Moscow Mathematical Papyrus,

Moscow Mathematical Papyrus,

ALMEDINA Srednja ekonomska škola, Sarajevo

The Moscow Mathematical Papyrus is an ancient Egyptian mathematical papyrus, also called the Golenishchev Mathematical Papyrus, after its first owner outside of Egypt, Egyptologist Vladimir Golenishchev. Golenishchev bought the papyrus in 1892 or 1893 in Thebes. It later entered the collection of the Pushkin State Museum of Fine Arts in Moscow, where it remains today.

Date 13th dynasty, Second Intermediate Period of Egypt Place of origin Thebes Language(s) Hieratic Size Length: 5.5 metres (18 ft) Width: 3.8 to 7.6 cm (1.5 to 3 in)

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Moscow Mathematical Papyrus,

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Based on the palaeography and orthography of the hieratic text, the text was most likely written down in the 13th Dynasty and based on older material probably dating to the Twelfth Dynasty of Egypt, roughly 1850 BC.[1] Approximately 5½ m (18 ft) long and varying between 3.8 and 7.6 cm (1.5 and 3 in) wide, its format was divided into 25 problems with solutions by the Soviet Orientalist Vasily Vasilievich Struve[2] in 1930.[3] It is a well-known mathematical papyrus, usually referenced with the Rhind Mathematical Papyrus. The Moscow Mathematical Papyrus is older than the Rhind Mathematical Papyrus, while the latter is the larger of the two.[4]

Moscow Mathematical Papyrus

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Exercises contained in the Moscow Papyrus

Ship's part problems

Exercises contained in the Moscow Papyrus 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. The papyrus is well known for some of its geometry problems. Problems 10 and 14 compute a surface area and the volume of a frustum respectively. The remaining problems are more common in nature.[1]

Problems 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 of a ship's mast given that it is 1/3 + 1/5 of the length of a cedar log originally 30 cubits long.[

Moscow Mathematical Papyrus

Aha problems

Baku problems

Geometry problems

Aha problems involve finding unknown quantities (referred to as Aha) if the sum of the quantity and part(s) of it are given. The Rhind Mathematical Papyrus also contains four of these type of problems. Problems 1, 19, and 25 of the Moscow Papyrus are Aha problems. For instance problem 19 asks one to calculate a quantity taken 1 and ½ times and added to 4 to make Aha problems involve finding unknown quantities (referred to as Aha) if the sum of the quantity and part(s) of it are given. The Rhind Mathematical Papyrus also contains four of these type of problems. Problems 1, 19, and 25 of the Moscow Papyrus are Aha problems. For instance problem 19 asks one to calculate a quantity taken 1 and ½ times and added to 4 to make

Seven of the twenty-five problems are geometry problems and range from computing areas of triangles, to finding the surface area of a hemisphere (problem 10) and finding the volume of a frustum (a truncated pyramid).

Problems 11 and 23 are Baku problems. These calculate the output of workers. Problem 11 asks if someone brings in 100 logs measuring 5 by 5, then how many logs measuring 4 by 4 does this correspond to? Problem 23 finds the output of a shoemaker given that he has to cut and decorate sandals.[1

Moscow Mathematical Papyrus

Volume of frustum of square pyramid

The fourteenth problem of the Moscow Mathematical calculates the volume of a frustum. Problem 14 states that a pyramid has been truncated in such a way that the top area is a square of length 2 units, the bottom a square of length 4 units, and the height 6 units, as shown. The volume is found to be 56 cubic units, which is correct.[1] The text of the example runs like this: "If you are told: a truncated pyramid of 6 for the vertical height by 4 on the base by 2 on the top: You are to square the 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" [6] The solution to the problem indicates that the Egyptians knew the correct formula for obtaining the volume of a truncated pyramid:

Moscow Mathematical Papyrus,

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Pefsu problems

A higher pefsu number means weaker bread or beer. The pefsu number is mentioned in many offering lists. For example problem 8 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.