2/pq = (2/A) x (A/p) where A = (p + 1).
Akhmim P. Eves suggests n/pq = 1/pr + 1/qr, where r = (p + q)/n
The rule seems to have been used in the RMP for the unusual 2/35 and 2/91 series, as given by:
2/35, r = (12/2) = 6, or 2/35 = 1/30 + 1/42
2/91, r = (20/2) = 10, or 2/91 = 1/70 + 1/130
2/21 = 2/8 × ((7+1)/21) = 1/4 × (1/3 + 1/21) = 1/12 + 1/84
Fraction 
Ahmes (Rhind) Papyrus, 1/a + 1/b as a + b 
2/p = 1/A + (2A  p)/Ap, with A = (p+1)/2 
Other representations of interest 
2/3  2 + 6  2 + 6 

2/5  3 + 15  3 + 15 

2/7  4 + 28  4 + 28 
6 + 14 + 21 
2/9  6 + 18  5 + 45 

2/11  6 + 66  6 + 66 

2/13  8 + 52 + 104  7 + 91 
10 + 26 + 65 
2/15  10 + 30  8 + 120 
12 + 20 
2/17  12 + 51 + 68  9 + 153 

2/19  12 + 76 + 114  10 + 190 

2/21  14 + 42  11 + 231 
15 + 35 
2/23  12 + 276  12 + 276 

2/25  15 + 75  13 + 325 

2/27  18 + 54  14 + 378 

2/29  24 + 58 + 174 + 232  15 + 435 

2/31  20 + 124 + 155  16 + 496 

2/33  22 + 66  17 + 561 

2/35  30 + 42  18 + 630 
20 + 140 
2/37  24 + 111 + 296  19 + 703 

2/39  26 + 78  20 + 780 
52 + 60 + 65 
2/41  24 + 246 + 328  21 + 861 

2/43  42 + 86 + 129 + 301  22 + 946 

2/45  30 + 90  23 + 1035 
36 + 60 
2/47  30 + 141 + 470  24 + 1128 

2/49  28 + 196  25 + 1225 
42 + 98 + 147 
2/51  34 + 102  26 + 1326 

2/53  30 + 318 + 795  27 + 1431 

2/55  30 + 330  28 + 1540 
40 + 88 
2/57  38 + 114  29 + 1653 

2/59  36 + 236 + 531  30 + 1770 

2/61  40 + 244 + 488 + 610  31 + 1891 

2/63  42 + 126  32 + 2016 
56 + 72 
2/65  39 + 195  33 + 2145 
45 + 117 
2/67  40 + 335 + 536  34 + 2278 

2/69  46 + 138  35 + 2415 

2/71  40 + 568 + 710  36 + 2556 

2/73  60 + 219 + 292 + 365  37 + 2701 

2/75  50 + 150  38 + 2850 
60 + 100 
2/77  44 + 308  39 + 3003 
63 + 99 
2/79  60 + 237 + 316 + 790  40 + 3160 

