THE ALGEBRA OF EGYPT
In our discussion on the Rhind 2/n Table, we discovered some algebra implicitly necessary for the egyptians to do "elementary" computations. In our discussions of multiplication and division, we discussed how the egyptians knew the unique decomposition of each integer as sums of powers of 2. Here we continue our discussions, and hope to convince the reader of the egyptain knowledge of perfect numbers and of arithmetic and geometric series. 
Rhind papyrus Problem 72. 100 loaves of pesu10 are to be exchanged for a certain number of loaves of pesu45. What is the number? The pesu number of a bread determined its strength in inverse order; in particular, pesu10 is stronger/better than pesu45 with the number determining some kind of percent of something undesired. Thus the problem reads in our modern sense as "What is (45/10)100?" Obviously its 450. Here's the egyptian solution: First find the excess of 45 over 10. You get 35. Divide 35 by 10. You get 3 + 1/2. Now multiply 3 + 1/2. You get 350. Ad 100 to 350, and get 450. The above may seem odd, but it is agebraically correct. Suppose we have x loaves of pesup and y loaves of x loaves of pesu p is to be exchanged forpesu q. 
Rhind papyrus Problem 24. A quantity (any) plus oneseventh of it becomes 19. What is the quantity? Says "x + x(1/7) = 19. What is x?" Restated, divide
19 by 1 + 1/7. Of course, we would write
Now if we double once more we get a number, 36 + 1/2 + 1/14,
greater than the numerator, so the Egyptians asked the question,
"What can be added to 18 + 1/4 + 1/28 to get 19?" They
deduced the answer to be 1/2 + 1/7 + 1/14. Then they asked, "What
must be multiplied by 1 + 1/7 to get 
Rhind papyrus Problem 28. A quantity together with its twothirds has one third its sum taken away to yield 10. What is the quantity? Says "x + (2/3)x  (x + (2/3)x)/3 = 10. What is x?" Restated, divide 10 by 1 + 2/3  (1 + 2/3)/3, or 10 by 1 + 1/3 + 1/6 + 1/18 (see 2/n table).
Now if we double once more we get a number, 12 + 1/3 + 1/9,
greater than 10, so the Egyptians asked the question, "What
can be added to 6 + 1/6 + 1/18 to get 10?" They deduced
the answer to be 3 + 1/2 + 1/6 + 1/9. Then they asked, "What
must be multiplied by 1 + 1/3 + 1/6 + 1/18 to get 3 + 1/2 + 1/6
+ 1/9?" This was deduced to be 2 + 1/4 + 1/28. So the solution
is 
Rhind papyrus Problem 33. The sum of a certain quantity together with its twothird, its half, and its onesventh becomes 37. What is the quantity? Says "x + x(2/3) + x(1/2) + x(1/7) = 37. What is x?" Restated, divide 37 by 1 +2/3 + 1/2 + 1/7.

Berlin Papyrus Problem 1. You are told the area of a square of 100 square cubits is equal to that of two smaller squares, the side of one square is 1/2 + 1/4 of the other. What are the sides of the two unknown squares. In modern terms we would express this as x^{2} + y^{2}
= 100 and x = (3/4)y. What are x and y? However, most translators believe the egyptians viewed this problem the way we do the simultaneous equations
Here was their solution. Assume the square of the first side
(y) to be 1 cubit. Then the other side (x) will be 1/2 + 1/4.
Then y^{2} = 1, and using egyptian
multiplication we determine x^{2} with
So x^{2} = 1/2 + 1/16. Thus, x^{2} + y^{2} = 1 + 1/2 + 1/16. Now (1 + 1/2 + 1/16)^{1/2} = 1 + 1/4 and (100)^{1/2} = 10 (we will discuss square roots later). Divide 10 by 1 + 1/4 and you get 8 (see the method of problem 24). So we get y=8. The Berlin Papyrus contains damage here so we can at best assume the solution for x was to divide 8 by 1/2 + 1/4 (as in the method of problem 24) to achieve x=6. 
Berlin Papyrus Problem 2. You are told the area of a square of 400 square cubits is equal to that of two smaller squares, the side of one square is 1/2 + 1/4 of the other. What are the sides of the two unknown squares. This is analogous to problem 1, so you solve it. 
Traditionally, arithmetic and geometric series are attributed to the Greek Pythagoras; however, the traditionalists have erred, for nearly 1500 years earlier we have:
Rhind papyrus Problem 64. Divide 10 hekats of barley among 10 men so that the common difference is 1/8 of a hekat of barley. Their solution is as follows: Average the value 10/10 = 1.
The total number of differences is then 101=9. Find half the
common difference, (1/2)(1/8) = 1/16. Multiply 9 by 1/16: 1/16
+ (8 by 16) = 1/2 + 1/16. Add this to the average value to get
the largest share 1 + 1/2 + 1/16. Subtract the common difference,
1/8, nine times [9*(1/8) = 1 + 1/8] to get the lowest share {s, s+d, s+2d, ... , s+d+2d+^{...}+(n1)d}, is sn + (1/2)dn(n1), or as applied above S/n = s + d(n1)/2. 
Rhind papyrus Problem 40. Divide 100 hekats of barley among 5 men so that the common difference is the same and so that the sum of the two smallest is 1/7 the sum of the three largest. This problem uses both arithmetic
series and simultaneous
equations.The common difference is d and s is the starting
number. 20 =100/5 = s + d(51)/2 = s+2d. The sum of the two smallest
is the sum of the three largest, so 3s +9d = 7*(2s + d) = 14s
+ 7d. So 2d = 11s and 20 = s+2d = 12s. So s = 1 + 1/2 + 1/6 and
2d = 11s = 18 + 1/3. So d = 9 + 1/6. The men get respectively. 
Rhind papyrus Problem 79. There are seven houses; in each house there are seven cats; each cat kills seven mice; each mouse has eaten seven grains of barley; each grain would have produced seven hekat. What is the sum of all the enumerated things.
The first two columns leads to the sum (in the bottom row) of the five terms of the geometric sequence with ratio 7 beginning with 7: 7+7^{2}+7^{3}+7^{4}+7^{5}. While the second to colums is the usual method for multiplying 7*2801. Finally, it is observed that the former sum equals the latter product. To an archeologist the table above and the relationship between the two columns may be meaningless, and several have said this. However, to an arithmetician, the relationship between the two columns is clear since (we know) the formula for the geometric series of the first n terms of a geometric series {1, r, r^{2}, ^{...}, r^{n}} of ratio r and beginning with 1 is 1+r+r^{2}+^{...}+r^{n} = (r^{n}1)/(r1). We have 7 times this value with r=7 and n=4. Thus, 7*(7^{4}1)/(71) = 7*(168071)/6 = 7*16806/6 = 7*2801 = 19607. Thus, problem79 is a table exhibiting the formula for the sum of a geometric series!!! 
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