Page:The Rhind Mathematical Papyrus, Volume I.pdf/88

72 $1/undefined$ also he can take without difficulty. For 13 he has $1/undefined$ of 10 equal to 1, and $1/undefined$ of 3, by the table that precedes Problem 1, is $1/undefined$$1/undefined$. Finally, $1/undefined$ of $1/undefined$ is $1/undefined$.

In the original our author seems to say that we must multiply $1/undefined$ by $1/undefined$ to get $2/3$$1/undefined$. It may be that he did not mean to put a dot (the sign for fraction) over 23, but meant to say, Multiply $1/undefined$ by 23 to get $2/3$$1/undefined$. This is a correct statement but does not explain his solution. Following the method given in the other problems of this group, he should have said, Multiply $1/undefined$ by $2/3$$1/undefined$ to get $1/undefined$.

Problem 31

A quantity, its $2/3$, its $1/undefined$, and its $1/undefined$, added together, become 33. What is the quantity?

Multiply 1$2/3$$1/undefined$$1/undefined$ so as to get 33.

Total 14$2/3$. 14$1/undefined$ times 1$1/undefined$$1/undefined$$1/undefined$ makes 32$1/undefined$ plus the small fractions $1/undefined$$1/undefined$$1/undefined$$1/undefined$$1/undefined$. 32$1/undefined$ from 33 leaves the remainder $1/undefined$ to be made up by these fractions and a further product by a number yet to be determined.

$1/undefined$$1/undefined$$1/undefined$$1/undefined$$1/undefined$

taken as parts of 42 are

65$1/undefined$31$1/undefined$1$1/undefined$

making in all 17$1/undefined$, and requiring 3$2/3$$1/undefined$ more to make 21,$1/undefined$ of 42.

Take 1$1/undefined$$1/undefined$$1/undefined$ as applying to 42:

That is, 1$1/undefined$$1/undefined$$1/undefined$ applied to 42 gives 97 in all. $1/undefined$ of 42, or 1, will be $1/undefined$ of this, and 3$1/undefined$$1/undefined$ will be 3$1/undefined$$1/undefined$ times as much. Therefore we multiply $1/undefined$ by 3$1/undefined$$1/undefined$.