2/81  54 + 162  41 + 3321 

2/83  60 + 332 + 415 + 498  42 + 3486 

2/85  51 + 255  43 + 3655 
55 + 187 
2/87  58 + 174  44 + 3828 

2/89  60 + 356 + 534 + 890  45 + 4005 

2/91  70 + 130  46 + 4186 
7 + 637 
2/93  62 + 186  47 + 4371 

2/95  60 + 380 + 570  48 + 4560 

2/97  56 + 679 + 776  49 + 4753 

2/99  66 + 198  50 + 4950 
90 + 110 
2/101  101 + 202 + 303 + 606  51 + 5151 

http://compgeom.cs.uiuc.edu/~jeffe/wagon/s848.html
Famous unsolved problem (see [Guy, Unsolved Problems in Number Theory, Springer; or Klee & Wagon, Old and New Unsolved Problems in Plane Geometry and Number Theory, MAA]: Does the odd greedy algorithm for getting a representation of m/n (n odd) as a sum of distinct odd unit fractions terminate always? ("Greedy" means: at each step chose the largest 1/m that will fit, with m odd and different from previous choices.)
My PoW was motivated by my discovery a year ago that 3/179 gives quite interesting results on the odd greedy algorithm! It yields 19 terms, beginning with:
63, 2731, 5963959, . . .The last term has almost 500,000 digits! More amazing, the numerators of the remainders (which decrease in the unrestricted greedy algorithm), are
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 2, 3, 4, 1Quite a pattern! What is going on?
Now, some correspondents last year found better representations: Kevin Brown (Seattle) found: 63, 1611, 3759 and Milo Gardner (Sacramento) found: 63, 1253, 11277.
Several of you (Ron Hardin, AT&T Labs, Research; John Guilford, HP, Everett, Wash.; David Eppstein, UC Irvine Dept. of Information & Computer Science; Charles Rees, Univ. of New Orleans) found Brown's answer, so it might well be the case that Brown's solution is the simplest 3term representation there is. [It is. See below.] There is no 2term rep'n (exercise).
Some of you might be interested in the question of what algorithms the Egyptians used when computing Egyptian fractions. Many of them are immortalized on the Rhind papyrus, and Milo Gardner has made an extensive investigation. Contact him for details at gardnerm@ecs.csus.edu.
Eppstein also found that 3/179 =
1/171 + 1/209 + 1/285 + 1/895 + 1/1611 + 1/1969 + 1/2685which might be the smallest maxdenominator. [It is.]
Eppstein's home page has misc. Egyptian fraction stuff (http://www.ics.uci.edu/~eppstein/numth/egypt/) and his article in Mathematica in Education and Research [vol 4, no. 2, 1995, pp 515] discusses Mathematica implementations of a variety of algorithms. I guess I should throw my own code down. It implements the odd greedy method, together with an option for seeing the numerators of the remainders, which might, if a proof is ever found, play a role in the proof that this halts. The code is from a revision of my book Mathematica in Action (Springer).
Now, Neil Sloane and Ron Hardin (AT&T) conjecture that every 3/b (b not divisible by 3) has a 3term representation as a sum of distinct odd unit fractions. This is clearly true in the unrestricted case (greedy algorithm, numerators of remainders decrease!). Their conjecture nicely complements famous conjectures in this area:
David Bailey (NASA Ames), using customized highprecision arithmetic on an SGI work station and aware of my desire to "stump" the odd greedy algorithm, found the following monster: 3/2879. The odd greedy rep'n has 21 terms; the last term has 3,018,195 digits; and the sequence of numerators of the remainders is
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 1
I believe with this computation David gets the record for the largest Egyptian fraction computation ever done. I think my 3/179 was a record when I did it last year. Nice!
Eppstein then found: 1/963 + 1/308053 + 1/2772477 and 1/1615 + 1/5355 + 1/14395 + 1/20153 + 1/25911 + 1/43185 + 1/48943 + 1/54701 + 1/60459 for this example, the latter being the unique example with minmaxdenominator.
It would be nice to have some heuristics to predict where the bad examples for the odd greedy algorithm are. In particular, I would like one that stumps David Bailey's software in the sense that he will be unable to tell whether the algorithm halts.
Late News: Boy, #847 generated a lot of correspondence! David Bailey reports that 5/5809 stumps his program, which is limited to about 15,000,000,000digit precision. He gets:
1/163 + 1/1125979 + [21 terms omitted] + 1/s + ...where the integer s is approximately 6.875 x 1012565952. So here we have an example of a conceptually very simple algorithm, and a specific input, and we have no idea, and may never have any idea, whether the algorithm halts. This is very nice indeed.
Of course, I have no doubt that nice rep'ns exist. The issue is that odd greedy is very mysterious.
The remaindernumerators for 5/5809 look like: 5, 6, 7, 8,.... 20, 21, which is as far as I checked. So I guess a weaker conjecture than the main one here is:
Can one disprove the possibility that the numerators of the remainders in an odd greedy instance look like 5,6,7,8,.........all the way to infinity?
Eppstein is working on a Mathematica package that incorporates a variety of methods. David Bailey (and I) continue to wonder about 5/5809's behavior under odd greedy. His program checks to 60,000,000 digits, but it did not halt. So, if this halts, it leads to numbers having more than 60,000,000 digits.
Eppstein's package has no trouble finding reps. of 5/5809 such as: 1/897 + 1/6739 + 1/20217 and 1/2051 + 1/2925 + 1/5567 + 1/5985 + 1/7325. Eppstein adds that his small maxdenominator result reported in previous mailing is provably optimal. Readers interested in his generalpurpose Egyptian fractions package should contact him: eppstein@euclid.ics.uci.edu.
NEW! David Eppstein, using modular reduction (continually reduce the denominator modulo, say, 30!) was able to prove that the sequence does indeed halt, and that the numerators of the remainders look like 5, 6, 7, ..., 29, 30, 1. (David actually used the modulus 37943838567204570000 = 30!/224·310·53·7.) This is very clever and completely avoids the ridiculously large numbers one gets without the reduction. Bailey got up to 60,000,000 digits in the 24th denominator, and then was forced to stop. But now we now that a few more steps does indeed lead to termination in this one case. Nice work, David.
Jeff Erickson forwarded me your abovetitled message on odd Egyptian fraction representations of 3/179. You asked to be sent any solutions.
As I believe you know, I have a large collection of algorithms for constructing Egyptian fractions, published in Mathematica in Educ. & Res., and online at http://www.ics.uci.edu/~eppstein/numth/egypt/. Most of these methods produce some even denominators, but a few do not. (Beware, I probably introduced some typos transferring these from my Mac to my Sun.)
EgyptOddGreedy[3/179] didn't terminate in a reasonable amount of time. I assume this is why you chose this particular fraction to ask about...
The proofs I know of that odd representations exist are I think pretty similar to the method I implemented as EgyptRemainder, which takes as a second argument a "powerful" number (with lots of divisors so that there are lots of ways to form sums of divisors). If the powerful number is odd, you get odd representations, modulo some recombination that happens because of possible repeated terms. EgyptRemainder[3/179,3*3*3*5*5*7] yielded the following:
1/105 + 1/189 + 1/525 + 1/33831 + 1/93795EgyptShort[3/179,3], a closetobruteforce search for representations with few terms (the second argument is the number of terms) produced a number of threeterm representations, among them the odd
1/105 + 1/189 + 1/525 + 1/31325 + 1/120825
1/105 + 1/175 + 1/675 + 1/33831 + 1/93795
1/105 + 1/175 + 1/675 + 1/31325 + 1/120825
1/75 + 1/525 + 1/675 + 1/33831 + 1/93795
1/75 + 1/525 + 1/675 + 1/31325 + 1/120825
1/75 + 1/315 + 1/4725 + 1/33831 + 1/93795
1/75 + 1/315 + 1/4725 + 1/31325 + 1/120825
1/63 + 1/1575 + 1/4725 + 1/33831 + 1/93795
1/63 + 1/1575 + 1/4725 + 1/31325 + 1/120825
1/61 + 1/2745 + 1/491355These are therefore the only odd threeterm representations. The last of these is particularly succinct. There is only one twoterm representation of 3/179, not odd.
1/63 + 1/1253 + 1/11277
1/63 + 1/1611 + 1/3759
A small modification of EgyptSmallMult (another brute force method, which combines small multiples of the original fraction in order to get something that differs from the input by a fraction not divisible by the original denominator) shows that
3/179 = 1/895 + 1/1611 + 1/1969 + 1/2685 + 7/495,and that any other set of odd unit fractions having a difference from 3/179 that is not itself a multiple of 1/179 must include a higher denominator. Calling EgyptShort on 7/495 produces the representations
3/179 = 1/179 + 1/537 + 1/895 + 1/1253 + 1/2327 + 1/2685 + 3/455,
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/77 + 1/1485 + 1/2079which have the minimum max denominator (2685) of any odd representation of 3/179. Unless I missed one in going through the results, all other representations with the same max denominator have more terms.
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/77 + 1/1575 + 1/1925
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/77 + 1/1683 + 1/1785
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/891 + 1/1485
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/935 + 1/1377
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/561 + 1/1683
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/693 + 1/1071
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/765 + 1/935
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/87 + 1/495 + 1/1595
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/91 + 1/585 + 1/693
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/95 + 1/495 + 1/627
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/275 + 1/2475
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/285 + 1/1881
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/297 + 1/1485
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/315 + 1/1155
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/385 + 1/693
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/429 + 1/585
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/105 + 1/315 + 1/693
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/105 + 1/385 + 1/495
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/117 + 1/195 + 1/2145
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/117 + 1/209 + 1/1235
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/117 + 1/221 + 1/935
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/135 + 1/165 + 1/1485
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/135 + 1/189 + 1/693
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/135 + 1/209 + 1/513
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/143 + 1/195 + 1/495
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/165 + 1/225 + 1/275
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/171 + 1/209 + 1/285

