Looking at the pefsu problem of the Moscow Mathematical Papyrus here I don't understand why the algorithm takes $1/2$ of the calculated grain measure to produce beer. Why aren't the $5$ heqats multiplied by $4$ to get $20$ quantities of beer which would be a better deal in exchange?


1 Answer 1


The short answer, it seems, is that the type of beer referenced in the problem ("1/2 1/4 malt-date beer") is a weaker beer, and thus only uses 1/2 as much grain as "regular" beer. So instead of 20 jars of regular beer using 5 heqat of grain (equaling 4 pefsu), you get 10 jars of weaker beer using 2.5 heqat (still equaling 4 pefsu).

Clagget discusses this in his book Ancient Egyptian Science: Ancient Egyptian Mathematics (free in Google Books). See page 229, note 7 regarding Problem 5, which is also a pefsu problem from the MMP. Clagget says:

...the name of the weak beer only serves to indicate that this is a weaker beer like that of beer made from a mixture of malt and dates whose components are somehow represented by the fractions 1/2 and 1/4. The usual pefsu of stronger beer is 2. Hence the beer of pefsu 4 needs only half the quantity of that of pefsu 2...Therefore, in determining the jugs of beer here, the author takes 1/2 the heqat measure of meal that produced the 100 loaves of bread with pefsu 20 in order to produce 10 jugs of beer of pefsu 4.

Clagget's translation of Problem 5 (page 214 in the same book) sheds a little light on this as well. The last two lines indicate you'd solve the problem for "regular" beer, then take half that answer for the weaker malt-date variety.

So, in Problem 8, the scribe calculates that 100 loaves of bread of pefsu 20 must use 5 heqat of grain. As you note in your question, using 5 heqat of grain would yield 20 jars of "regular" beer. But the problem concerns the weaker malt-date beer, so the scribe cuts the heqat in half to get 2.5 heqat. To maintain the overall pefsu of 4, only 10 jars of this weaker beer can be made.

  • $\begingroup$ Looking at their cooking ratio it doesn't make sense though. If $1/2$ a grain is enough for $1$ jug of beer of pefsu $4$ then $4 \neq \frac{1}{1/2}$? That doesn't hold. $\endgroup$ Aug 20, 2021 at 10:28
  • $\begingroup$ They're not using 1/2 heqat per jug - they're using half as much heqat as they normally would for a stronger beer. The ratio they end up with is 10 jugs per 2.5 heqat, which is still 4 pefsu, and equates to 1 jug per 1/4 heqat. $\endgroup$
    – Brant
    Aug 20, 2021 at 12:36
  • $\begingroup$ what you say makes sense, but what doesn't make sense is what about the other $2.5$ heqats left over from the initial $5$? They should still get $20$ beers, because they have $5$ heqat in total from the bread quantity. $\endgroup$ Nov 28, 2021 at 11:00
  • $\begingroup$ I see your point, however that's not how the solution to Problem 8 is presented, presumably because we're making weaker beer. Yes, 100 loaves of bread of pefsu 20 means we must have used 5 heqats of grain. But in Problem 8 the scribe specifically directs us to halve the 5 heqats (since we need to make the malt-date beer), then he notes we end up with 10 jugs, which he states is correct. This is all a hypothetical math exercise, so there isn't actually 2.5 heqats left over. I'm not sure what else to say; I'm relying on Clagett's translation and notes, and the Problem 8 text. $\endgroup$
    – Brant
    Dec 17, 2021 at 21:55
  • $\begingroup$ I should also note that Clagett does reference some other discussions regarding the "1/2 1/4 malt-date beer". When I have a moment I will see if I can locate and review those, maybe that will shed more light on the subject. $\endgroup$
    – Brant
    Dec 17, 2021 at 22:00

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