into the hekat-measure' and returned filled, but clearly it is the hekat-measure that is filled.
In Problem 36 the word hekat is omitted and the problem is stated and solved as a simple numerical problem. It is worded, however, in the same way as the other three except for the omission of the word hekat and it is solved in the same way. Also the proof is the same as the proof of the numerical result in each of the other three. In the three hekat problems the answer obtained as an ordinary fractional part of a hekat and proved as such is then reduced to ro and proved for the number of ro; and, finally, it is reduced to the "Horus eye" fractions as far as possible, and proved for this form of expression (see Introduction, page 31).
The method of solution of these problems is that of false position and the solution of Problem 35 is explained in full in the Introduction, page 11.
For an account of the hekat and its subdivisions see Introduction pages 31-33.
Problem 35
I have gone three times into the hekat—measure, my 1⁄3 has been added to me,and I return having filled the hekat—measure. What is it that says this?
Do it thus: Assume 1. Multiplying by 31⁄3 we have
| \ | 1 | 1 |
| \ | 2 | 2 |
| \ | 1⁄3 | 1⁄3 |
| Total | 31⁄3. | |
Get 1 by operating on 31⁄3
| 1 | 31⁄3 | |
| 1⁄10 | 1⁄3 | |
| 1⁄5 | 2⁄3 | |
| Total | 1, | |
The answer is 1⁄51⁄10
Proof.
| \ | 1 | 1⁄51⁄10 |
| \ | 2 | 1⁄21⁄10 |
| \ | 1⁄3 | 1⁄31⁄10 |
| Total | 1. | |