David Eppstein UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
http://compgeom.cs.uiuc.edu/~jeffe/wagon/s848.html

http://www.ics.uci.edu/~eppstein/numth/egypt/
Nowadays, we usually write noninteger numbers either as fractions (2/7) or decimals (0.285714). The floating point representation used in computers is another representation very similar to decimals. But the ancient Egyptians (as far as we can tell from the documents now surviving) used a number system based on unit fractions: fractions with one in the numerator. This idea let them represent numbers like 1/7 easily enough; other numbers such as 2/7 were represented as sums of unit fractions (e.g. 2/7 = 1/4 +1/28). Further, the same fraction could not be used twice (so 2/7 = 1/7 + 1/7 is not allowed). We call a formula representing a sum of distinct unit fractions an Egyptian fraction.
This notation is cumbersome and difficult to compute with, so the Egyptian scribes made large tables so they could look up the answers to arithmetic problems. However there is also some interesting mathematics associated with the problem of converting modern fraction notation into the Egyptian form. A number of famous mathematicians have looked at this problem, and invented different ways of doing this conversion process. Each of these methods has advantages and disadvantages in terms of the complexity of the Egyptian fraction representations it produces and in terms of the amount of time the conversion process takes. There are still some unsolved problems about whether some of these processes finish, or whether they get into an infinite loop.
To investigate some of these questions, I wrote a Mathematica notebook, now called "Algorithms for Egyptian Fractions", which as the title implies implements on the computer some of these computation methods. A version of this notebook was published as "Ten Algorithms for Egyptian Fractions" in Mathematica in Education and Research, vol. 4, no. 2, 1995, pp. 515, available online through MathSource.
Since then I have made several changes including improvements to the binary remainder method and two new sections on reverse greedy methods and brute force searches. I am making this updated version available here.
1  1  1  1  1  1  1  1  1  1  1  1  
1 = 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

