It is common to suppose that in ancient Egypt all fractions other than 2/3 were expressed as sums, where each element was a

Principles How They Reckoned -

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

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(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

[(4/10) + (3/15) + (9/30)] = [(12 + 6 + 9)/30] = 27/30

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."

"... the ancient Egyptians(') ... mathematics was based on two very elementary concepts. ... their complete knowledge of the

"If one-fifth ... was cut off from one end of a loaf of bread, the Egyptian did not seem to think of what was left as being 4/5 of the original loaf, but rather as the loaf reduced; if a fifth of this reduced loaf was cut off, the two "fifths" thus removed would of course be unequal." ... "the scribes ... carefully avoided writing