Chapter 1

Represent 375 and 4856 in Egyptian hieroglyphics and Babylonian cuneiform.

Use Egyptian techniques to multiply 34 by 18 and to divide 93 by 5

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.)

Multiply \(7 \overline{2} \overline{4} \overline{8}\) by \(12 \overline{3}\) using the Egyptian multiplication technique. Note that it is necessary to multiply each term of the multiplicand by \(\overline{\overline{3}}\) separately.

Solve by the method of false position: A quantity and its \(1 / 7\) added together become 19. What is the quantity? (problem 24 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)

Calculate a quantity such that if it is taken two times along with the quantity itself, the sum comes to 9. (problem 25 of the Moscow 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?

Convert the sexagesimal fractions \(0 ; 22,30,0 ; 08,06\), \(0 ; 04,10\), and \(0 ; 05,33,20\) to ordinary fractions in lowest terms.

In the Babylonian system, multiply 25 by 1,04 and 18 by 1,21 . Divide 50 by 18 and 1,21 by 32 (using reciprocals). Use our standard multiplication algorithm modified for base \(60 .\)

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 \(\sqrt{3}\).)

Convert the Babylonian approximation \(1 ; 24,51,10\) to \(\sqrt{2}\) to decimals and determine the accuracy of the approximation.

Use the assumed Babylonian square root algorithm of the text to show that \(\sqrt{3} \approx 1 ; 45\) by beginning with the value 2 . Find a three- sexagesimal-place approximation to the reciprocal of \(1 ; 45\) and use it to calculate a three-sexagesimalplace approximation to \(\sqrt{3}\).

The scribe of Plimpton 322 did not use the value \(v+\) \(u=2 ; 18,14,24\), with its associated reciprocal \(v-u=\) \(0 ; 26,02,30\), in his work on the tablet. Find the smallest Pythagorean triple associated with those values.

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 Babylonian problem taken from a tablet found at Susa: Let the width of a rectangle measure a quarter less than the length. Let 40 be the length of the diagonal. What are the length and width? Use false position, beginning with the assumption that 1 (or 60 ) is the length of the rectangle.

. 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 \(60 .\) )

Solve the following Babylonian problem about a concave square: The sum of the area, the arc, and the diagonal is \(1 ; 16,40\left(=1 \frac{5}{18}\right) .\) Find the length of the arc. (Recall that the coefficient of the area is \(4 / 9\) and the coefficient of the diagonal is \(11 / 3\)-see Exercise 23.)

Solve the following problem from BM \(13901:\) I added onethird of the square- side to two-thirds of the area of the square, and the result was \(0 ; 20(=1 / 3)\). Find the squareside.

Solve the following problem from tablet AO 8862: Length and width. I combined length and width and then I built an area. I turned around. I added half of the length and a third of the width to the middle of my area so that it was \(15 . \mathrm{I}\). returned. I summed the length and width and it was 7 . What are the length and width?

Construct two or three real-life division problems where giving the answer using just unit fractions, rather than other common fractions, makes sense.