Katz, chapter 1:

 

2. 93/5              1                     5

                        Ô2                     10Õ

                        4                      20

                        8                      40

                        `16                   80Õ    Still need 3.

                        `1/5                  1Õ

                        Ô1/3                  1 2/3Õ

                        Ô1/15                1/3Õ                              Answer: 18 + 1/3 + 1/5 + 1/15

 

4. 2 /11            1                      11

                        ½                      5 ½

                        1/3                   3 2/3

                        `1/6                  1 ½ 1/3Õ    Still need 1/6.

                        `1/66                1/6Õ                              Answer:  1/6 + 1/66

 

2/23

1                                 23

½                      11 ½

1/3                   7 2/3

1/6                   3 ½ 1/3

`1/12                1 ½  ¼ 1/6Õ  Still need 1/12.

`1/276              1/12Õ                            Answer: 1/12 + 1/276           

 

6,  Assume 7.   7 + 1/7 x 7 = 8. So multiply by 19/8 to get 133/8.

  Express this number in Egyptian form.

133/8               1                      8

2                                 16

4                      32

8                      64

`16                   128Õ     Still need 5.

`1/2                  4`

¼                      2

`1/8                  1Õ                     Answer: 16 + ½ + 1/8

 

8.  Assume 12.  12 + 12/4 + 12/3 = 19.  Multiply by 2/19 to get 24/19.

  Express this number in Egyptian form.

24/19               Ô1                     19Õ     Still need 5.

                        ½                      9 ½

                        1/3                   6 1/3

                        Ô1/4                  4 ½ ¼Õ   Still need ¼.

                        `1/76                ¼Õ                     Answer: 1 + ¼ + 1/76

 

 

 

12.      Divide 1; 0, 0, 0, 0, É by 7, using long division algorithm in base 60.

a)     1 divided by 7 is 0 with remainder 1. Bringing down the 0 multiplies by 60.

b)    60 divided by 7 is 8 with remainder 4. Bring down the 0 to get 240.

c)     240 divided by 7 is 34 with remainder 2. Bring down the 0 to get 120.

d)    120 divided by 7 is 17 with remainder 1. Bring down the 0 to get 60.

e)     We now have 60 divided by 7 again so the answer repeats.

 

18.   1 + 24/60/ 51/3600 + 10/ 216000  =

1 + .4 + .00141667 + .0000463      =  1.41421296297

The square root of 2 to 6 decimal places is 1.414214, so the approximation is accurate to 5 decimal places.

 

24.        How would we solve this pair of equations?

a)  One way is to take a guess.  Since we have multiples of 5 and division by 3,  we might try x = 30 and y = 25. This was a lucky guess and is the correct answer.

 x = 15 and y = 10. Then the sum of the squares is 225 + 100 = 325.  This  is between 4 and 5 times the desired result, so try doubling x to 30. Then y = 25 and we have the correct answer again.

b) If we apply the quadratic formula to the equation x^2 + (2/3 x + 5) ^2 Ð 1500 = 0 and take the positive solution, we get x = 30.

c) How might the Mesopotamians  have solved this problem?  They could put it into the form x^2 + 180/39 x = 1500x9/13 and used their  method for equations of this form to get  x = ( (90/39) ^2 = 1500x9/13)^1/2 Ð 90/39, again getting x = 1170/39 = 30, y = 25.

 

29.     Draw a large square with side x and mark off a rectangle b by x. The remaining rectangle is c.

Then divide the bx rectangle into two rectangles of width b/2 and height x. Move one of these rectangles to the bottom of the figure. The resulting figure is a gnomon with side

x Ð b/2.      If we add the square b/2 by b/2, we get a square of side x Ð b/2.

This gives ( x Ð b/2) ^2 = (b/2)^ 2 + c . So x Ð b/2 =    sqrt ((b/2)^ 2 + c) and finally

x = b/2 + (b/2)^ 2 + c.