5.2: BM85200 + VAT 6599 #6

Obv. I

9 An excavation. So much as the length, that is the depth. 1 the dirt I have torn out. My ground and the dirt I have heaped, \(1^{\circ} 10^{\prime}\). Length and width, \(50^{\prime}\). Length, width, what?

10 You, \(50^{\prime}\) to 1, the conversion, raise, \(50^{\prime}\) you see. \(50^{\prime}\) to 12 raise, 10 you see.

11 Make \(50^{\prime}\) confront itself, \(41^{\prime} 40^{\prime \prime}\) you see; to 10 raise, \(6^{\circ} 56^{\prime} 40^{\prime \prime}\) you see. Its igi detach, \(8^{\prime} 38^{\prime \prime} 24^{\prime \prime \prime}\) you see;

12 to \(1^{\circ} 10^{\prime}\) raise, \(10^{\prime} 4^{\prime \prime} 48^{\prime \prime \prime}\) you see, \(36^{\prime}\), \(24^{\prime}\), \(42^{\prime}\) are equals.

13 \(36^{\prime}\) to \(50^{\prime}\) raise, \(30^{\prime}\), the length. \(24^{\prime}\) to \(50^{\prime}\) raise, \(20^{\prime}\), the width; \(36^{\prime}\) to 10 raise, 6, the depth.

14 The procedure.

This is a problem of the third degree, coming from a tablet that has been broken into two parts, one of which is in London and one in Berlin (whence the composite name). It deals with a parallelepipedal “excavation,” of length \(\ell\) [\(\mathrm{NINDAN}\)], width \(w\) [\(\mathrm{NINDAN}\)] and depth \(d\) [\(mathrm{kùš}]. The length is equal to the depth, but because of the use of different metrologies in the two directions that means that \(d=12 \ell\).

Figure \(5.5\): The excavation extended 1 kùš downwards.

Further the sum of the length and the width is \([\ell+w=]\) \(50^{\prime}\), and the sum of the volume of dirt that has been "torn out," that is, dug out ² and the “ground” (the base) is \([\ell \cdot w \cdot d+\ell \cdot w=]\) \(1^{\circ}10^{\prime}\). This latter equation can be transformed into \(\ell \cdot w \cdot(d+1)=1^{\circ} 10^{\prime}\)—that is, if the excavation had been dug 1 kùš deeper, its volume would have equalled \(1^{\circ}10^{\prime}\) [\(\mathrm{NINDAN}^{2} \cdot \mathrm{kùš}\)] (Figure 5.5).³

The solution is based on a subtl variant of the false position (in its proper form this method would not serve, since the problem is not homogeneous—see note 7, page 50). "The position" consists in the construction of a "reference cube" with the side \(\ell + w\). In horizontal measure, its side is \(1 \cdot 50^{\prime}=50^{\prime}\) [\(\mathrm{NINDAN}\), since "the conversion" of \(\mathrm{NINDAN}\) asks for a multiplication by 1. In vertical measure, it is \(12 \cdot 50^{\prime}=10\) kùš, since "the conversion" of \(\mathrm{NINDAN}\) into kùš implies a multiplication by 12 (both conversions take place in line 10).

Lines 11-12 find the volume of the reference cube to be \(6^{\circ} 56^{\prime} 40^{\prime \prime}\). This volume is contained \(10^{\prime} 4^{\prime \prime} 48^{\prime \prime \prime}\) times in the extended excavation.

We should now imagine that the sides of the extended excavation are measured by the corresponding sides of the reference cube. If \(p\) is the number of times the length \(\ell\) is measured by \(50^{\prime} \mathrm{NINDAN}\), \(q\) the number of times the width \(w\) is measured by \(50^{\prime} \mathrm{NINDAN}\), and \(r\) the number of times the depth \(d+1\) kùš is measured by \(10\) kùš (\(=50^{\prime} \mathrm{NINDAN}\)), then

\(p \cdot 50^{\prime}+q \cdot 50^{\prime}=\ell+w=50^{\prime}\),

and therefore

\(p + q =1\);

further

\(r \cdot 10=d+1=12 \ell+1=12 \cdot p \cdot 50^{\prime}+1=10 p+1\)

whence

\(r=p+\frac{1}{10}=p+6^{\prime}\)

and finally

\(p \cdot q \cdot r=10^{\prime} 4^{\prime \prime} 48^{\prime \prime \prime}\)

We therefore have to express \(10^{\prime} 4^{\prime \prime} 48^{\prime \prime \prime}\) as the product of three factors \(p\), \(q\) and \(r\) that fulfiil these conditions. That is what the text does in line 12, wehre the factors appear as the "equals" \(36^{\prime}\), \(24^{\prime}\) and \(42^{\prime}\). Afterwards, line 13 finds \(\ell\), \(w\) and \(d\).

