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Egyptian fractions for 2/105
Mark Dominus · 2026-06-13 · via The Universe of Discourse

Fri, 12 Jun 2026

Egyptian fractions for 2/105

!!\def\u#1{\frac1{#1}}!!

The ancient Egyptians had a terrible notation for fractions. They had notations for !!\u n!! for each !!n!!, for !!\frac23!!, but everything else was written as a sum of these, with repeats forbidden, so that for example !!\frac25!! had to be written as !!\u3 + \u{15}!!. (Wikipedia)

In an older article about Egyptian fractions and the Rhind Mathematical Papyrus, I said:

Getting the table of good-quality representations of !!\frac2n!! is not trivial, and requires searching, number theory, and some trial and error. It's not at all clear that !!\frac2{105}=\u{90} + \u{126}!!.

I think I see now where this comes from. !!105 = 3·7·5!!, so two of the summands must have denominators divisible by !!5!! and by !!7!! respectively. The first thing you should do is consider $$\u5 + \u7 = \frac{12}{35} = \frac{36}{105}.$$

But you don't want !!\frac{36}{105}!!, you want !!\frac{2}{105}!!, so you multiply by !!\u{18}!!:

$$\u{18}\left(\u5 + \u7\right) = \u{90}+\u{126} = \frac 2{105}$$

and there it is.

Why pick !!\u5!! and !!\u7!! rather than, say, !!\u3!! and !!\u5!!? I suspect the answer is probably: Ahmes (or someone earlier) tried it both ways and picked the result they liked best. Remember Ahmes is compiling a reference table here, so he does these calculations once, writes down the best result, and throws the others away.

If you do the same trick with !!3!! and !!5!! instead you get !!\u3+\u5 = \frac8{15} = \frac{56}{105}!!. Then you multiply everything by !!\u{28}!! producing $$\u{84} + \u{140} = \frac2{105}$$ which seems a little worse than the other one. Using the !!3!! and the !!5!! produces $$\u{75} + \u{175} = \frac2{105}$$ which seems much worse.

Of course this only works when the denominator is composite.

Here's another approach, which doesn't work too well in this case but might be useful for other examples. Consider that !!\frac23 = \u2 + \u6!!. We want !!\frac2{105} = \u{35}\cdot\frac23!!. So

$$ \begin{align} \frac2{105} & = \u{35}\cdot\frac23 \\ & = \u{35}\left(\u2+\u6\right) \\ & = \u{70} + \u{210} \end{align} $$

The denominators here are a lot bigger than the first expansion, but they do at least have the advantage of being multiples of !!10!!. The Egyptians like this because they, like us, often need to multiply numbers by !!10!!, and whereas a fraction like !!\u{126}!! is hard for them to multiply by !!10!!, it's trivial to multiply !!\u{210}!! by !!10!!.

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