Proof.
| \ | 1 | 222⁄31⁄71⁄21 |
| \ | 2 | 451⁄31⁄41⁄141⁄281⁄42 |
| Total | 80. | |
3 1⁄1/2 hekat makes 1120 ro, for
| \ | 1 | 320 |
| \ | 2 | 640 |
| \ | 1⁄2 | 160 |
| Total | 1120. | |
Therefore multiply 80 so as to get 1120.
Do it thus:
| 1 | 80 | |
| \ | 10 | 800 |
| 2 | 160 | |
| \ | 4 | 320 |
| Total | 14. | |
That is, one of the loaves contains 14ro, or 1⁄32 hekat 4 ro, of meal.
Proof.
| 1⁄32 | hekat | 4 | ro | ||
| 2 | 1⁄161⁄64 | " | 3 | " | |
| 4 | 1⁄81⁄321⁄64 | " | 1 | " | |
| 8 | 1⁄41⁄161⁄32 | " | 2 | " | |
| \ | 16 | 1⁄21⁄81⁄16 | " | 4 | " |
| 32 | 11⁄41⁄81⁄64 | " | 3 | " | |
| \ | 64 | 21⁄21⁄41⁄321⁄64 | " | 1 | " |
It makes 31⁄2 hekat of meal for the 80 loaves.
The author asks two questions, first, what is the amount of meal in 1 loaf of bread, and second, how many loaves of bread one hekat of meal will make. In the solution, however, he determines first the number of loaves that one hekat will make, which he finds to be 222⁄31⁄71⁄21. This is the pefsu. Before determining the amount of meal in one loaf he reduces 31⁄2 hekat to 1120 ro ,so that he can say, If 80 loaves take 1120ro, 1 loaf will take 14 ro, for 80 times 14, or, what is the same thing, 14 times 80 makes 1120. The 14 ro is written in standard from as 1⁄32 4 ro, and this is the answer to the first question.
In the first multiplication it would seem as if the easiest way to get 1⁄21 would be from 1⁄7 by halving and taking 2⁄3. Perhaps originally it was obtained in that way. In the second line of the first proof there are two applications of the table at the beginning of the papyrus, the double of 1⁄7 being 1⁄41⁄28 and the double of 1⁄211⁄141⁄42.
