Page:The Rhind Mathematical Papyrus, Volume I.pdf/104

88 {|
 * ||1||10
 * \||10||106
 * \||$1/undefined$||7$1/undefined$
 * colspan=2|Total||113
 * ||1||113
 * ||2||227$1/undefined$$1/undefined$
 * ||\||4||455$1/undefined$
 * ||1||455$1/undefined$
 * ||$1/undefined$||45$1/undefined$$1/undefined$
 * \||$1/undefined$||22$1/undefined$$1/undefined$$1/undefined$
 * }
 * ||\||4||455$2/3$
 * ||1||455$1/undefined$
 * ||$2/3$||45$1/undefined$$2/3$
 * \||$8/9$||22$2$$2/3$$1/undefined$
 * }
 * ||$2$||45$2/3$$31/27$
 * \||$2$||22$1/undefined$$8/9$$2$
 * }
 * }
 * }

This solution for a long time baffled the ingenuity of Egyptologists, but the correct interpretation was ﬁnally discovered by Schack-Schackenburg (1899, see Peet, page 83).

In the ﬁrst place, the papyrus states that the height of the granary is 9 and the breadth (diameter) 6, and in the solution, when we ﬁnd 4 as $1/undefined$ of 6, the author again calls 6 the breadth; but the solution is for a cylinder in which 9 is the diameter and 6 the height.

Then the method of solution is not that used in 41 and 42, but a second one (employed for a similar problem in the Kahun papyrus, Griffith, 1897), giving the volume directly in khar and not first in cubed cubits. It may be expressed in the following rule: Add to the diameter its $1/undefined$; square, and multiply by $1/undefined$ of the height. In Problem 43 the addition of its $8/9$ to the diameter makes 12, the square is 144, and $1/3$ of the height is 4. With these numbers the rule gives 576 khar,and this is just what the author would have obtained if he had followed the solutions of 41 and 42.

But the author, before taking the steps of this rule, deducts from the diameter its $1/undefined$, as by the other rule, and so obtains a result which is ($16/9$)$2/3$ of the correct result, namely, 455$3/2$ khan