
Re: unitary (Egyptian) fractions
Posted:
Mar 30, 2000 9:41 AM


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 >

