Ed Pegg Jr., June 21, 2004
Imagine that you are overseeing a farm. There are 29 equal sacks of grain to be distributed evenly, to 45 people. There is only one table, so you can't divide up more than one sack at a time. How do you make sure each person gets a fair share?
Figure 1. Dividing 29 sacks equally between 45 people. Each
person gets .
This problem dates back over 4000 years. There was no algebra or geometry. No decimals. Whole numbers existed, but not fractions like 29/45. The ancient egyptians would indicate a "part" number by a symbol that looked like a mouth, somewhat. Thus, "part 7" would indicate 1/7. An egyptian fraction would be something like part 3 + part 5 + part 9. In other words, 1/3 + 1/5 + 1/9 = 29/45. As seen above, this system was useful in dividing sacks of grain.
How smart were the egyptians, 4000 years ago? You can see if you can solve one of their problems. Find three different whole numbers, all less than 600, so that their
reciprocals (the "parts") add up to 2/95. Consider 48 and
4560. As reciprocals, 1/48 + 1/4560 = 2/95. That doesn't
solve the problem, since 4560 exceeds 600.
As mentioned earlier, this problem was solved 4000 years ago, in around 2000 BC,
without algebra, number theory, or computers. A large table of similar solutions was
recorded for posterity. Later, in 1600 BC, a scribe by the name Ahmes found the
document, and recorded the "ancient" results. He also wrote down a number of math problems to be solved just for fun, which makes him the earlier known recreational mathematician. Much later, in 1858 AD, Ahmes' papyrus scroll was found in Thebes, where it was bought by Henry Rhind. Eventually, the scroll went to the British Museum. The first two columns of the table below are the numbers scribed by Ahmes.
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 | |
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 |
Accurate reckoning: the entrance into knowledge of all existing things and all obscure secrets. -- Ahmes, 1600 BC
What method did the ancient egyptians use to come up with these? In answer to the earlier question, 2/95 = 1/60 + 1/380 + 1/570. The Ahmes table has many "best" solutions, with either the lowest largest denominator, or other optimizations.
One method, suggested by Akhmim P. Eves, uses n/pq = 1/(p(p + q)/n) + 1/(q(p + q)/n). For example:
2/35, p = 5, q = 7, or 2/(5 7) = 1/(5(5+7)/2) + 1/(7(5+7)/2) = 1/30 + 1/42
2/91, p = 7, q = 13, or 2/(7 13) = 1/(7(7+13)/2) + 1/(13(7+13)/2) = 1/70 + 1/130
Note, as well, the solution for 2/101, namely 2/p = 1/p + 1/2p + 1/3p + 1/6p. Ahmes likely knew of some kind of proto-number theory, and seemed well-versed in it. Every composite entry in the table is based on a decomposition of n into its prime factors.
Mathematica Code:
Code used for this column can be found in the Mathematica Information Center, at the Ten Algorithms for Egyptian Fractions entry, by David Eppstein.
References:
David Eppstein, Ten
Algorithms for Egyptian Fractions, Mathematica in Education and Research,
1995, p5-15.
Bruce Friedman, mathorigins.com, http://www.mathorigins.com/image
grid/BRUCE OLD_005.htm.
Milo Gardner, blah, http://www.math.buffalo.edu
Milo Gardner http://www.math.buffalo.edu/mad/Ancient-Africa/best-egyptian-fraction.html
J J O'Connor and E F Robertson, Egyptian numerals, http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Egyptian_numerals.html.
http://www.math.buffalo.edu/mad/Ancient-Africa/mad_egyptian-fractions.html
http://mathforum.org/epigone/historia_matematica/stangtroastrimp
http://www.math.buffalo.edu/mad/Ancient-Africa/best-egyptian-fraction.html
http://www.mathpages.com/home/rhind.htm
http://www.mathpages.com/home/rhindapp.htm
Comments are welcome. Please send comments to Ed Pegg Jr. at ed@mathpuzzle.com.
Ed Pegg Jr. is the webmaster for mathpuzzle.com. He works at Wolfram Research, Inc. as the administrator of the Mathematica Information Center.