
Re: Integer pairs in sum of reciprocals
Posted:
Jan 4, 2001 2:39 PM


Surely you are joking???
Brian Evans wrote:
> From what I can deduce the answer is actually none. > > Starting at the boundary condition for where a= b : > 1/4002 + 1/4002 = 1/2001 > Which reduces to > 1/2 + 1/2 = 1 > Now for us to move off the boundary condition one > of these must grow and the other shrink. Problem > is there is no 1/x greater than 1/2 so one can not > get any bigger. > > Brian Evans > > <saxon970@yahoo.com> wrote ... > > Hello. I have some questions about how to approach the problem below: > > > > How many pairs of positive integers a and b are there such that a < b > > and > > 1/a + 1/b = 1/2001 ? > > > > End of problem. > > > > Is trial and error and making a manual list the way to go?

Jan Kristian Haugland http://home.hia.no/~jkhaug00

