Proof of the result as expressed in ro
| \ | 1 | 90 |
| \ | 2 | 180 |
| \ | 1⁄3 | 30 |
| \ | 1⁄3 of 1⁄2 | 10 |
| \ | 1⁄9 | 10 |
| Total | 320. | |
It amounts in grain to 1⁄41⁄32 hekat.
Proof of the result expressed in this form.
| \ | 1 | 1⁄41⁄32 |
| \ | 2 | 1⁄21⁄16 |
| \ | 1⁄3 | 1⁄161⁄32 |
| \ | 1⁄3 of 1⁄3 | 1⁄32 |
| \ | 1⁄9 | 1⁄32 |
| Total 1⁄21⁄81⁄41⁄8 | ||
In this problem the hekat is to be divided by 31⁄21⁄18. The author expresses this number in the peculiar form 31⁄31⁄3 of 1⁄31⁄9. but when he assumes 1 and multiplies 1 by this number he gets the simpler form, and this he uses in determining that it must be multiplied by1⁄41⁄18 to get 1. In his proof, however, instead of taking 31⁄21⁄16, he very properly goes back to his original fractions as given in the statement of the problem, multiplying his answer by 31⁄31⁄3of1⁄31⁄9.
The second operation is to multiply 31⁄21⁄18 so as to get 1. He proceeds by halving to obtain a series of fractional expressions, from which he finds at once a combination that makes exactly 1 (see Introduction, page 5). This would almost seem to indicate that the problem was made up from the answer. We find, however, that his procedure was most natural, and in no way does it indicate that he knew what would be the result. He wishes to get 1 or some number nearly equal to 1. The first two partial products are too large, but the third, 1⁄21⁄41⁄8 is less than 1 and can be used. Having selected this he might have proceeded at once to find the remainder by the process of completion, but he seems to have chosen to carry his multiplication further in the hope of getting still nearer 1. Since 1⁄21⁄41⁄8, even without the 1⁄72, require only 2⁄8 to make 1, any further expression that he could use must be less than 1⁄8, and so he was compelled to pass over the next two partial products and the first one that he could try was the one beginning with 1⁄16. With this he finds that he gets exactly 1 and he does not have to proceed further. See notes to Problem 34.
After obtaining and proving the result expressed in the ordinary form as a fractional part of a hekat he reduces it to ro and then to the "Horus eye" fractions and proves for both of these forms that it is correct. As the denominators of the fractions that he first obtains are themselves powers of 2 he has the same fractions in his last expression, but he writes them in the "Horus eye" forms, and in his proof he has a slightly different combination because he takes 1⁄2 of 1⁄41⁄32 as 1⁄161⁄32 instead of 1⁄121⁄96., which the notation would not permit.
At the very end, instead of writing 1 hekat as the result of the multiplications of his proof, he writes its equivalent as 1⁄21⁄81⁄41⁄8.
