Re: Egyptian arithmetic and its remaidners
- From: Milo Gardner <milogardner@xxxxxxxx>
- Date: Fri, 30 Dec 2005 11:27:36 EST
There is a simple form of remainder arithmetic
found in the Reisner Papyri. It is also reported
several times in the RMP, such as problems
24- 38. It says that quotients and remainders
were found whenever a vulgar fraction appeared.
That is, RMP # 65 asked 100 to be divided by 13,
with a quotient 7 and a remainder 9/13 mentally
cited by Ahmes. The proof is given in the Egyptian
fraction series 2/3 1/19 which equals 9/13.
Additional proof is given by RMP 31 and 33 where
the unknown value x was asked to be found
#31 (97/42)x = 33 and
#33 (97/42)x = 37
#33 is clearly 16 + 2/19 with
2/19 - 1/56 = (112 - 97)/(56*97)
or,
2/97 = 1/56 + (8 + 7)/(56*97)
= 1/56 + 1/674 + 1/776
More interesting is
#31 where x = 14 + 28/97
which Ahmes solved by
x = 14 2/97 + 26/97
with
26/97 - 1/4 = (104 -97)/(4*97)
or,
26/97 = (4 + 2 + 1)/(4*97)
= 1/97 + 1/194 + 1/388
There is more, if anyone is interested.
Look up :
http://egyptianmath.blogspot.com
and several links will appear with the
basic discussion of this subject.
Happy New Year to all,
Milo Gardner
.
- Follow-Ups:
- Re: Egyptian arithmetic and its remaidners
- From: Dave Rusin
- Re: Egyptian arithmetic and its remaidners
- Prev by Date: Re: cant solve this problem...
- Next by Date: Re: GCD(0,0)
- Previous by thread: Absolute continuity, another question,
- Next by thread: Re: Egyptian arithmetic and its remaidners
- Index(es):