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Show that the solution to the problem of dividing 7 loaves among 10 men is that each man gets \(\overline{\overline{3}} \overline{30}\). (This is problem 4 of the Rhind Mathematical Papyrus.)

Solve by the method of false position: A quantity and its \(2 / 3\) are added together and from the sum \(1 / 3\) of the sum is subtracted, and 10 remains. What is the quantity? (problem 28 of the Rhind Mathematical Papyrus)

A quantity, its \(1 / 3\), and its \(1 / 4\), added together, become 2 . What is the quantity? (problem 32 of the Rhind Mathematical Papyrus)

Problem 72 of the Rhind Mathematical Papyrus reads " 100 loaves of pesu 10 are exchanged for loaves of pesu 45 . How many of these loaves are there? The solution is given as, "Find the excess of 45 over \(10 .\) It is 35 . Divide this 35 by 10. You get \(3 \overline{2}\). Multiply \(3 \overline{2}\) by 100. Result: 350. Add 100 to this 350 . You get 450 . Say then that the exchange is 100 loaves of pesu 10 for 450 loaves of pesu \(45 . "^{18}\) Translate this solution into modern terminology. How does this solution demonstrate proportionality?

Various conjectures have been made for the derivation of the Egyptian formula \(A=\left(\frac{8}{9} d\right)^{2}\) for the area \(A\) of a circle of diameter \(d\). One of these uses circular counters, known to have been used in ancient Egypt. Show by experiment using pennies, for example, whose diameter can be taken as 1, that a circle of diameter 9 can essentially be filled by 64 circles of diameter 1. (Begin with one penny in the center; surround it with a circle of six pennies, and so on.) Use the obvious fact that 64 circles of diameter 1 also fill a square

Convert the fractions \(7 / 5,13 / 15,11 / 24\), and \(33 / 50\) to sexagesimal notation. (Do not worry about initial zeros, since the product of a number with its reciprocal can be any power of \(60 .\) ) What is the condition on the integer \(n\) that ensures it is a regular sexagesimal, that is, that its reciprocal is a finite sexagesimal fraction?

Show that the area of the Babylonian "barge" is given by \(A=(2 / 9) a^{2}\), where \(a\) is the length of the arc (one-quarter of the circumference). Also show that the length of the long transversal of the barge is \((17 / 18) a\) and the length of the short transversal is \((7 / 18) a\). (Use the Babylonian values of \(C^{2} / 12\) for the area of a circle and \(17 / 12\) for \(\sqrt{2}\).)

Show that the area of the Babylonian "bull's-eye" is given by \(A=(9 / 32) a^{2}\), where \(a\) is the length of the arc (one-third of the circumference). Also show that the length of the long transversal of the bull's-eye is \((7 / 8) a\), whereas the length of the short transversal is \((1 / 2) a\). (Use the Babylonian values of \(C^{2} / 12\) for the area of a circle and \(7 / 4\) for \(\sqrt{3}\).)

Solve the problem from the Old Babylonian tablet BM 13901: The sum of the areas of two squares is 1525 . The side of the second square is \(2 / 3\) that of the first plus 5 . Find the sides of each square.

. Solve the following problem from tablet YBC \(6967: \mathrm{A}\) number exceeds its reciprocal by \(7 .\) Find the number and the reciprocal. (In this case, that two numbers are "reciprocals" means that their product is \(60 .\) )