February
3, 2003

Presented by Bruce Friedman

Last modified: October 10, 2004 as per H. Vymazalova.

There are three versions of tables on these tablets with the thirteens.

Two include thirteenths and these
two are the most careless of all!

The Thirteens table reads:

3 10 1

30 100 10

6 20 2

2 50 4

The first line should be read as:

One [times 13] is 13

The second line should be read as:

Ten [times 13] is 130

*Note that the 30 has been worn away and looks very
much like a 10

The third line should be read as:

Two [times 13] is 26

*Note that the table actually shows 28 instead of 26
but that this particular peculiarity [error] occurs frequently on these tables and that I believe the
scribe understood the correct value.

The fourth line should be read as:

Four [times 13] is 52

Note that this single stroke representation of the
number 50 is identical to that found on the Moscow Mathematical Papyrus.
This is unlike the number 50 as it appears on the EMLR. See the **EMLR**.

On part of tablet CG.25.367 this table [of THIRTEENS]
is repeated and extended to include THIRTEENTHS.

The analysis of this extended table starting from line seven follows:

Line Seven: 2_/104_/52_/8

Read as: 2/13ths = /104 + /52 + /8 = 16/104 = 2/13

This is correct and is the same identity found on the RMP!

.

Line Eight: 4_/52_/26_/4

Read as: 4/13ths = /52 + /26 + /4 = 16/52 = 4/13

This is correct.

.

Line Nine: 8_/26_/13_/2

Read as: 8/13ths = /26 + /13 + /2 = 16/26 = 8/13

This is correct.

.

Line Ten (actual with error): /26_/13_/2_4 ro_/64_/16_1

Read as: 1/13th hk3t = /16 + /64 + 4 ro + [(/2 + /13 + /26) * (/320)]

In terms of ro = 20 + 5 + 4 + (/2 + /13 + /26) = 29 8/13 ro

This is incorrect as 13 times 29 8/13 ro exceeds 1 hk3t.

The excess of 5 ro stems from the accidental inclusion of /64.

With this /64 removed the identity is equal to 24 8/13 ro and would be correct.

.

Line Eleven (actual with error): /104_/52_/13_/8_4 ro_/64_/32_/8_2

Read as: 2/13th hk3t = /8 + /32 + /64 + 4 ro + [(/8 + /13 + /52 + /104) * (/320)]

In terms of ro = 40 + 10 + 5 + 4 + (/8 + /13 + /52 + /104) = 59 3/13 ro

This is incorrect as 13 times 59 3/13 ro exceeds 2 hk3t.

The excess of 10 ro stems from the accidental inclusion of /32.

This error appears to compound the previous error in line ten.

With this /32 removed the identity is equal to 49 3/13 ro and would be correct.

.

Line Twelve (actual with error): 3 ro_/64_/32_/16_/4_4

Read as: 4/13th hk3t = /4 + /16 + /32 + /64 + 3ro

In terms of ro = 80 + 20 + 10 + 5 + 3 ro = 118 ro

This line is incorrect and incomplete.

The correct number of ro comprising 4/13 hk3t is 98 6/13.

To correct this we could double the repaired identity in line eleven as follows:

*Line Twelve (corrected): /104_/13_/8_/4_3 ro_/64_/32_/4_4

Read as: 4/13th hk3t = /4 + /32 + /64 + 3 ro + [(/4 + /8 + /13 + /104) * (/320)]

In terms of ro = 80 + 10 + 5 + 3 + (/8 + /13 + /52 + /104) = 98 48/104

= 98 6/13 ro

This is correct as 13 times 98 6/13 ro equals 4 hk3t.

.

The workings for 8/13 do not appear on this table.

If the scribe maintained and doubled the correct workings from line twelve he would have shown 8/13 hk3t as follows.

/104_/26_/8_/4_/2_1 ro_/64_/32_/16_/2_8

Read as: 8/13th hk3t = /2 + /16 + /32 + /64 + 1 ro + [(/2 + /4 + /8 + /26 + /104) * (/320)]

In terms of ro = 160 + 20 + 10 + 5 + 1 + (/2 + /4 + /8 + /26 + /104) = 196 12/13

This is correct as 13 times 196 12/13 ro equals 8 hk3t.

This entry does actually seem
to appear on tablet C.G. 25.367 but is very worn and unclear. Within the
images provided by Hana Vymazalova this entry appears complete and correct.
I have not posted the image of this entry.