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Topic: Kahun Papyrus and its Arithmetic Progression
Replies: 8   Last Post: Mar 18, 2007 2:42 PM

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 Milo Gardner Posts: 1,105 Registered: 12/3/04
Kahun Papyrus and its Arithmetic Progression
Posted: Mar 3, 2007 8:41 AM

Dear Listmembers,

Today a partially unsolved ancient Egyptian problem is being offered for discussion. We know a great deal about the problem, but not the exact method(s) used by Ahmes in RMP 64 and the scribe that wrote the Kahun Papyrus (KP).

The basic facts of the KP cite two columns of data. Column 11, per Gillings is reproduced on page 176 of Mathematics in the "Time of the Pharaohs". Gillings cites the multiplication of 9 times 5/12, or 45/12, Egyptian fractions 3 2/3 1/12, as needed to write column 12.

Column 12 cites an arithmetic progression of 10 members that differ by 5/6, that sum to 100. For comparision reasons Gillings cited RMP 64 working a closely related problem inserting the sum 110, rather than the correct 110 KP sum. This insertion error has been long corrected by others.

The problem remains of finding all of the arithmetic progression logic used by Ahmes and the KP scribe.

In the KP using modern logic as suggested by Gauss, an arithmetic progression of 5/6 that sums to 100 is created by matching 5 sums of 20, as Gauss summed the counting numnbers from 1 to 100 by matching 1 + 100 = 101 50 times finding 5050 as the answer.

The KP data asks the inverse question to Gauss' formula,
as created, and confirmed, by Gaussian logic by:

2x + 5/6 = 100/5 = 20,
such that 2x = 19 1/6,
or x6 = 9 5/12, and
x5 = 9 5/12 + 5/6 = 10 1/4.

Note that 10 was added to column 11's 3 2/3 1/12 to begin column 12. Can anyone explain the difference 10, as related to the sum of the series 100, divided by the number of its members 10, 100/10 = 10? Is there any other source of the 10 added to column 11 answer?

That is, citing all 10 members of the KP arithmetic progression, the following data is cited by Gillings:

13 2/3 1/12 (165/12) - x1
12 2/3 1/6 1/12 (155/12) - x2
12 1/12 (145/12) - x3
11 1/6 1/12 (135/12) - x4
10 1/3 1/12 (125/12) - x5
9 1/3 1/6 1/12 (115/12) - x6
8 2/3 1/12 (105/12) - x7
7 2/3 1/6 1/12 (95/12) -x8
7 1/12 (85/12) - x9
6 1/6 1/12 (75/12) - x10

with a Gillings comment that 33, 31, 29, 27, 25, 23, 21, 19, 17 and 15, was multiplied by 12 throughout, as a clue to finding the scribal logic. Again, can anyone cite the the overall logic used by the scribe?

As associated information, RMP 64 data will be posted shortly.

Best Regards,

Milo Gardner

Date Subject Author
3/3/07 Milo Gardner
3/3/07 Milo Gardner
3/3/07 Milo Gardner
3/3/07 Milo Gardner
3/5/07 Milo Gardner
3/5/07 Milo Gardner
3/6/07 Milo Gardner
3/6/07 Milo Gardner
3/18/07 Milo Gardner