Page:The Rhind Mathematical Papyrus, Volume I.pdf/44

28 that is, the parts that came from the whole number 16 of his answer, he has 36 $2/3$ $1/undefined$ $1/undefined$, which is the same number that he got at the beginning as 16 times the given fractional number. The remaining fractions of his multiplication are so complicated that he seems to have some difficulty in carrying through his proof. In the first place, he applies all of his fractions to 5432, here again taking the largest number whose reciprocal is found among them, and he places under each of the three fractions of the sum already obtained, as well as under each of the fractions that he has not yet added, the number that it makes as a part of 5432. The numbers placed under $2/3$, $1/undefined$, and $1/undefined$ make $5173 1/3$. This subtracted from 5432 leaves $258 2/3$, and the proof would be completed if he should show that this is the sum of the numbers placed under the fractions not yet added. Instead, he writes after $2/3$ $1/undefined$ $1/undefined$ that there remains $1/undefined$ $1/undefined$ (although in the ﬁrst part of the solution he had the simpler form $1/undefined$ for this remainder), and that the two fractions of this remainder make 194 and $64 2/3$ when taken as parts of 5432.

Problems 35—38 are numerically the same kind as the preceding. I have explained 35 on page 11 and the other three are solved in the same way.

In Problem 67 it is required to ﬁnd how many cattle there are in a herd when $1/undefined$ $1/undefined$ of them makes 70, the number due as tribute to the owner. The numerical work of the solution is exactly like that for 35-38. We can suppose that the Egyptian, without going through all of the reasoning, recognized that it was a problem of the same kind, to be solved by the same process.

Four of the miscellaneous problems, 62, 63, 65, and 68, are problems of separation in given proportions.

Problem 62 is about a bag with pieces of gold, silver, and lead in it. The meaning is not very clear, but the numerical problem is simply to divide 84 into three parts proportional to 12, 6, and 3. These numbers added together make 21, 84 is 4 times 21, and therefore the parts into which 84 is to be divided are 4 times the numbers 12, 6, and 3.

Problem 63 is exactly the same kind of problem and is solved in the same way. 700 loaves are to be divided among four men in the proportion of the numbers $2/3$, $1/undefined$, $1/undefined$, and $1/undefined$. The sum of these fractions is 1 $1/undefined$ $1/undefined$. Instead of dividing 700 by 1 $1/undefined$ $1/undefined$ the author divides 1 by this