# Papiro rhynd

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Recreational Mathematics in Ancient Egypt

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roblem 79 of the Rhind Papyrus says (ﬁg. 3):1 A house inventory: 2,801 5,602 11,204 19,607 houses cats
a

7 49 343 2,301b 16,807 19,607

mice spelt hekat Total

Total
a

The Egyptian word for “cat” is myw; when the missing vowels are inserted, this becomes meey’a uw. b Obviously Ahmes made a mistake here. The correct entry shouldbe 2,401.

What is the meaning behind this cryptic verse? Clearly we have before us a geometric progression whose initial term and common ratio are both 7, and the scribe shows us how to ﬁnd its sum. But as any good teacher would do to break the monotony of a routine math class, Ahmes embellishes the exercise with a little story which might be read like this: There are seven houses; in eachhouse there are seven cats; each cat eats seven mice; each mouse eats seven ears of spelt; each ear of spelt produces seven hekat of grain. Find the total number of items involved. The right hand column clearly gives the terms of the progression 7 72 73 74 75 followed by their sum, 19,607 (whether the mistaken entry 2,301 was Ahmes’s own error in copying or whether it had already been in the originaldocument, we shall never know). But now Ahmes plays his second card: in the left-hand column he shows us how to obtain the answer in a shorter, “clever” way; and in following it we can see the Egyptian method of multiplication at work. The Egyptians knew that any integer can be represented as a sum of terms of the geometric progression 1 2 4 8 and that the representation is unique (this isprecisely the representation of an integer in terms of the base 2, the coefﬁcients, or “binary digits,” being 0 and 1). To multiply, say, 13 by 17, they only had to write one of the multipliers, say 13, as a sum of powers of 2, 13 = 1 + 4 + 8,

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RECREATIONAL MATHEMATICS

Fig. 3. Problem 79 of the Rhind Papyrus.

multiply each power by the other multiplier, and add the results: 13 × 17 = 1 ×17 + 4 × 17 + 8 × 17 = 17 + 68 + 136 = 221. The work can be conveniently done in a tabular form: 17 × 1 = × 2 = × 4 = 17 * 34 68 *

× 8 = 136 * The astrisks indicate the powers to be added. Thus the Egyptians could do any multiplication by repeated doubling and adding. In all the Egyptian mathematical writings known to us, this practice

RECREATIONAL MATHEMATICS

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is always followed; itwas as basic to the Egyptian scribe as the multiplication table is to a pupil today. So where does 2,801, the ﬁrst number in the left-hand column of Problem 79, come from? Here Ahmes uses a property of geometric progressions with which the Egyptians were familiar: the sum of the ﬁrst n terms of a geometric progression with the same initial term and common ratio is equal to the common ratiomultiplied by one plus the sum of the ﬁrst n − 1 terms; in modern + an = a 1 + a + a 2 + + an−1 . notation, a + a2 + a3 + This sort of “recursion formula” enabled the Egyptian scribe to reduce the summation of one geometric progression to that of another one with fewer (and smaller) terms. To ﬁnd the sum of the progression 7 + 49 + 343 + 2 401 + 16 807, Ahmes thought of it as 7 × 1 + 7 + 49 + 343 + 2401 ; since the sum of the terms inside the parentheses is 2,801, all he had to do was to multiply this number by 7, thinking of 7 as 1 + 2 + 4. This is what the left-hand column shows us. Note that this column requires only three steps, compared to the ﬁve steps of the “obvious” solution shown in the right-hand column; clearly the scribe included this exercise as an example in creative thinking.One may ask: why did Ahmes choose the common ratio 7? In his excellent book, Mathematics in the Times of the Pharaohs, Richard J. Gillings answers this question as follows: “The number 7 often presents itself in Egyptian multiplication because, by regular doubling, the ﬁrst three multipliers are always 1, 2, 4, which add to 7.”2 This explanation, however, is somewhat unconvincing, for it would...

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