The factorization seems to be drawn from the teacher-magician's sleeves, and that is probably how it has actually been produced, just like the various square roots and quotients. Since the solution was known beforehand, that would be easy. but it is also possible to find it by systematci reasoning, beginning with simple numbers—one must just express \(10``4`48\) (\(=2^{6} \cdot 3^{4} \cdot 7\)) as the product of three numbers \(P\), \(Q\) and \(R\) where \(P + Q = 60\), \(R = P + 6\).<sup>4</sup> Knowing the general character of Old Babylonian mathematics we may even claim that the text can only allow itself to draw the answer from the sleeves because it would be possible (albeit somewhat laborious) to find it without magic. Let us first assume that \(P = 1\); then, since \(P + Q = 60\), \(Q\) will be \(59\), which is impossible; the hypotheses \(P = 2\) and \(P = 3\) can be rejected for analogous reasons; \(P = 4\) gives \(R = 10\), which is also excluded— \(10``4`48\) contains no factor 5; \(P = 5\) is impossible in itself; \(P = 6\) gives \(Q = 54\) and \(R = 12\), which must be rejected, both because the factor 7 is missing and because control shows the product not to be what is the same reasons; \(P = 18\) is impossible because the product is only around half of what is needed. \(P = 24\) and \(P = 30\) must be rejected for the same reasons as \(P = 6\). Finally we arrive at \(P = 36\), a value that fits. If we had counted prime factors it would have been even easier, but nothing suggests that the Babylonians knew that technique.

It must be emphasized, however, that this method only works because a simple solution exists. Thereby the problem differs fundamentally from those of the second degree, where a good approximation to that which "is equal" would give an almost correct solution (and the Babylonians were thus not able to solve cubic problems in general as they could solve second-degree problems—for that, one had to wait for the Italian algebraist of the sixteenth century ce.

Our text speaks of three "equals" which are not even equal. This usage evidently represents a generalization of an idea coming from the sides of the square and the cube there is nothing strange in such a generalization—our own notion of the "roots" of an equation comes in the same way form early Arabic algebra, where the fundatmental equations were formulated in terms of an amount of money and its square root. As this origin was forgotten the word came to be understood as a designation for the value of the unknown that satisfies the equation.

Other problems from the same tablet speak of a single "equal"; that is the case when the volume of the excavation measured by the reference parallelepiped (not always a cube) must be factorized as \(p^{3}\) or as \(p^{2} \cdot(p+1)\). Tables indeed exist for these two funcitons, and in these \(p\) appears precisely as "the equal;" the latter table had the name "equal, 1 joined"—see page 126.

As in the second-degree algebra, the treatment of the third-degree problems is analytic—what we have just looked at is a typical representative of the category: one presupposes that a solution exists and draws the consequences from what can tehn be stated. In the same way every solution by means of a false position is analytic—it begins by the hypothesis of a solution.

Apart from that, only rather peripheral characteristics connect the second and the third degree: the terminology for operations, the use of tables, the fundamental arithmetical operations.

Other problems on the same tablet (all dealing with parallelepipedal "excavations") are reduced to problems of the second or even the first degree. These are solved by the techniques we already know, and never by factorization. The Babylonians were thus aware of possessing another (and in their opinion, as we see) better technique, and they knew perfectly the difference between problems that can be solved by their algebraic techniques and those which do not yield to such attacks. But they seem not to have seen this difference between problems that can be solved by their algebraic techniques and those which do not yield to such attacks. but they seem not to have seen this difference as fundamental—the mathematical genre that is defined by the contents of the tablet is rather "excavation problems," just as the genre defined by BM 13901 must be understood as "square problems" even though one of the problems is reduced to a rectangle problem. Once more, the distinction between "algebra" and "quasi-algebra" seems to be secondary, less important than the classification of problems according to the object they consider.