6  7  8  9  10  14  15  18  20  24  28  30 
1  1  1  1  1  
k = 

+ 

+ 

+ 

+ ··· + 

2  3  4  5  n 
(Hint: consider the largest power of two less than or equal to n.)
Some of the papyrus scrolls date back to about 2000 BC, around the time of the construction of the larger Egyptian pyramids. Because there are deserts on either side of the Nile, papyrus scrolls have been well preserved in the dry conditions.
So what was on them? How to preserve a body
as a mummy? Maybe it was
how to construct the extensive system of canals used for irrigation
across Egypt or on storage of grain in their great storage
granaries? Perhaps they tell how about navigation and how to build
boats (again out of papyrus reeds which float very well, and pictures of
these boats have been found in many Egyptian tombs)?
The surprising answer is that the oldest ones are about mathematics!
The hieroglyphs (picturewriting) on the papyrus were only deciphered in 1842 (and the Babylonian claytablet cuneiform writing was deciphered later that century).
It starts off by saying that the scribe "Ahmes" is writing it about 1600 BC but that he had copied it from "ancient writings" so it probably goes back to at least 2000BC and probably further. The picture is also a link so click on it to go to the St Andrews MacTutor biography of Ahmes.
Since early civilizations would need to predict the start of spring accurately in order to sow seeds, then a large part of such mathematical writing has applications in astronomy. Also, calculations were needed for surveying (geometry) and for building and for accounting. However, quite a lot of the problems in the RMP are arithmetic puzzles  problems posed just for the fun of solving them!
On this page we will look at how the Egyptians of 5000 years ago worked with
fractions.
Instead, they were able to write any fraction as a sum of unit fractions where all the unit fractions were different.
For example,
6/7 = 1/2 + 1/3 + 1/42
First Fatima sees that they all get at least half a sack of grain, so she gives all
8 of them half a sack each, with one sack left.
Now it is easy to divide one sack into 8, so they get an extra eighth of a sack each
and all the sacks are divided equally between the 5 workers. On the picture
here they each receive one red part
(1/2 a sack) and one green part (1/8 of a sack):
and
5/8 = 1/2 + 1/8
We could use decimals so that 3 /4 =0.75 whereas 4/5 =0.8 so we can see that 4/5 is bigger.
Could you do this without converting to decimals?
What common fraction could we convert both 3/4 and 4/5 into?
20 would do so that
3/4 = 15/20 whereas
4/5 = 16/20
so again we can easily see that 4/5 is larger
than 3/4.
Using Egyptian fractions we write each as a sum of unit fractions:
3/4 = 1/2 +
1/4
4/5 = 1/2 +
3/10
and, expanding 3/10 as 1/4 + 1/20
we have
4/5 = 1/2 + 1/4 +
1/20
We can now see that 4/5 is the larger  by exactly
1/20.
Here are some results that mathematicians have proved:
You can skip over these two subsections if you like.
Remember that
The method is to find the biggest unit fraction we can and take it from T/B. With what is left, we repeat the process. We will show that this series of unit fractions always decreases, never repeats a fraction and eventually will stop. Such processes are now called algorithms and this is an example of a greedy algorithm since we (greedily) take the largest unit fraction we can and then repeat on the remainder.
Let's look at an example before we present the proof: 521/1050.
521/1050 is less than onehalf (since 521 is less than a half of 1050)
but it is bigger than onethird. So the largest unit fraction we can take away from
521/1050 is 1/3:
Now let's see how we can show this is true for all fractions T/B. We want
What is the remainder?
It is
That completes the proof that
However, the egyptian fraction produced by the greedy method may not be the shortest
such fraction. Here is an example:
by the greedy method, 4/17 reduces to
Here is the complete list of all the shortest representations of T/B for B up to 11. We use a list notation here to make the unit fractions more readable. For instance, above we saw that:
2/3 = [2,6]8/11 has an unusually large number of different (shortest) reprentations!
2/5 = [3,15]
2/7 = [4,28]
2/9 = [5,45] = [6,18]
2/11 = [6,66]
3/4 = [2,4]
3/5 = [2,10]
3/7 = [3,11,231] = [3,12,84] = [3,14,42] = [3,15,35] = [4,6,84] = [4,7,28]
3/8 = [3,24] = [4,8]
3/10 = [4,20] = [5,10]
3/11 = [4,44]
4/5 = [2,4,20] = [2,5,10]
4/7 = [2,14]
4/9 = [3,9]
4/11 = [3,33]
5/6 = [2,3]
5/7 = [2,5,70] = [2,6,21] = [2,7,14]
5/8 = [2,8]
5/9 = [2,18]
5/11 = [3,9,99] = [3,11,33] = [4,5,220]
6/7 = [2,3,42]
6/11 = [2,22]
7/8 = [2,3,24] = [2,4,8]
7/9 = [2,4,36] = [2,6,9]
7/10 = [2,5]
7/11 = [2,8,88] = [2,11,22]
8/9 = [2,3,18]
8/11 = [2,5,37,4070] = [2,5,38,1045] = [2,5,40,440] = [2,5,44,220]
= [2,5,45,198] = [2,5,55,110] = [2,5,70,77] = [2,6,17,561] = [2,6,18,198]
= [2,6,21,77] = [2,6,22,66] = [2,7,12,924] = [2,7,14,77] = [2,8,10,440]
= [2,8,11,88]
9/10 = [2,3,15]
9/11 = [2,4,15,660] = [2,4,16,176] = [2,4,20,55] = [2,4,22,44] = [2,5,10,55]
10/11 = [2,3,14,231] = [2,3,15,110] = [2,3,22,33]
Here is a table of the lengths of the shortest Egyptian Fractions for all fractions T/B (Top over Bottom) where the denominator B takes all values up to 30:
Shortest Egyptian Fractions lengths for fraction T/BSo the "smallest" fraction that needs three terms is T=4 B=5 i.e. 4/5
KEY:  means the fraction T/B is not in its lowest form (eg 9/12).
Find its lowest form P/Q (9/12=3/4) and then look up that fraction
. means the fraction T/B is bigger than 1. Try B/T instead!
Otherwise, the number is the *minimum* number of unit fractions
that are needed to sum to T/B.
Find T down the side and B across the top
\B: 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3
T\ 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
2 2  2  2  2  2  2  2  2  2  2  2  2  2  2 
3 . 2 2  3 2  2 2  3 2  2 2  3 2  2 2  2 2  2 2 
4 . . 3  2  2  2  3  2  3  2  2  2  3  2  3 
5 . . . 2 3 2 2  3 2 3 2  2 3 2 2  2 3 3 2  2 2 2 2 
6 . . . . 3    2  3    2  3    2  2    2 
7 . . . . . 3 3 2 3 2 2  3 3 3 2 3 2  3 3 2 3 2 2  3 2
8 . . . . . . 3  4  3  2  4  3  2  2  3  3  3 
9 . . . . . . . 3 4  3 2  2 2  4 2  3 3  3 2  2 3 
10 . . . . . . . . 4  3    3  2  2  3    2  2 
11 . . . . . . . . . 3 3 3 3 3 3 2 3 2 2  3 2 3 3 3 2 3 2
12 . . . . . . . . . . 4    3  3    2  4    4 
13 . . . . . . . . . . . 4 3 3 3 3 3 3 3 2 3 2 2  3 3 3 2
14 . . . . . . . . . . . . 3  4  4    3  3  2  4 
15 . . . . . . . . . . . . . 4 4  4   3 4   2  2 2 
16 . . . . . . . . . . . . . . 5  3  3  4  3  3  3 
17 . . . . . . . . . . . . . . . 3 4 3 3 3 4 3 3 3 3 3 3 2
18 . . . . . . . . . . . . . . . . 4    4  3    4 
19 . . . . . . . . . . . . . . . . . 3 3 3 4 3 3 4 3 3 4 3
20 . . . . . . . . . . . . . . . . . . 4  4    4  4 
21 . . . . . . . . . . . . . . . . . . . 4 5  3 4   4 
22 . . . . . . . . . . . . . . . . . . . . 5  4  4  3 
23 . . . . . . . . . . . . . . . . . . . . . 3 4 4 3 3 4 3
24 . . . . . . . . . . . . . . . . . . . . . . 4    4 
25 . . . . . . . . . . . . . . . . . . . . . . . 4 4 3 4 
26 . . . . . . . . . . . . . . . . . . . . . . . . 4  4 
27 . . . . . . . . . . . . . . . . . . . . . . . . . 4 5 
28 . . . . . . . . . . . . . . . . . . . . . . . . . . 5 
29 . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
4/5 = 1/2 + 1/4 + 1/20
and
4/5 = 1/2 + 1/5 + 1/10
i 2/(2i+1)=Let's concentrate on the first sum on each line since some of the fractions above have more than one form as a sum of two unit fractions.
1 2/3 = 1/2 + 1/6
2 2/5 = 1/3 + 1/15
3 2/7 = 1/4 + 1/28
4 2/9 = 1/5 + 1/45 = 1/6 + 1/18
5 2/11 = 1/6 + 1/66
6 2/13 = 1/7 + 1/91
7 2/15 = 1/8 + 1/120 = 1/9 + 1/45 = 1/10 + 1/30 = 1/12 + 1/20
8 2/17 = 1/9 + 1/153
9 2/19 = 1/10 + 1/190
It looks as if
2/(2i+1) = 1/(i+1) + 1/XCan you spot how we can use (2i+1) and i to find the missing X?
1 2/3 = 1/2 + 1/6Yes! Just multiply them together!
2 2/5 = 1/3 + 1/15
3 2/7 = 1/4 + 1/28
4 2/9 = 1/5 + 1/45 = 1/6 + 1/18
5 2/11 = 1/6 + 1/66
6 2/13 = 1/7 + 1/91
7 2/15 = 1/8 + 1/120 = 1/9 + 1/45 = 1/10 + 1/30 = 1/12 + 1/20
8 2/17 = 1/9 + 1/153
9 2/19 = 1/10 + 1/190
2/(2i+1) = 1/(i+1) + 1/[(i+1)(2i+1)]We can check it by simplifying the fraction on the right and seeing if it reduces to the one on the left:
1/(i+1) + 1/[(i+1)(2i+1)]So algebra has shown us that the formula is always true.
= (2i+1 + 1)/[(i+1)(2i+1)]
= (2i+2)/[(i+1)(2i+1)]
= 2(i+1)/[(i+1)(2i+1)]
= 2/(2i+1) which is the fraction we wanted!!
2/15 = 1/8 + 1/120so it has four representations. In the table below, under T/B=2/15 we have the entry 4:
= 1/9 + 1/45
= 1/10 + 1/30
= 1/12 + 1/20
NUMBER of Shortest Egyptian Fractions:
\B 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
T\ 2: 1  1  1  2  1  1  4  1  1 
3: . 1 1  6 2  2 1  8 3  2 1  8 4
4: . . 2  1  1  1  4  3  4  1 
5: . . . 1 3 1 1  3 2 5 1  1 5 2 1 
6: . . . . 1    1  1    1  2 
7: . . . . . 2 2 1 2 2 1  6 5 3 1 5 2
8: . . . . . . 1  15  3  2  21  2 
9: . . . . . . . 1 5  1 1  1 1  14 1
10: . . . . . . . . 3  1    3  1 
11: . . . . . . . . . 2 1 1 2 2 1 1 3 1
12: . . . . . . . . . . 3    1  1 
13: . . . . . . . . . . . 6 2 1 1 2 1 5
14: . . . . . . . . . . . . 1  2  5 
15: . . . . . . . . . . . . . 5 4  3 
16: . . . . . . . . . . . . . . 39  1 
17: . . . . . . . . . . . . . . . 1 3 2
18: . . . . . . . . . . . . . . . . 1 
19: . . . . . . . . . . . . . . . . . 1
The above is just a resource for you to experiment with and to check your own results.
Egyptian Fractons
You may find it amazing that fractions, as we know them, barely existed, for the european civilization, until the 17'th century. Even in the 19th century, a method called russian peasant fractions, was the same used by the Europeans since they met the African, and the Egyptians at least since 4000BC in Egypt. As the method was found on several papyrus, we now call this technique egyptian fractions. Let us agree to call a number a unit fraction if it has form 1/n , where n is a positive integer. An egyptian fraction is an expression of the sum of unit fractions
1/a + 1/b + 1/c + ... , where the denominators a,b,c, ... are increasing.
An egyptian number is any number equal which can be expressed as the sum of an integer plus the sum of an Egyptian fraction. Here are some egyptian fractions:1/2 + 1/3 (so 5/6 is an egyptian number), 1/3 + 1/11 + 1/231 (so 3/7 is an egyptian number), 3 + 1/8 + 1/60 + 1/5280 (so 749/5280 is an egyptian number). The egyptians also made note of the fraction 2/3.
1/5 + 1/37 + 1/4070 and 1/6 + 1/22 + 1/66 are regarded as different egyptian fractions even though the sum of each is 5/22.
The earliest records of egyptian fractions date to nearly 3900 years ago in the papyrus copied by Ahmes (sometimes called Ahmos  ref1, ref2) purportedly from records at least 300 years earlier. It is conjectured that the mysterious, so called, meaningless, egyptian triple 13, 17, 173 actually means
3 + 1/13 + 1/17 + 1/173 = 3.141527
which approximatesto
4 places!!!
(considerably better than the usual 3.16 credited to the
egyptians.)
However, I have not been able to locate a credible
reference for the triple 13, 17, 173 in this context!!!)
However, to the victor goes the spoils; i.e., prior to the "discovery" of the Rhind papyrus, egyptian fractions were thought, by european mathematicians, to come from the Greeks. Even the name pi is Greek.
The rule of Egyptian fractions requires us
to write only unit fractions, integers, and their sums. So if
a duke is awarded 3/7'th of the conquered land, the quanity might
be represented as (1/4 + 1/7 + 1/28)'th of the conquered land,
which is a bit better than
(1/3 + 1/11 + 1/231)'th of the conquered land, but still awkward.
Until the 18'th century, when our present method, from India and
also thousands of years old, of writing any integer in the numerator
became popular in Europe, Egyptian fractions were the primary
method of writing noninteger numbers.
THEOREM. Every rational number is an egyptian number.
The modern proof of the Theorem was discovered in 1880, but European's have known how to compute Egyptian numbers since Fibonacci in the 12'th century. Before exhibiting the rule we make a convention. Given a noninteger r, let [r] denote the smallest integer > r.
Suppose p/q < 1 is written in lowest terms then there is an egyptian fraction with at most p terms and whose sum is p/q (see the proof below). Let r/s = p/q 1/[q/p]. So p/q = 1/[q/p] + r/s. If r=1, we are done; otherwise, repeat the process. Here are some examples:
1. Consider p/q = 4/23. Since 23/4 = 5.65, [23/4] = 6. Compute 4/23  1/6 as a fraction, and get 1/138. Thus, 4/23 = 1/6 + 1/138.
2. Consider 5/22. [22/5] = 5. 5/22  1/5 =
3/110. Now [110/3] = 37. But
3/110  1/37 = 1/4070. So 5/22 = 1/5 + 1/37 + 1/4070.
3. With a little ingenuity, you can determine other egyptian fractions whose sum is 5/22. For example, let's start, for no special reason, with 1/6 instead of 1/5. 5/22  1/6 = 2/33. [33/2] = 17 and 2/33  1/17 = 1/561. Thus, 5/22 = 1/6 + 1/17 + 1/561.
4. In the last expression for 5/22, keep 1/6 but exchange 1/17 for 1/22. This means we compute 2/33  1/22 = 1/66. Thus, 5/22 = 1/6 + 1/22 + 1/66. This expression is more satisfactory since the denominators are not as large as in the proceeding two cases..
Note: There is significant interest in determining which expression is "best" or what the egyptians would have used. We discuss this on the page "The Best Egyptian Fraction."
About the proof of the theorem:
Notice that when 0 < p < q are integers with only 1 as a common divisor (i.e., p/q < 1 is written in lowest terms) the construction in the 2nd paragraph after the theorem gives, via mathematical induction, the proof of the theorem, since for p[q/p]q < p.
In spite of the Theorem, there is very little interest in egyptian fractions (or even their modernized version  continued fractions) today, and only infinite series, a topic of elementary calculus could indicate their passage (recall from calculushttp://www.math.buffalo.edu/mad/AncientAfrica/mad_egyptianfractions.html

