Egyptian MathematicsShorter ... and Visually
The problem of dividing nine loaves of bread among ten people can be
solved in a seemingly-fair way ... or as the Egyptians would have done
it. [Here dividing ... among means giving equal portions of the nine
loaves to all ten people involved. The task also requires distributing the
entire initial amount of bread.]
We would say that each person is satisfied with exactly the same
quantity of bread. Hence nine would receive nine-tenths a loaf. However
that is a substantial hunk, one that would likely be envied by the tenth
person who receives nine one-tenth loaves.
Egyptian mathematics used only unit fractions
(numerators must be one). Their solution to dividing items equally among
individuals has merit even today. The key observation is that their
results diplomatically distress none.
To solve the problem as they would have, recognize that thirds, fifths,
ninths, tenths,
fifteenths, and thirtieths are all fractions with denominators that are
factors of the product of nine and ten. (A simplified version of their
approach eliminates fifths, thirds and ninths. It assigns a-many tenths,
b-fifteenths, and c-thirtieths, to be added, then multiplied by ten to yield nine:
[a(1/10) + b(1/15) +c(1/30)]10 = 9
Contemporary notation would rewrite this as:
a + b(2/3) +c(1/3) = 9
We allow nonunity numerators, so chosing a four, b three and
c nine satisfies the relationship. Then:
This is the amount of one loaf that can be given to each of the ten
people. [Clearly ten times 27/30 is simply nine.])
Why the Egyptians would only work with unit fractions is not known.
Neither is their rational for allowing a single exception: 2/3, for
which they had a separate symbol. Nevertheless they tabulated enormous
representations akin to what we would write as:
2/15 = 1/30 + 1/10 [ = (1 + 3)/30]
Framing the problem as in the words above it is fairly simple to derive
this acceptable division:
2/3 + 1/5 + 1/30 [ = (20 + 6 + 1)/30]
Hence take the nine loaves and create nine 2/3 of a loaf pieces. Store
two of the remaining 1/3 pieces: they will be used to compose the tenth
2/3 sized major unit. Divide one of the remaining seven 1/3 pieces into
tenths, yielding the required ten 1/30 units. There are now six 1/3
pieces. We require ten 1/5 units. But 1/3 is 10/30 and that fact can be used.
Just apportion each 10 into a 6 and 4. We immediately get six of size
6/30 = 1/5. Combine the remaining parts to find [6(4) = 24]/30 and cut
that quantity in fourths to complete the fair division.
Paul Stelling read the above and "wondered about the solution that satisfies the
following conditions:
a) all portions are identical (not just in size, but also in makeup;
b) each portion has the minimum number of pieces possible; and
c) subject to a) and b), the size of the smallest piece is maximized.
d) [alternative to c)] subject to a) and b), the number of cuts is minimized.
The solution that I came up with that meets this criteria is to divide 5 loaves
into halves, and the remainder into fifths, so that each portion consists of
1/2 + 1/5 + 1/5.
It is easy to see (by examining alternatives) that this solution satisfies all
the conditions."
For further information please see Gillings, Richard J., Mathematics in the Time of the
Pharaohs, Cambridge, MA: The MIT Press, 1972; NY: Dover Publications,
Inc., 1982, ISBN 0-486-24315-X.