Date: Mar 30, 2000 9:41 AM
Author: Tapio Hurme
Subject: Re: unitary (Egyptian) fractions

Milo Gardner <milogardner@juno.com> kirjoitti

viestissÃÂÃÂ¤:b3wkpo4yc6wr@forum.mathforum.com...

> Dear David, Joe, and others:

>

>

> I can agree that with the first partitions 1/2, 1/3, 1/7 that four

> additional partitions are needed to solve the 732/733 problem. The

> algebraic identities needed to solve this problem include onlu

> 366/733 = 1/2 - 1/1466

> 245/733 = 1/3 + 2/2199

>

How about 732=4*183= 4*3*61

What is 4/733= 1/x +1/y +1/z (a tip: there is a answer.)

What is 183/733= 1/a +1/b +1/c

Count: (4/733)*(183/733) and simplify.

Tapio

>

> Milo Gardner

>