread of pefsu 20
(6) The result is 5 heqat. Then reckon what you need for a des-jug of beer like the beer called 1/2 1/4 malt-date beer
(7) The result is 1/2 of the heqat measure needed for des-jug of beer made from Upper-Egyptian grain.
(8) Calculate 1/2 of 5 heqat, the result will be 2 1/2
(9) Take this 2 1/2 four times
(10) The result is 10. Then you say to him:
(11) Behold! The beer quantity is found to be correct. [1]


Problem 10 MMP. Computes the area of a hemisphere. [2]

Problem 11 MMP asks if some one brings in 100 logs measuring 5 by 5, then how many logs measuring 4 by 4 does this correspond to?

Problem 14 MMP. The base is a square of side 4 cubits, the top is a square of side 2 cubits and the height of the truncated pyramid is 6 cubits.

Last part of problem 14 from the Moscow Mathematical Papyrus. The real inscription is in hieratic; this is the hieroglyphic transcription of the text. The glyphs start at: You are to add the 16 and the 8 and the 4 - See below.

The text reads:

" Example of calculating a truncated pyramid. If you are told: a truncated pyramid of 6 for the vertical height by 4 on the base by 2 on top:

You are to square this 4; result 16. You are to double 4; result 8.

You are to square this 2; result 4. You are to add the 16 and the 8 and the 4; result 28.

You are to take 1/3 of 6; result 2. You are to take 28 twice; result 56. See it is of 56.

You will find (it) right." [2]


Problem 17 MMP. Given a triangle with area 20 (setjat) and a length that is 1/3+1/5 of its breadth; Find the length and breadth. (The answer is length 10 and breadth 4) [1]

Problem 18 MMP. Find the area of a length of garment-cloth measuring 5 cubits 5 pals by 2 palms. (The answer is 80 [square palms]) [1]

Problem 19 MMP. Calculate a quantity taken 1 and ½ times and added to 4 to make 10. [1] In other words, in modern mathematical notation we are asked to solve 3/2 \times x + 4 = 10

Problem 23 MMP. finds the output of a shoemaker given that he has to cut and decorate sandals.

References

  1. 1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11
  2. 2.0 2.1 2.2 2.3 R.C. Archibald Mathematics before the Greeks Science, New Series, Vol.71, No. 1831, (Jan. 31, 1930), pp.109-121
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