I was intrigued by this description of Sumerian mathematics in the Sargonic (Akkadian) era.

Here’s a typical problem:

3600 + 5×60 nindan minus 1 ‘seed-cubit’ is the side of a square. Find the area.

And the answer, as every schoolchild knows, is:

2 šar-gal, 2 šar-u, 4 bur-u, 9 bur 5 1/8 iku, 5 1/2 sar 1 gin 2/3 še

Are you smarter than a Sumerian 5th grader? To start breaking this down, we need to get our heads around some basic units. For length the ninda (or “rod”) is about 6m. A “seed-cubit” seems to be a 2 cubits, or about 1m, a sixth of a ninda. So we envision a huge field, more than 20km on each side, with some part of its area of unusable, perhaps a path along two perpendicular edges. I have to admit, the problem feels a little contrived, as school word problems often are, even today.

The basic unit for area is a sar, which is a square ninda, or about 36 m^2. In our fancy decimal notation with fractions, we can calculate

(3900 – 1/6)^2 = 15,208,700 + 1/36

Some more area units:

- iku = 100 sar
- bur = 18 iku = 1800 sar
- bur-u = 10 bur (the -u prefix is pretty consistently x10) = 18,000 sar
- šar-u = 60 bur-u (note the hacek on the š, it’s significant) = 1,080,000 sar
- šar-gal = 6 šar-u = 6,480,000 sar

If you work backwards from this, it breaks down in the obvious way (to a point)

2 šar-gal = 12,960,000

+ 2 šar-u = 2,160,000

+ 4 bur-u = 72,000

+ 9 bur = 16,200

These terms alone add up to 15,208,200, so the answer is already accurate within a five significant digits. These tablets don’t give any indication of how exactly these solutions were worked out, but some evidence suggests that they had significant skill in working with polynomial expressions and the quadratic formula. Most likely they knew that (a-b)^2 could be solved as a^2 + b^2 – 2ab.

Anyway, from here on out it gets weird. Most likely I don’t understand the subtlety of the system, though the possibility always exists that the student who wrote the tablet made a mistake. From our fancy decimal calculation, we ‘know’ that we need 500 + 1/36 and we’re done. The next term

5 1/8 iku = 512.5 (?)

+ 5 1/2 sar = 5.5

The extra 1/8 iku seems unnecessary, and in fact makes the calclulation overshoot. The extra five-and-a-half sar make the overshoot add up to an even 18 sar, for reasons that are completely unclear. (And if the overshoot of 18 was intentional, why not write it as 5 iku 18 sar?)

For the final terms, 1 gin 2/3 še, I have yet to find a convincing source on what these units mean, but they seem to indicate the final 1/36 sar (or 1 m^2) as part of the solution. (Note in particular that if the smaller units are 1/60 sar (or 0.6 m^2), then you would need 1 and 2/3 to get that final fraction on the nose, though the calculus geek in me wants to ignore this 2nd order term…)

These were incredibly clever people. -m

[…] found a reference to the preceding problem in the book A Remarkable Collection of Babylonian Mathematical Texts, in appendix 6 on page […]