http://mathforum.org/epigone/historia_matematica/stangtroastrimp
To discuss the later time period, I like Howard Eves' explanation of a
general Egyptian fraction rule. Eves reported that the 400 AD Akhmim P.
contained such a general rule. I first came across this rule in 1962, in
a history of math class, as Eves stated in the simple form:
n/pq = 1/pr + 1/qr
where r = (p + q)/n
The rule seems to have been used in the RMP for the unusual 2/35 and
2/91 series, as given by:
2/35, r = (12/2) = 6, or 2/35 = 1/30 + 1/42
2/91, r = (20/2) = 10, or 2/91 = 1/70 + 1/130
For myself, I doubt that Ahmes thought in this manner. I suspect two
other algebraic identity ideas as being more likely to have computed
2/35 and 2/91. One is the product of the arithmetic mean = (p + q)/2
and the geometric mean = pq^1/2 , as Howard Eves discusses (in the
same paragraph that he reviews the Moscow Mathematical Papyrus
and the apparent use of the general formula V = h(a^2 + ab+ b^2)/3).
The second algebraic identity may have been a trail and error form of
thinking (as Ahmes knew by his use of the method of false position).
Finally the 400 AD Akhmim P. offers other clues to reading Egyptian
fraction in historical ways. The AP n/3  n/32 tables were published by
Knorr in 1982. I have condensed the AP to its n/17 and n/19 tables
since its. conversions of 4/17, 8/17, 3/19, 7/19 and 9/19 are
interesting, using methods that improve upon the RMP optimal series
Kevin Brown, a number theorist, was a major contributor
http://www.ecst.csuchico.edu/~atman/Misc/horuseye.html
Thank you all. once again, for considering Egyptian fractions, not in
their usual modern recreational math context, or as a step child of
Greek math, or limited to hieroglyphic script, but vividly read by royal
higher math, protonumber theory. My examples are sources three Middle
Kingdom mathematical texts and one 400 AD Coptic/Greek mathematical text,
restated as should have been compared and jointly analyzed, beginning
130 140 years ago. Comments are welcomed.
Best Regards to all,
Milo Gardner
Sacramento, CA
http://mathforum.org/epigone/historia_matematica/stangtroastrimp
http://www.math.buffalo.edu/mad/AncientAfrica/bestegyptianfraction.html
The Best Egyptian Fractons
It is our intention to use some of the recent work of Milo Gardner and others to understand Egyptian Fractions. Below is a list, found on the Rhind (Ahmes) Papyrus, used for 2/n where n is an odd n umber from 3 to 101.
2/3 = 1/2 + 1/6 
2/5 = 1/3 + 1/15 
2/7 = 1/4 + 1/28 
2/9 = 1/6 + 1/18 
2/11 = 1/6 + 1/66 
2/13 = 1/8 + 1/52 + 1/104 
2/15 = 1/10 + 1/30 
2/17 = 1/12 + 1/51 + 1/68 
2/19 = 1/12 + 1/76 + 1/114 
2/21= 1/14 + 1/42 
2/23 = 1/12 + 1/276 
2/25 = 1/15 + 1/75 
2/27 = 1/18 + 1/54 
2/29 = 1/24 + 1/58 + 1/174 + 1/232  
2/31 = 1/20 + 1/124 + 1/155 
2/33 = 1/22 + 1/66 
2/35 = 1/25 + 1/30 + 1/42 
2/37 = 1/24 + 1/111 + 1/296 
2/39 = 1/26 + 1/78 
2/41 = 1/24 + 1/246 + 1/328 
2/43 = 1/42 + 1/86 + 1/129 + 1/301 
2/45 = 1/30 + 1/90  
2/47 = 1/30 + 1/141 + 1/470 
2/49 = 1/28 + 1/196 
2/51 = 1/34 + 1/102 
2/53 = 1/30 + 1/318 + 1/795 
2/55 = 1/30 + 1/330 
2/57 = 1/38 + 1/114 
2/59 = 1/36 + 1/236 + 1/531 
2/61 = 1/40 + 1/244 + 1/488 + 1/610  
2/63 = 1/42 + 1/126 
2/65 = 1/39 + 1/195 
2/67 = 1/40 + 1/335 + 1/536 
2/69 = 1/46 + 1/138 
2/71 = 1/40 + 1/568 + 1/710 
2/73 = 1/60 + 1/219 + 1/292 + 1/365 
2/75 = 1/50 + 1/150 
2/77 = 1/44 + 1/308 
2/79 = 1/60 + 1/237 + 1/316 + 1/790 
2/81 = 1/54 + 1/162 
2/83 = 1/60 + 1/332 + 1/415 + 1/498  
2/85 = 1/39 + 1/195 
2/87 = 1/58 + 1/174 
2/89 = 1/60 + 1/356 + 1/534 + 1/890 
2/91 = 1/70 + 1/130 
2/93 = 1/62 + 1/186 
2/95 = 1/60 + 1/380 + 1/570 
2/97 = 1/56 + 1/679 + 1/776 
2/99 = 1/66 + 1/198 
2/101 = 1/101 + 1/202 + 1/303 + 1/606 
from Milo Gardner:
I would like to add a simpler 'Occam's Razor'
view of 2/pq series found in the Rhind
Mathematical Papyrus [RMP]. Rather than 2/pq = (1/q + 1/pq)2/(p
+ 1) as used for all but four series, I suggest that: 2/pq = 2/A
x A/pq where A = (p + 1) and (p + q) covers all but 2/95. As discussed
previously, 2/95 is simple a mod 5 version of 2/p, where 2/p =
1/A + (2A p)/Ap with
p = 19, A = 12, with aliquot parts of A ,3 + 2, used as follows:
2/19 = 1/12 + (3 + 2)/(12*19) = 1/12 + 1/76 + 1/114
such that: 2/95 = 1/5 x (1/12 + 1/76 + 1/114) = 1/60 + 1/380 +
1/570 as Ahmes may have mentally computed.
In conclusion, I would like to add that Fibonacci
used a parametric version of the 2/p rule, extending Ahmes' range
for A, from p/2 < A , p, with A being a highly composite number,
to p/n < A < 2p for solving n/p series.
Regards,
Milo Gardner 12/16/99
http://www.math.buffalo.edu/mad/AncientAfrica/bestegyptianfraction.html

