Page:The Rhind Mathematical Papyrus, Volume I.pdf/87

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SECTION VI

Problem 30

If the scribe says, What is the quantity of which as $1/undefined$$1/undefined$ will make 10, let him hear.

Multiply $1/undefined$$1/undefined$ as to get 10.

Total 13. 13 times $1/undefined$$1/undefined$ makes 9 and the fractions $1/undefined$,$2/3$, $1/undefined$, $2/3$, and $1/undefined$. The remainder is $2/3$. Take 30. $1/undefined$$1/undefined$ of 30 is 23. Therefore $1/undefined$ of 30, or 1, will be $1/undefined$ of this. 13$1/undefined$ is the required number.

For proof we multiply 13$1/undefined$ by $2/3$$1/undefined$.

To get the remainder $2/3$ the author could apply the fractions of his partial products, $1/undefined$. $1/undefined$, etc., to 30, their values taken this way making 29 in all, and requiring 1 part more to make the full 30, so that in order to make a full 10 he would require, in addition to what he already has, $1/undefined$.

In the multiplication of his proof we may notice that$1/undefined$ of 13 is given at once as 8$1/undefined$ and that $2/3$ of $1/undefined$ is given by the rule in Problem 61 (see Introduction, page 25).