Page:The Rhind Mathematical Papyrus, Volume I.pdf/32

16 until he gets a result a little less than 2. Most frequently he begins with $2/3$ and then halves a sufficient number of times. Sometimes he halves without taking $2/3$, and sometimes he takes $1/undefined$ or $1/undefined$. In some cases with these fractional multipliers he gets a whole number and can use the reciprocal of this for a multiplier and get the result more quickly. He always does this with a multiple of 3, except for 3 itself, 9, and 15, because $2/3$ of it is a number whose reciprocal gives him at once $1 1/2$. When the number is a multiple of 5 or 7, the fraction $1/undefined$ or $1/undefined$ gives him a whole number. In the Egyptian mind the case of $2/3$ is similar to $1/undefined$ and $1/undefined$. Finally, a special method seems to have been employed for 35 and 91, and still another method for 101. These cases will be explained below.

In reproducing the table, I have marked the different cases A, B, AD, BD, C, and E; that is, I have used:

A when the author ﬁrst takes $2/3$;

B when he simply halves;

D along with A or B when he also uses $1/undefined$ or $1/undefined$;

C when at some step he gets a whole number and uses its reciprocal as a multiplier;

E for the three special cases of 35, 91, and 101.

In accordance with the foregoing rules I have marked each number with the letter indicating the kind of multiplier used, and it seems desirable to give the details of one problem of each case.

For A, I will take the case of 17, which is the ﬁrst number for which the method is fully given. Beginning with the number itself the Egyptian writes:

What remains to make 2 is the problem that constitutes the second step of this solution. This step is not explained in any of the examples that make up the table, but the method, as we have seen, is explained in Problems 21–23. In this case the amount needed is $2/3$ $1/undefined$.