http://www.mathpages.com/home/rhind.htm
The Rhind Papyrus 2/N Table
One of the most puzzling episodes in the history of human thought is
the 2000year reign of Egyptian unit fractions. We can, at least
in part, reconstruct the arithmetical manipulations involved, but
the underlying reason or motive for expressing fractional quantities
as sums of unit fractions remains mysterious. Was it simply a
cumbersome style of writing that persisted for so many centuries just
out of deference to traditional forms, or did it express an actual way
of thinking that has since been forgotten?
At the beginning of almost every general history of mathematics we
find a description of how the ancient Egyptians operated with fractions
almost exclusively in terms of UNIT fractions. For example, instead
of saying 2/5 of my land was flooded, they would say 1/3 + 1/15 of
my land was flooded. One of the earliest written records from
ancient Egypt (transcribed circa 1650 BC from a source believed to
date from around 1850 BC or earlier) is known as the Rhind Mathematical
Papyrus, and contains a table expressing fractions of the form 2/n as
sums of two, three, or four unit fractions with distinct denominators.
The table covers 2/n for n up to 101, although the fractions with
"even" denominators, e.g., 2/4, 2/6, etc, are omitted, showing that
they clearly perceived the obvious equivalence of these with the
reduced forms 1/2, 1/3, etc.
The first entry in the table is 2/3, to which they assigned the
expression 1/2 + 1/6. Every other table entry of the form 2/(3k)
is assigned the expression 1/2k + 1/6k, which suggests they
consciously treated all denominators divisible by 3 as a single
family, just as all denominators divisible by 2 were implicitly
treated as a single family.
Of the remaining table entries, the next is 2/5, to which they
assigned the expression 1/3 + 1/15. All but one of the remaining
denominators in the table that are divisible by 5 are assigned a
simple multiple of this expression, i.e., for 2/(5k) they used
1/3k + 1/15k. Similarly they assigned 1/4 + 1/28 to the table
entry 2/7, and then "seived out" all the remaining denominators
divisible by 7 using expressions of the form 1/4k + 1/28k.
Finally, they assigned 1/6 + 1/66 to the table entry 2/11 and
then used 1/6k + 1/66k for 2/11k with k=5.
The prime 11 seems to be where they stoped this procedure, which
is consistent with that fact that the table extended only to
denominators up to 101, so all the composites are seived out by
the primes less than 11. It's remarkable that the Egyptians of
1850 BC (and probably much earlier) had already developed this
crude version of the "Sieve of Eratosthenes", and seemed to
have a grasp of the difference between prime and composite
numbers.
Admittedly the seive is not perfect, at least not according to our
present understanding. For one thing, the number 55 should have
been seived out as a multiuple of 5, but for some reason they
chose to treat it as a multiple of 11. Also, the composite
numbers 35, 91, and 95 were evidently not treated as composites,
but were assigned unique representations. Nevertheless, the
overall impression is very strong that they consciously seived
out the multiples of the smaller primes up to the square root of
the largest denominator in the table, and then treated the
remaining primes with unique representations.
As we've seen, for each of the small primes 3,5,7,11 the Egyptians
expressed 2/p as a sum of two unit fractions using the simple formula
2 1 1
 =  +  (1)
p (p+1)/2 p(p+1)/2
(The same formula also applies to the expression they assigned to
2/23, although it may be coincidental.) Once these primes, and
their multiples, have been resolved, the table entries for the
remaining prime denominators suggest that the Egyptians determined
the representations by using the identity
2 1 2a  p
 =  +  (2)
p a ap
where "a" is just some nice round number a > p/2. To find the
remaining terms, you partition the quantity 2a  p into one, two,
or three distinct parts such that each part is a divisor of a. (That's
why it's good to choose a nice round number for a, so it has lots of
divisors.) For example, with n=89 we chose a=60, which gives the
difference 31. Thus, we need to express 31 as a sum of three or fewer
distinct integers each of which divides 60. One such partition is
31 = 15 + 10 + 6, which leads to the representation that appears in
the Rhind Papyrus for 2/89:
2 1 1 1 1
 =  +  +  + 
89 60 356 534 890
On this basis, it's possible to summarize the 2/n table in the
Rhind Papyrus by giving the values of a,b,(c,(d)) for each prime
p such that 2/p = 1/a + 1/b + (1/c + (1/d)). These values are
presented in the table below.
TABLE 1: Summary of Rhind Papyrus 2/n Representations
p 2ap a b c d Also covers these
      
