Proof.
| \ | 1 | 11⁄61⁄121⁄1141⁄228 |
| 2⁄3 | 2⁄31⁄91⁄181⁄1711⁄342 | |
| \ | 1⁄3 | 1⁄31⁄181⁄361⁄3421⁄684 |
| 1⁄2 | 1⁄21⁄121⁄241⁄2281⁄456 | |
| \ | 1⁄4 | 1⁄41⁄241⁄481⁄4561⁄912. |
The total is 11⁄21⁄4 and a series of smaller fractions. 11⁄21⁄4 taken from 2 leaves a remainder of 1⁄4. Apply the smaller fractions to 912. The fractions are:
1⁄11⁄1141⁄2281⁄181⁄301⁄3421⁄6841⁄241⁄481⁄4561⁄912.
and as parts of 912 they are equal to
7684502⁄3251⁄322⁄311⁄3381921,
which together make 228, or 1⁄4 of 912. For
| 1 | 912 | |
| 1⁄2 | 456 | |
| 1⁄4 | 228. |
In this problem the fractional number is 11⁄31⁄4 and the product to be obtained is 2. In order to add the partial products when multiplying 11⁄31⁄4 so as to get 2 the author applies the fractions to 144, getting this number apparently as 12 times 12, although he might have taken it as the largest number whose reciprocal occurs among the fractions to be added, which is his usual procedure.
In his first multiplication he finds that the multiplier 11⁄61⁄12 gives him very nearly 2. It may be noticed that he checks the 2⁄3 and 1⁄3 instead of the first line. To produce exactly 2 he has to take as additional multipliers 1⁄226 and 1⁄114, getting 1⁄228 by the same process of reasoning that gave him 1⁄97 in the preceding problem and 1⁄18 in Problem 30.
He gives the answer as 11⁄61⁄121⁄1141⁄228 and proceeds to prove that it is correct. Going back to his original point of view, he multiplies it by the given fractional number 11⁄31⁄4, taking 2⁄3 and halving to get 1⁄3. and halving twice to get 1⁄4.
Problem 33
A quantity, its 2⁄3, its 1⁄2, and its 1⁄7, added together, become 37. What is the quantity?
Multiply 12⁄31⁄21⁄7 so as to get 37.
| 1 | 12⁄31⁄21⁄7 | |
| 2 | 41⁄31⁄41⁄28 | |
| 4 | 91⁄61⁄14 | |
| 8 | 181⁄31⁄7 | |
| \ | 16 | 362⁄31⁄41⁄28 |
