Therefore 1⁄51⁄15 is what is to be added to the given number.
For proof add them all together, namely,
2⁄31⁄51⁄151⁄15, making 1;
for, applied to 15, these fractions are equal to
10311, making 15.
After obtaining the reminder. 4, our author has to determine what fractions, taken as parts of 15, make 4. But this is the same as to say, what should multiply 15 to make 4—and so he takes 1⁄10, 1⁄5, and 1⁄15 of 15, and from the last two gets his result. The same process is used in the next problem, but in Problem 23 the fractions are given without explanation.
The solutions of this problem and the next are proved by adding together the fractions of the given expression and the fractions of the answer.
In the papyrus at the end of this solution are placed the words "Another, 1⁄51⁄10 to be added." This has no connection with Problem 21, but the number 1⁄51⁄10 is the answer to the problem that immediately follows.
Problem 22
Complete 2⁄31⁄30 to 1.
Applied to 30, 2⁄31⁄30 is 21. 30 exceeds 21 by 9. Multiply 30 so as to get 9.
| 1 | 30 | |
| \ | 1⁄10 | 3 |
| \ | 1⁄5 | 6 |
| Total | 9 | |
Therefore 1⁄51⁄10 is to be added to make the completion.
For proof add them all together, namely,
2⁄31⁄51⁄101⁄30, making 1;
for, applied to 30, these fractions are equal to
2031, making 30.
Problem 23
Complete1⁄41⁄81⁄101⁄301⁄45 to 2⁄3.
Applied to 45 these are equal to
111⁄4 51⁄21⁄8 41⁄2 11⁄2 1
which requires 61⁄8 more to make up 2⁄3 of 45, or 30. 61⁄8 is equal to 1⁄91⁄40 of 45. Therefore 1⁄91⁄40 is to be added to the given number to make 2⁄3.
For proof add them all together, namely,
1⁄41⁄81⁄91⁄101⁄301⁄401⁄45,