3 1 2 6 all multiples of 3
5 1 3 15 25, 65, 85
7 1 4 28 49, 77
11 1 6 66 55
23 1 12 276
13 3 8 52 104
17 7 12 51 68
19 5 12 76 114
31 9 20 124 155
37 11 24 111 296
41 7 24 246 328
47 13 30 141 470
53 7 30 318 795
59 13 36 236 531
67 13 40 335 536
71 9 40 568 710
97 15 56 679 776
29 19 24 58 174 232
43 41 42 86 129 301
61 19 40 244 488 610
73 47 60 219 292 365
79 41 60 237 316 790
83 37 60 332 415 498
89 31 60 356 534 890
exceptional cases:
35 25 30 42
91 49 70 130
95 25 60 380 570
101 1111 606 101 202 303
This table raises two obvious questions. First, assuming the
Egyptians used something like formula (2) to determine their general
unit fraction representations for 2/p where p is a "large" prime, how
did they select the value of "a" and the partition of 2a  p from
the available possibilities? Remarkably, if you examine all the
possibilities using a computer, and limit yourself to just the three
and fourterm representations where the smallest number x in the
partition of 2ap is greater than 1, then in most cases the expression
appearing in the Rhind Papyrus is the one for which a/x is minimized.
For example, the only possible solutions for p=43 are
partition of 2np
p a 2ap x y z a/x
      
