Page:The Rhind Mathematical Papyrus, Volume I.pdf/132

These are problems, about the feed for fattening geese. First the author determines that it takes 100 hekat to feed 10 fattening geese for 40 days. Then he asks how much grain is required to produce in grinding 100 hekat.

He begins by mentioning two kinds of grain. The first appears to be spelt. The second is wheat. Adding to 100 hekat ²⁄₃ of 100 hekat, which would make 166 ²⁄₃ hekat, he says that this is the amount of spelt that has to be ground to produce 100 hekat of meal. Then he simply takes ²⁄₃ of 100 hekat, namely, 66²⁄₃ hekat, and says that this is the amount of wheat that would be required. In the next line he takes ¹⁄₁₀ of the latter quantity, says that this is the amount to be taken away, and subtracts it from 100 hekat, presenting the remainder as the solution of the problem. In other words, whatever the two lines about spelt and wheat may mean, we may suppose that grinding increases the bulk of the grain (wheat perhaps), and that the amount required to produce 100 hekat of meal was to be determined by taking away from 100 hekat of grain ¹⁄₁₀ of ¹⁄₂ of it. This may have been a rule that had been established by experiment and was well-known, or the author may have made up an empirical rule to determine somewhat roughly the smaller amount of grain that he knew would be sufficient to make a given amount of meal. See notes to Problem 53, page 94.

In Problem 82B, which Eisenlohr included in the previous problem, we have a similar series of calculations for geese, assuming this time that it takes just half as much to fatten them. All the numbers are half as large as in Problem 82, and so all of the steps are omitted and only the last line is given.

The quantities in these solutions are expressed in the form which uses the "Horus eye" fractions of a hekat, and for larger quantities writes the number of times 100 hekat, and writes 50 hekat and 25 hekat as ¹⁄₂ and ¹⁄₄ of 100 hekat. See the notes to Problem 76, page 111, and Introduction, pages 31-32. In this notation ²⁄₃ of 100 hekat, or 66 ²⁄₃ hekat would be written as ¹⁄₂ of 100 hekat 16¹⁄₂¹⁄₈¹⁄₃₂ hekat 3¹⁄₃ro, and this is the way in which the author writes it first, in the expression for 1+²⁄₃ times 100 hekat, but in the next expression he writes ¹⁄₃, and then for the other third just a half of these numbers. I am inclined to think that the ¹⁄₃ was a little irregular, and that the scribe became confused in trying to write down ²⁄₃. ¹⁄₁₀ of ²⁄₃ of 100 hekat, or 6+²⁄₃ hekat is written correctly in the papyrus. 6+²⁄₃ from 100 leaves 93+²⁄₃, and 93+¹⁄₃ hekat reduces without difficulty to the expression given. We can perform these various operations directly with the forms used by the Egyptian.

We may notice that we have in this problem the double hekat and its parts with the double ro. The double hekat is also mentioned in Problem 84. See Introduction, page 32.

Problem 83

Estimate the feed necessary for various kinds of birds.

If the feed of four geese that are cooped up is 1 hînu of Lower-Egyptian grain, the portion of one of the geese is ¹⁄₆₄ hekat 3 ro.

If the feed of a goose that goes into the pond is ¹⁄₁₆¹⁄₃₂ hekat 2 ro, it is 1 hînu for 1 goose.

For 10 geese it takes 1 hekat of Lower-Egyptian grain.

For 10 days 10 hekat.

For a month ¼ of 100 hekat 5 hekat.