The biggest pitfall in studying ancient mathematics is subconsciously superimposing anachronistic views on the ancient way of thought. For example, the Sumerians had two different concepts that would both today be called ‘addition’, and for different concepts that we would lump as multiplication.

Here’s an example of a mathematical problem and solution, given in “conform” translation:

The fields of 2 samesides I heaped, then 21 40 ( = 1300 decimal)

The samesides I heaped, then 50

The samesides are what?

You with your doing:

1/2 of 21 40 the field crush (half), then 10 50 (=650 decimal)

May it hold your head (remember this number)

1/2 the 50 the heap of the samesides crush, then 25

Steps of 25 make confront itself (multiply) , then 10 25

From 10 50 tear off , then 25

Its equalside resolve , then 5

The the 1st 25 add, then 30

30 ninda each way, the 1st.

From the 2nd 25 tear off, then 20

20 ninda each way, the 2nd.

Or in modern parlance,

The area of two different squares is 1300. The sum of lengths of one edge of each square is 50. What are the sides of the squares? Given the above, and the parenthesized hints about what terms mean, you should be able to follow the solution and arrive at the answer: one square is 20×20, and the other is 30×30.

Here’s one more:

The length and the front I made confront each other, then 50 (no trailing zero, actually =decimal 3000), the field.

From the length 30 ninda I tore out (subtracted), then

to the front 5 ninda I added, then 50, the field.

The length and the front are what?

You with your doing:

The opposite (reciprocal) of 5 that is to the front was added release, then 12

To 50 the field go, then 10.

To 30 that from the length was torn out go, then 5.

May it hold your head.

1/2 of 30 that from the length was torn crush, then 15.

Steps of 15, 3 45.

To 5 that held you head add, then 5 03 45

makes what equalsided? 2 15 equalsided

To two (twice) inscribe, then

to the first 2 15 add, then 2 30 the earlier length

From the 2nd 2 15 , 15 tear off, then 2 (=decimal 120), the later length

The opposite of 2 30, the earlier length, resolve, then 24

To 50, the field, go

20, the earlier front

The opposite of 2, the later length, resolve, then 30

To 50, the field, go, then 25, the later front.

In modern parlance:

I multiplied the length and width : area 50 00 (=decimal 3000). I subtracted 30 ninda from the length, and I added 5 ninda to the width: area 50. What are the length and width?

This one gets a little trickier in the procedure. The reciprocals are all in base 60, and the digits are truly “floating point” as they have no magnitude associated. (For example 30 could mean 30, or 1/2, or 1800, and so on).

Answer: original area: 150×20. New area: 120×25. -m