43 24 5 2 3 12
43 28 13 2 4 7 14
43 30 17 2 15 15
43 30 17 2 5 10 15
43 36 29 2 9 18 18
43 42 41 6 14 21 7
and the representation appearing the the Rhind Papyrus is the one
with a/x = 7. In all, the Egyptians used the solution with the
minimum a/x for the "large" primes
13, 17, 19, 29, 31, 37, 41, 43, 59, 67, 73, 79, 83, 97
whereas they missed it for the primes
47, 53, 61, 71, 89
In these "missed" cases they missed the minimums by 2, 6, 1, 3, and 1
respectively.
Another interesting fact that appears from a review of all the
possible representations for each prime is that p=29 is the first
prime for which there is no threeterm representation of 2/p (with
the restrictions noted above). Thus, it's not surprising that 2/29
is the first entry in the Rhind Papyrus where a fourterm
representation is used.
The second major question raised by Table 1 is how to explain the
four exceptional cases. The first three are the composites 35,
91, and 95, that for some reason were not seived out like the rest
of the composites. From out point of view the case 2/95 = 2/(5*19)
should have been seived out by the small prime p=5, giving it a
representation of 1/3k + 1/15k with k=19. Instead, we find that
its representation was evidently based on the "large" prime p=19,
i.e., it is of the form 1/12k + 1/76k + 1/114k with k=5.
The cases 2/35 and 2/91 are even more unusual, and in a sense
these are the most intriguing entries in the table. These are the
only two composites whose representations are not simple multiples
of the representations of one of their prime factors. Remarkably,
in these two cases it appears the Egyptians reverted from the normal
multiplicative decomposition to what might be called a "harmonic
airthmetic" decomposition.
Recall that the ancient Greeks had definitions for various kinds
of "means", including the
Arithmetic Mean: A(p,q) = (p+q)/2
Geometric Mean: G(p,q) = sqrt(pq)
Harmonic Mean: H(p,q) = 2/(1/p + 1/q)
It's believed the Greeks inherited these definitions from the
Babylonians, but it's certainly possible they were also known to
the Egyptians. In particular, the Harmonic Mean certainly LOOKS
Egyptian, given their affinity for unit fractions.
In any case, notice that G(p,q) is not only the geometric mean of
p and q, it's also the geometric mean of A(p,q) and H(p,q). In
other words, for any p,q we have
G = sqrt(pq) = sqrt(AH)
which follows simply because pq = AH. In other words, AH gives
an alternative decomposition of the composite number pq. This
leads to the formula
2 2 2 / 1 1 \
 =  =  (  +  ) (3)
pq A(p,q) H(p,q) p + q \ p q /
where of course the leading factor on the right is a unit fraction
because p+q is even. This formula yields the Rhind Papyrus
representations
2 1 / 1 1 \ 1 1
 =  (  +  ) =  + 
5*7 6 \ 5 7 / 30 42
and
2 1 / 1 1 \ 1 1
 =  (  +  ) =  + 
7*13 10 \ 7 13 / 70 130
Thus we can say that every composite entry in the Rhind Papyrus
2/n table is based on a decomposition of n into its prime factors.
In most cases the simple geometric decomposition pq was used, but
in two cases they used the arithmeticharmonic decomposition A*H.
(As to why the numbers 35, 91, (and 95) might have been singled out
for special treatment, see Appendix I )
This leaves only the final entry in the 2/n table, which gives
2 1 1 1 1
 =  +  +  + 
101 101 202 303 606
This entry can actually be constructed by formula (2) with a=606 and
the partition 1111 = 202 + 303 + 606, but it seems to stand out from
the other table entries due to the fact that it's a simple multiple
of 1/n. This entry may have been just a formality, suggesting that for
any n not covered in the table (i.e., larger than 100), we can use the
fourterm expansion
2/n = 1/n + 1/2n + 1/3n + 1/6n (4)
so this effectively "completes" the table, allowing us to say that
it provides a unit fraction representation of 2/n for ALL integers n.
Interestingly, formula (4) can be seen as an illustration of the
"perfectness" of the number 6, in the sense that the sum of the
divisors equals double the number, i.e., 1+2+3+6 = 12 = 2*6.
In summary, the 2/n table of the Rhind Papyrus, which dates from more
than a thousand years before Pythagoras, seems to show an awareness
of prime and composite numbers, a crude version of the "Sieve of
Eratosthenes", a knowledge of the arithmetic, geometric, and harmonic
means, and of the "perfectness" of the number 6. This all seems to
suggest a greater numbertheoretic sophistication than is generally
credited to the ancient Egyptians. Whether they originated these
ideas or borrowed them, perhaps from the Babylonians, is unclear.
(We shouldn't overlook the possibility that the Babylonians borrowed
them from the Egyptians.)
http://www.mathpages.com/home/rhind.htm
